Maharashtra Board Class 10 Science Chapter 4 Effects of electric current Solutions

Std 10 Science Part 1 Chapter 4 Effects Of Electric Current Question Answer Maharashtra Board

Class 10 Science Part 1 Chapter 4 Effects Of Electric Current Question Answer Maharashtra Board

Class 10 Science 1 Chapter 4 Effects Of Electric Current Exercise

Question 1. Tell the odd one out. Give proper explanation.
a. Fuse wire, bud conductor, rubber gloves, generator.
Answer:
Generator. It converts mechanical energy into electric energy, the remaining three do not.
In simple words: The generator is the odd one out because it produces electricity, while the others are related to electrical safety or conduction.

🎯 Exam Tip: Identify the core function of each item to differentiate it from the group. Focus on the primary role (e.g., energy conversion vs. safety/conduction).

 

b. Voltmeter, Ammeter, gulvanometer, thermometer.
Answer:
Thermometer. It measures temperature, the remaining three measure electrical quantities.
In simple words: The thermometer is different because it measures temperature, whereas the other instruments measure electrical properties.

🎯 Exam Tip: Group instruments by the physical quantity they measure (e.g., electrical quantities, temperature) to correctly identify the outlier.

 

c. Loud speaker, microphone, electric motor, magnet.
Answer:
Magnet. It exerts a force on a magnetic material, the remaining three convert one form of energy into another.
In simple words: The magnet is the odd one out as it creates a magnetic force, while the other devices convert energy (e.g., electrical to sound, sound to electrical, electrical to mechanical).

🎯 Exam Tip: Understand the energy conversion or primary function of each device to categorize and find the item that doesn't fit the pattern.

 

4 Effects Of Electric Current Exercise

Question 2. Explain the construction and working of the following. Draw a neat diagram and label it.
a. Electric motor
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक इलेक्ट्रिक मोटर की संरचना को दर्शाता है। इसमें एक आयताकार तांबे की तार की कुंडली ABCD को एक शक्तिशाली चुंबक के उत्तरी और दक्षिणी ध्रुव के बीच रखा गया है, ताकि शाखाएं AB और CD चुंबकीय क्षेत्र की दिशा के लंबवत हों। कुंडली के सिरे स्प्लिट रिंग्स X और Y के दो हिस्सों से जुड़े हैं, जिनकी भीतरी सतहों पर प्रतिरोधी परत होती है और वे एक धुरी पर कसकर फिट होते हैं। एक्स और वाई के बाहरी चालक सतहें दो स्थिर कार्बन ब्रश E और F के संपर्क में होती हैं।
Working:
1. When the circuit is completed with a plug key or switch, the current flows in the direction E \( \rightarrow \) A \( \rightarrow \) B \( \rightarrow \) C \( \rightarrow \) D \( \rightarrow \) F. As the magnetic field is directed from the north pole to the south pole, the force on AB is downward and that on CD is upward by Fleming's left hand rule. Hence, AB moves downward and CD upward. These forces are equal in magnitude and opposite in direction. Therefore, as observed from the side AD, the loop ABCD and the axle start rotating in anticlockwise direction.
2. After half a rotation, X and Y come in contact with brushes F and E respectively and the current flows in the direction EDCBAF. Hence the force on CD is downward and that on AB is upward. Therefore, the loop and the axle continue to rotate in the anticlockwise direction.
3. After every half rotation, the current in the loop is reversed and the loop and the axle continue to rotate in anti-clockwise direction. When the current is switched off, the loop stops rotating after some time.
In simple words: An electric motor uses the principle that a current-carrying coil in a magnetic field experiences a force, causing it to rotate continuously. This rotation is achieved by reversing the current direction in the coil every half rotation using split rings and brushes.

🎯 Exam Tip: For explaining construction and working of devices like electric motors, clearly describe each component's role and then detail the step-by-step process, especially how direction changes or continuous motion is achieved.

 

b. Electric Generator (AC)
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक एसी इलेक्ट्रिक जनरेटर की संरचना को दर्शाता है। इसमें तांबे के तार की एक कुंडली ABCD को एक शक्तिशाली चुंबक के ध्रुवीय टुकड़ों (N और S) के बीच रखा गया है। कुंडली के सिरे दो चालक रिंगों R1 और R2 से कार्बन ब्रश B1 और B2 के माध्यम से जुड़े होते हैं। रिंग्स धुरी पर स्थिर होते हैं और रिंगों के बीच एक प्रतिरोधी परत होती है, जिससे वे धुरी से विद्युत रूप से पृथक रहते हैं। स्थिर ब्रश एक गैल्वेनोमीटर से जुड़े होते हैं जो परिपथ में धारा की दिशा को दर्शाता है।
Working:
When the axle is rotated with a machine from outside, the coil ABCD starts rotating. Suppose the coil rotates in clockwise direction, as observed from the side AD. Then as the branch AB moves upward, the branch CD moves downward. By Fleming's right hand rule, the induced current flows in the direction A \( \rightarrow \) B \( \rightarrow \) C \( \rightarrow \) D and in the external circuit, it flows from B2 to B1 through the galvanometer. The induced current is proportional to the number of turns of the copper wire in the coil.
After half a rotation, AB and CD interchange their places. Hence, the induced current flows in the direction D \( \rightarrow \) C \( \rightarrow \) B \( \rightarrow \) A. As AB is always in contact with B1 and CD is in contact with B2, the current in the external circuit flows from B1 to B2 through the galvanometer. Thus, the direction of the current is the external circuit is opposite to that in the previous half rotation. The process goes on repeating and alternating current is generated.
In simple words: An AC generator converts mechanical energy into electrical energy by rotating a coil within a magnetic field, inducing an alternating current. The direction of this current reverses periodically as the coil rotates, creating an AC output.

🎯 Exam Tip: For AC generators, emphasize the role of continuous rings and how they lead to the periodic reversal of current direction, unlike split rings in DC motors/generators.

 

4 Effects Of Electric Current

Question 3. Electromagnetic induction means
a. Charging of an electric conductor.
b. Production of magnetic field due to a current flowing through a coil.
c. Generation of a current in a coil due to relative motion between the coil and the magnet.
d. Motion of the coil around the axle in an electric motor.
Answer: (c) Generation of a current in a coil due to relative motion between the coil and the magnet.
In simple words: Electromagnetic induction is the process where a moving magnet or a changing magnetic field near a coil generates an electric current within that coil.

🎯 Exam Tip: Remember that electromagnetic induction is about *inducing* current, which requires relative motion or a changing magnetic field, not simply producing a magnetic field or charging a conductor.

 

Electric Current

Question 4. Explain the difference: AC generator and DC generator.
Answer:
AC generator:
1. In an AC generator, the rings used are not split.
2. The direction of the current produced reverses after equal intervals of time.
DC generator:
1. In a DC generator, split rings are used.
2. The current produced flows in the same direction all the time.
In simple words: The main difference between AC and DC generators lies in their commutator design: AC generators use slip rings, causing the current to periodically reverse direction, while DC generators use split rings (commutators) to ensure the current always flows in a single direction in the external circuit.

🎯 Exam Tip: When differentiating AC and DC generators, focus on the type of rings (slip vs. split) and the resulting current direction (alternating vs. unidirectional) as key distinguishing features.

 

Question 5. Which device is used to produce electricity? Describe with a neat diagram.
(1) Electric motor
(2) Galvanometer
(3) Electric generator (DC)
(4) Voltmeter
Answer: (3) Electric generator (DC)

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक डीसी जनरेटर की संरचना को दर्शाता है। इसमें एक आयताकार आर्मेचर कुंडली ABCD को एक शक्तिशाली चुंबक के उत्तरी (N) और दक्षिणी (S) ध्रुवों के बीच रखा गया है। कुंडली के सिरे स्प्लिट रिंग्स R1 और R2 से जुड़े हैं, जो धुरी पर लगे होते हैं। कार्बन ब्रश B1 और B2 इन स्प्लिट रिंग्स के संपर्क में रहते हैं, और बाहरी परिपथ में एक बल्ब (या गैल्वेनोमीटर) से जुड़े होते हैं। चुंबक के ध्रुव और आर्मेचर कुंडली को भी दर्शाया गया है।
Working: The axle is rotated with a machine from outside. When the armature coil of the generator rotates in the magnetic field, electric potential difference is produced in the coil due to electromagnetic induction. This produces a current as shown by the glowing of the bulb or by a galvanometer. The direction of the current depends on the sense of rotation of the coil.
In a DC generator, one brush is always in contact with the arm of the coil moving up while the other brush is in contact with the arm of the coil moving down in the magnetic field. Hence, the flow of the current in the circuit is always in the same direction and the current flows so long as the coil continues to rotate in the magnetic field.
In simple words: A DC electric generator produces electricity by rotating a coil in a magnetic field, inducing a current that always flows in one direction in the external circuit, thanks to its split-ring commutator.

🎯 Exam Tip: When asked to describe a device used to produce electricity, remember that generators convert mechanical energy into electrical energy. For DC generators, highlight the role of split rings in ensuring unidirectional current flow.

 

Question 6. How does the short circuit form? What is its effect?
Answer:
If a bare live wire (phase wire) and a bare neutral wire touch each other (come in direct contact) or come very close to each other, the resistance of the circuit becomes very small and hence huge (very high) electric current flows through it. This condition is called a short circuit or short circuiting.
In this case, a large amount of heat is produced and the temperature of the components involved becomes very high. Hence, the circuit catches fire.
In simple words: A short circuit occurs when a live wire directly touches a neutral wire, drastically reducing circuit resistance and causing a massive, unsafe flow of current, which can generate extreme heat and lead to fire.

🎯 Exam Tip: Focus on the cause (low resistance, high current) and the major effect (excessive heat, fire) when explaining short circuits.

 

Question 7. Give scientific reasons:
a. Tungsten is used to make a solenoid type coil in an electric bulb.
Answer:
1. The intensity of light emitted by the filament of a bulb depends on the temperature of the filament. It increases with the temperature.
2. The melting point of the material used to make the filament of a bulb should be very high so that the filament can be heated to a high temperature by passing a current through it, without melting it. This enables us to obtain more light. The melting point of tungsten is very high.
Hence, tungsten is used to make a solenoid type coil (filament) in an electric bulb.
In simple words: Tungsten is used for electric bulb filaments because its extremely high melting point allows it to be heated to very high temperatures by current, producing bright light without melting.

🎯 Exam Tip: When explaining material choices for heating elements or filaments, always refer to properties like high melting point and resistivity as crucial factors.

 

b. In the electric equipment producing heat e.g. iron, electric heater, boiler, toaster, etc. an alloy such as Nichrome is used, not pure metals.
Answer:
1. The working of heating devices such as a toaster and an electric iron is based on the heating effect of electric current, i.e., conversion of electric energy into heat by passage of electric current through a metallic conductor.
2. An alloy, such as Nichrome, has high resistivity and it can be heated to a high temperature without oxidation, in contrast to pure metals. Therefore, the coils in heating devices such as a toaster and an electric iron are made of an alloy, such as Nichrome, rather than a pure metal.
In simple words: Nichrome is preferred in heating appliances because it has high resistivity, meaning it heats up efficiently, and it resists oxidation even at high temperatures, making it durable unlike pure metals.

🎯 Exam Tip: For heating elements, key reasons for using alloys like Nichrome are high resistivity (for efficient heat production) and resistance to oxidation (for durability at high temperatures).

 

c. For electric power transmission, copper or aluminium wire is used.
Answer:
1. Copper and aluminium are good conductors of electricity.
2. Copper, and aluminium have very low resistivity. Hence, when an electric current flows through a wire of copper or aluminium, heat produced is comparatively low. Therefore, for electric power transmission, copper or aluminium wire is used.
In simple words: Copper and aluminum are used for power transmission because they are excellent electrical conductors with very low resistivity, minimizing heat loss during current flow.

🎯 Exam Tip: Good conductors with low resistivity are essential for efficient power transmission, as they reduce energy loss as heat.

 

d. In practice the unit kWh is used for the measurement of electric energy, rather than the joule.
Answer:
(1) If an electric device rated 230 V, 5 A is operated for one hour, electric energy used
= Vlt = 230 V \( \times \) 5 A \( \times \) 3600 s = 4140000 joules.
(2) If this energy is expressed in kW.h, it will be \[ \frac{4140000}{3.6 \times 10^6} \] kWh = 1.15 kW\( \cdot \)h (more convenient).
Hence, in practice the unit kW\( \cdot \)h is used for the measurement of electric energy, rather than the joule.
In simple words: Kilowatt-hour (kWh) is used for measuring household electrical energy because it represents a much larger, more practical unit of energy consumption compared to the small joule, making billing and tracking more convenient.

🎯 Exam Tip: Understand that kWh is a practical unit for large-scale energy measurement (like in homes) due to its larger magnitude compared to the joule, simplifying calculations and billing.

 

Question 8. Which of the statements given below correctly describes the magnetic field near a long, straight current-carrying conductor?
(1) The magnetic lines of force are in a plane, perpendicular to the conductor in the form of straight lines.
(2) The magnetic lines of force are parallel to the conductor on all the sides of conductor.
(3) The magnetic lines of force are perpendicular to the conductor going radially outward.
(4) The magnetic lines of force are in concentric circles with the wire as the center, in a plane perpendicular to the conductor.
Answer: (4) The magnetic lines of force are in concentric circles with the wire as the center, in a plane perpendicular to the conductor.
In simple words: The magnetic field around a straight current-carrying wire forms concentric circles in a plane perpendicular to the wire, with the wire at the center.

🎯 Exam Tip: Recall the right-hand thumb rule: if the thumb points in the current direction, the curled fingers indicate the direction of concentric magnetic field lines around the conductor.

 

Question 9. What is a solenoid? Compare the magnetic field produced by a solenoid with the magnetic field of a bar magnet. Draw neat figures and name various components.
Answer:
When a copper wire with a resistive coating is wound in a chain of loops (like a spring), it is called a solenoid.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक धारावाही सोलेनोइड के कारण उत्पन्न चुंबकीय क्षेत्र रेखाओं को दर्शाता है। एक बेलनाकार सोलेनोइड दिखाया गया है जिसके चारों ओर एक तांबे का तार कुंडलित है। परिपथ में एक बैटरी (B), प्लग कुंजी (K) और धारा (I) प्रवाहित हो रही है। सोलेनोइड के सिरों पर उत्तरी (N) और दक्षिणी (S) ध्रुव बनते हैं, और चुंबकीय क्षेत्र रेखाएं इन ध्रुवों के बीच एक समान चुंबकीय क्षेत्र और बाहर घुमावदार रेखाएं दर्शाती हैं, जो एक बार चुंबक के समान होती हैं।

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक साधारण बार चुंबक के चारों ओर चुंबकीय क्षेत्र रेखाओं को दर्शाता है। इसमें एक आयताकार चुंबक है जिसके एक सिरे पर उत्तरी (N) ध्रुव और दूसरे पर दक्षिणी (S) ध्रुव अंकित है। चुंबकीय क्षेत्र रेखाएं उत्तरी ध्रुव से निकलकर दक्षिणी ध्रुव में प्रवेश करती हुई दिखाई गई हैं, जो चुंबक के बाहर घुमावदार मार्ग बनाती हैं, और चुंबक के अंदर दक्षिणी से उत्तरी ध्रुव की ओर सीधी रेखाएं होती हैं।
The magnetic field lines (magnetic lines of force) due to a current-carrying solenoid are similar to those of a bar magnet. One face of the coil acts as the south pole and the other face as the north pole.
[Note: A current-carrying coil, like a magnet, can be used to magnetise the rod of a given material such as carbon steel or chromium steel. With a strong megnetic field, permanent magnetism can be produced in these materials.]
In simple words: A solenoid is a coil of insulated wire that, when current passes through it, generates a magnetic field similar to a bar magnet, having distinct north and south poles.

🎯 Exam Tip: When comparing solenoid and bar magnet magnetic fields, highlight their similarity in field line patterns and the ability of a solenoid to act as an electromagnet with distinct poles.

 

Question 10. Name the following diagrams and explain the concept behind them.

ℹ️ चित्र व्याख्या (Diagram Explanation): ये चित्र फ्लेमिंग के दाएं हाथ के नियम को दर्शाते हैं। पहले चित्र में, दाहिने हाथ के अंगूठे, तर्जनी और मध्यमा को एक-दूसरे के लंबवत फैलाया गया है। अंगूठा चालक की गति की दिशा (Motion of the conductor) को दर्शाता है, तर्जनी चुंबकीय क्षेत्र की दिशा (Direction of the magnetic field) को दर्शाती है, और मध्यमा प्रेरित धारा की दिशा (Direction of the induced current) को दर्शाती है। यह नियम बताता है कि जब कोई चालक चुंबकीय क्षेत्र में गति करता है, तो उसमें प्रेरित धारा किस दिशा में उत्पन्न होगी।
(a) Fleming's right hand rule:
Answer:
Stretch the thumb, the index finger and the middle finger of the right hand in such a way that they are perpendicular to each other. In this position, the thumb indicates the direction of the motion of the conductor, the index finger the direction of the magnetic field, and the middle finger shows the direction of the induced current.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फ्लेमिंग के दाएं हाथ के नियम को अधिक स्पष्ट रूप से दर्शाता है। इसमें दाहिने हाथ के अंगूठे, तर्जनी (Index finger) और मध्यमा (Middle finger) को एक-दूसरे के लंबवत फैलाकर दिखाया गया है। अंगूठा चालक की गति की दिशा (Motion of the conductor) को इंगित करता है, तर्जनी चुंबकीय क्षेत्र की दिशा (Direction of the magnetic field) को, और मध्यमा प्रेरित धारा की दिशा (Direction of the induced current) को दर्शाती है।
[Note The induced current is maximum when the direction of motion of the conductor is at right angles to the magnetic field. ]
In simple words: Fleming's Right Hand Rule helps determine the direction of induced current when a conductor moves in a magnetic field; the thumb points to motion, the forefinger to the magnetic field, and the middle finger to the induced current.

🎯 Exam Tip: For Fleming's rules, clearly state which finger represents which physical quantity (motion, field, current) and remember that the rule is used for *induced* current direction in generators.

 

(b) Fleming's left hand rule: The left hand thumb, index finger, and the middle finger are stretched so as to be perpendicular to each other. If the index finger is in the direction of the magnetic field, and the middle finger points in the direction of the current, then the direction of the thumb in the direction of the force on the conductor.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फ्लेमिंग के बाएं हाथ के नियम को दर्शाता है। इसमें बाएं हाथ के अंगूठे, तर्जनी और मध्यमा को एक-दूसरे के लंबवत फैलाया गया है। तर्जनी चुंबकीय क्षेत्र की दिशा (Direction of the magnetic field) को दर्शाती है, मध्यमा धारा की दिशा (Direction of the current) को इंगित करती है, और अंगूठा चालक पर लगने वाले बल की दिशा (Force on the conductor) को दर्शाता है। यह नियम बताता है कि जब कोई धारावाही चालक चुंबकीय क्षेत्र में रखा जाता है, तो उस पर किस दिशा में बल लगेगा।
[Note: A magnetic field exerts a force on a current-carrying conductor. Electric current is the time rate of flow of electric charge. Thus, a magnetic field exerts a force on a moving charge. This property is used to accelerate charged particles such as protons, deuterons and alpha particles, as well as electrons, to very high energies. A machine used for this purpose is called a charged particle accelerator. It may be linear or circular in design and very big in size. Such high energy particles are used to study the structure of matter. ]
In simple words: Fleming's Left Hand Rule helps determine the direction of force experienced by a current-carrying conductor placed in a magnetic field; the thumb points to force, the forefinger to the magnetic field, and the middle finger to the current.

🎯 Exam Tip: For Fleming's rules, clearly state which finger represents which physical quantity (motion, field, current/force) and remember that this rule is used for *force* on a current-carrying conductor in motors.

 

Question 11. Identify the figures and explain their use.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र विद्युत सुरक्षा उपकरणों को दर्शाता है। चित्र (a) एक फ्यूज को दिखाता है, जो एक सर्किट में अतिरिक्त धारा प्रवाह को रोकने के लिए डिज़ाइन किया गया एक सुरक्षा उपकरण है। चित्र (b) कई लघु परिपथ वियोजक (MCBs) को दर्शाता है, जो आधुनिक घरों में विद्युत अधिभार या शॉर्ट-सर्किट की स्थिति में स्वचालित रूप से परिपथ को तोड़ने के लिए उपयोग किए जाते हैं। चित्र (c) में एक जनरेटर दिखाया गया है, जो यांत्रिक ऊर्जा को विद्युत ऊर्जा में परिवर्तित करता है, लेकिन इस संदर्भ में, यह स्पष्ट रूप से एक सुरक्षा उपकरण नहीं है।
(a) Fuse:
Answer:
A fuse protects electrical circuits and appliances by stopping the flow of electric current when it exceeds a specified value. For this, it is connected in series with the appliance (or circuit) to be protected. A fuse is a piece of wire made of an alloy of low melting point (e.g. an alloy of lead and tin). If a current larger than the specified value flows through the fuse, its temperature increases enough to melt it. Hence, the circuit breaks and the appliance is protected from damage.
[Note : The fuse wire is usually enclosed in a cartridge of an insulator such as glass or porcelain provided with metal caps. The current rating (such as 1 A, 2 A) may be printed on the cartridge. ]
In simple words: A fuse is a safety device that protects circuits by melting and breaking the circuit when current exceeds a safe limit, preventing damage to appliances and fire.

🎯 Exam Tip: Emphasize that a fuse is connected in series and works by melting due to excessive current (low melting point material) to break the circuit, ensuring safety.

 

(b) Miniature circuit breaker:
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र कई लघु परिपथ वियोजक (MCBs) को दर्शाता है। ये स्विच आमतौर पर घरों में उपयोग किए जाते हैं और जब परिपथ में धारा अचानक बढ़ जाती है (अधिभार या शॉर्ट-सर्किट के कारण) तो वे स्वचालित रूप से खुल जाते हैं, जिससे धारा का प्रवाह रुक जाता है। विभिन्न प्रकार के MCB उपलब्ध हैं, हालांकि पूरे घर के लिए सामान्यतः फ्यूज तार का उपयोग भी किया जाता है।
These days miniature circuit breaker (MCB) switches are used in homes. When the current in the circuit suddenly increases this switch opens and current stops. Different types of MCBs are in use. For the entire house, however the usual fuse wire is used.
In simple words: Miniature Circuit Breakers (MCBs) are modern safety switches that automatically cut off electrical current when there's an overload or short circuit, protecting appliances and wiring.

🎯 Exam Tip: Highlight that MCBs are automatic safety devices that trip (open) the circuit during overcurrent conditions, providing a reusable alternative to fuses.

 

(c) Figure shows the construction of a DC generator.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक डीसी जनरेटर की संरचना को दर्शाता है। इसमें एक आयताकार आर्मेचर कुंडली ABCD को एक शक्तिशाली चुंबक के उत्तरी (N) और दक्षिणी (S) ध्रुवों के बीच रखा गया है। कुंडली के सिरे स्प्लिट रिंग्स R1 और R2 से जुड़े हैं, जो धुरी पर लगे होते हैं। कार्बन ब्रश B1 और B2 इन स्प्लिट रिंग्स के संपर्क में रहते हैं, और बाहरी परिपथ में एक बल्ब (या एमीटर) से जुड़े होते हैं। चुंबक के ध्रुव और आर्मेचर कुंडली को भी दर्शाया गया है।
Here, an ammeter is shown instead of a bulb.
Working: The axle is rotated with a machine from outside. When the armature coil of the generator rotates in the magnetic field, electric potential difference is produced in the coil due to electromagnetic induction. This produces a current as shown by the glowing of the bulb or by a galvanometer. The direction of the current depends on the sense of rotation of the coil.
In a DC generator, one brush is always in contact with the arm of the coil moving up while the other brush is in contact with the arm of the coil moving down in the magnetic field. Hence, the flow of the current in the circuit is always in the same direction and the current flows so long as the coil continues to rotate in the magnetic field.
In simple words: A DC generator converts mechanical energy into electrical energy using electromagnetic induction, producing a direct current that flows in a consistent direction in the external circuit due to the action of split-ring commutators.

🎯 Exam Tip: Focus on how the DC generator uses external mechanical rotation to induce a current and how the split rings ensure this induced current always flows in one direction in the external circuit.

 

Question 12. Solve the following examples.
a. Heat energy is being produced in a resistance in a circuit at the rate of 100 W. The current of 3 A is flowing in the circuit. What must be the value of the resistance?
Answer:
Solution:
Data: P = 100 W, I = 3 A, R = ?, P = I\(^2\)R
\[ \therefore \text{Resistance, R} = \frac{P}{I^2} = \frac{100W}{(3A)^2} = \frac{100}{9} \Omega = 11.11 \Omega \]
In simple words: To find the resistance, we use the formula P = I\(^2\)R, where P is power, I is current, and R is resistance. By rearranging this, R = P/I\(^2\), which gives the resistance value.

🎯 Exam Tip: Remember to use the correct formula (P = I\(^2\)R) for calculating resistance when power and current are given, and ensure units are consistent for accurate results.

 

b. Two tungsten bulbs of wattage 100 W and 60 W power work on 220 V potential difference. If they are connected in parallel, how much current will flow in the main conductor?
Answer:
Solution:
Data: P\( _1 \) = 100 W, P\( _2 \) = 60 W, V = 220 V,
I = ?, \( \therefore \) I = \( \frac{P}{V} \)
P = VI
\( \therefore I_1 = \frac{P_1}{V} \) and \( I_2 = \frac{P_2}{V} \)
Current in the main conductor, I = I\( _1 \) + I\( _2 \) (parallel connection)
\[ = \frac{P_1}{V} + \frac{P_2}{V} = \frac{P_1+P_2}{V} = \frac{100 W + 60 W}{220 V} = \frac{160 A}{220} \]
= 0.727 A = nearly 0.73 A.
In simple words: When bulbs are connected in parallel, the total current is the sum of the individual currents drawn by each bulb. Each bulb's current is found by dividing its power by the applied voltage.

🎯 Exam Tip: For parallel connections, remember that the voltage across each component is the same, and the total current is the sum of currents through individual branches. Use P=VI to find individual currents.

 

c. Who will spend more electrical energy? 500 W TV set in 30 mins, or 600 W heater in 20 mins?
Answer:
Solution:
Data: P\( _1 \) = 500 W, t\( _1 \) = 30 min = \( \frac{30}{60} \) h = \( \frac{1}{2} \) h,
P\( _2 \) = 600 W, t\( _2 \) = 20 min = \( \frac{20}{60} \) h = \( \frac{1}{3} \) h
Electrical energy used = Pt
TV set : P\( _1 \)t\( _1 \) = 500 W \( \times \frac{1}{2} \) h = 250 W\( \cdot \)h
Heater: P\( _2 \)t\( _2 \) = 600 W \( \times \frac{1}{3} \) h = 200 W\( \cdot \)h
Thus, the TV set will spend more electrical energy than the heater.
In simple words: To compare energy consumption, calculate Power multiplied by time for both devices after converting time to hours. The device with the higher product consumes more energy.

🎯 Exam Tip: Always convert time to a consistent unit (like hours or seconds) before calculating energy consumption (Energy = Power x Time) to avoid errors in comparison.

 

d. An electric iron of 1100 W is operated for 2 hours daily. What will be the electrical consumption expenses for that in the month of April? (The electric company charges Rs.5 per unit of energy.)
Answer:
Solution:
Data: P = 1100 W, t = 2 \( \times \) 30 = 60 h,
Rs.5 per unit of energy, expenses = ?
\[ N = \frac{\text{Pt}}{1000 \text{ W-h/unit}} = \frac{1100 \text{ W} \times 60 \text{ h}}{1000 \text{ W-h/unit}} = 66 \text{ units.} \]
\( \therefore \) Electrical consumption expenses = 66 units \( \times \) Rs.5 per unit = Rs.330.
In simple words: To find the electricity bill, first calculate total energy consumed in Watt-hours, convert it to kilowatt-hours (units), and then multiply by the cost per unit.

🎯 Exam Tip: Remember that "unit" for electricity billing refers to kilowatt-hour (kWh). Always convert total Watt-hours to kWh by dividing by 1000 before calculating expenses.

N = \(\frac{\text{Pt}}{\text{1000 W-h/unit}}\) = \(\frac{\text{1100 W \(\times\) 60 h}}{\text{1000 Wh/unit}}\) = 66 units.

Electrical consumption expenses = 66 units \(\times\) Rs. 5 per unit = Rs. 330.

Project:

Do it your self.

Project 1.

Under the guidance of your teachers, make a 'free-energy generator'.

Can You Recall? (Text Book Page No. 47)

 

Question 1.How do we decide that a given material is a good conductor of electricity or is an insulator?
Answer:A material which has very low electrical resistance is called a good conductor of electricity. Examples: silver, copper, aluminium. A material which has extremely high electrical resistance is called an insulator of electricity. Examples: rubber, wood, glass.
In simple words: Materials with very low resistance are good conductors (e.g., metals), while those with very high resistance are insulators (e.g., rubber).

🎯 Exam Tip: Understanding the basic property of electrical resistance is crucial for classifying materials as conductors or insulators in exams.

 

Question 2.Iron is a conductor of electricity, but when we pick up a piece of iron resting on the ground, why don't we get electric shock?
Answer:When we pick up a piece of iron resting on the ground, we don't get electric shock because that piece does not carry any electric current at that time.
In simple words: You don't get a shock from a piece of iron on the ground because it's not part of an active circuit and has no current flowing through it.

🎯 Exam Tip: This question tests practical understanding of electric current flow. Emphasize the need for a closed circuit for current to flow and cause a shock.

Use Your Brain Power! (Text Book Page No. 48)

 

Question 1.If in the circuit, the resistor is replaced by a motor, in which form will the energy given by the cell get transformed into?
Answer:The energy given by the cell will get transformed into the kinetic energy of the copper coil in the motor.
In simple words: If a motor replaces a resistor, the cell's electrical energy converts into the motor's mechanical (kinetic) energy, causing it to spin.

🎯 Exam Tip: This question highlights energy transformation in electrical devices. Focus on the conversion of electrical energy into mechanical energy in a motor.

Use Your Brain Power! (Text Book Page No. 60)

 

Question 1.Draw the diagram of a DC generator. Then explain as to how the DC current is obtained.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक डीसी जनरेटर की संरचना को दर्शाता है। इसमें एक आयताकार कॉइल (ABCD) होती है जिसे एक मजबूत चुंबक के ध्रुवों (N और S) के बीच रखा जाता है। कॉइल के सिरे स्प्लिट रिंग्स (R1 और R2) से जुड़े होते हैं जो कार्बन ब्रश (B1 और B2) के माध्यम से बाहरी सर्किट (जिसमें एक बल्ब या एमीटर जुड़ा है) से संपर्क में होते हैं। पूरी व्यवस्था एक धुरी पर टिकी होती है, जो जनरेटर के यांत्रिक घूर्णन का केंद्र है। Working: The axle is rotated with a machine from outside. When the armature coil of the generator rotates in the magnetic field, electric potential difference is produced in the coil due to electromagnetic induction. This produces a current as shown by the glowing of the bulb or by a galvanometer. The direction of the current depends on the sense of rotation of the coil. In a DC generator, one brush is always in contact with the arm of the coil moving up while the other brush is in contact with the arm of the coil moving down in the magnetic field. Hence, the flow of the current in the circuit is always in the same direction and the current flows so long as the coil continues to rotate in the magnetic field.
In simple words: A DC generator works by rotating a coil in a magnetic field, inducing an electric current. Split rings ensure the current always flows in one direction in the external circuit, providing direct current.

🎯 Exam Tip: When explaining a DC generator, clearly state the principle of electromagnetic induction and the role of split rings in ensuring unidirectional current flow. A well-labeled diagram is highly recommended.

 

[Note In the case of a DC generator, the current is in the same direction during both the halves of the rotation of the coil. The magnitude of the current does vary periodically with time. In this respect, it differs from the current supplied by an electric cell.]

Fill In The Blanks And Rewrite The Completed Statements:

 

Question 1.Electric power = V2/.......
Answer:Electric power = \(\frac{\text{V}^2}{\text{R}}\)
In simple words: Electric power can be calculated as the square of voltage divided by resistance.

🎯 Exam Tip: Remember the different formulas for electric power: P = VI, P = I²R, and P = V²/R. Choose the appropriate one based on the given variables.

 

Question 2..= 1 joule/1 second.
Answer:1 watt = 1 joule /1 second.
In simple words: One watt is the unit of power, equivalent to one joule of energy expended per second.

🎯 Exam Tip: The definition of watt (Joule/second) is fundamental. Be precise with units and their relationships.

 

Question 3.1 kW.h =.........J.
Answer:1 kW.h = 3.6 x 10⁶ J.
In simple words: One kilowatt-hour, a common unit for electricity bills, is equal to 3.6 million joules of energy.

🎯 Exam Tip: This conversion is frequently asked. Memorize the value: 1 kWh = 3.6 × 10⁶ J. Show the derivation if asked.

 

Question 4.According to Joule's law, quantity of heat (H) produced by an electric current =.........
Answer:According to Joule's law, quantity of heat (H) produced by an electric current = I²Rt or Vlt or \(\frac{\text{V}^2}{\text{R}}\) t
In simple words: Joule's law states that the heat produced by an electric current is directly proportional to the square of the current, the resistance, and the time.

🎯 Exam Tip: Joule's law (H = I²Rt) is a key concept. Ensure you know all its forms (in terms of V, I, R, t) and the factors affecting heat production.

 

Question 5.Magnetic effect of electric current was dicovered by...........
Answer:Magnetic effect of electric current was dicovered by Hans Christian Oersted.
In simple words: Hans Christian Oersted discovered that electric currents create magnetic fields around them.

🎯 Exam Tip: Knowing the pioneers of electromagnetism, like Oersted, is important for historical context and sometimes tested in general knowledge questions.

 

Question 6............is expressed in oersted.
Answer:Intensity of magnetic field is expressed in oersted.
In simple words: Oersted is a unit used to measure the strength of a magnetic field.

🎯 Exam Tip: Units are critical in physics. Be familiar with both SI and CGS units for various physical quantities. Oersted is a CGS unit for magnetic field intensity.

 

Question 7.Electromagnetic induction was discovered by...........
Answer:Electromagnetic induction was discovered by Michael Faraday and independently by Joseph Henry.
In simple words: Electromagnetic induction, the generation of electric current by changing magnetic fields, was discovered by Michael Faraday and Joseph Henry.

🎯 Exam Tip: Faraday's and Henry's contributions to electromagnetic induction are foundational. Understand the principle and who discovered it.

 

Question 8.A galvanometer is used for.........
Answer:A galvanometer is used for detecting the presence of current in a circuit, as well as for some electrical measurements.
In simple words: A galvanometer is a sensitive device primarily used to detect small electric currents and their direction.

🎯 Exam Tip: Differentiate between a galvanometer (detects current), ammeter (measures current), and voltmeter (measures potential difference).

 

Question 9.In India, the frequency of alternating current is..........
Answer:In India, the frequency of alternating current is 50 Hz or 50 cycles per second.
In simple words: In India, alternating current (AC) changes direction 50 times per second, meaning its frequency is 50 Hertz.

🎯 Exam Tip: Be aware of the standard AC frequency in your region (e.g., 50 Hz in India, 60 Hz in some other countries) as it's a common practical detail.

 

Question 10.Electric motor converts electric energy into.........energy.
Answer:Electric motor converts electric energy into mechanical energy.
In simple words: An electric motor changes electrical energy into useful mechanical energy, causing motion.

🎯 Exam Tip: Clearly state the energy transformations for both motors and generators. Motor: electrical to mechanical. Generator: mechanical to electrical.

 

Question 11.Electric generator converts...........energy into electric energy.
Answer:Electric generator converts mechanical energy into electric energy.
In simple words: An electric generator takes mechanical energy and transforms it into electrical energy.

🎯 Exam Tip: Always remember that a generator performs the reverse energy conversion of a motor.

Rewrite the following statements by selecting the correct options:

 

Question 1.The device used for producing a current is called..........
(a) a voltmeter
(b) an ammeter
(c) a galvanometer
(d) a generator
Answer: (d) a generator
In simple words: A generator is the device specifically designed to produce electric current.

🎯 Exam Tip: Understand the function of each device: voltmeter measures voltage, ammeter measures current, galvanometer detects current, and generator produces current.

 

Question 2.At the time of short circuit, the current in the circuit.........
(a) increases
(b) decreases
(c) remains the same
(d) increases in steps
Answer: (a) increases
In simple words: During a short circuit, the electrical resistance drops dramatically, causing the current to surge dangerously high.

🎯 Exam Tip: Short circuit means very low resistance. According to Ohm's Law (I = V/R), if R decreases significantly, I increases significantly.

 

Question 3.The direction of the magnetic field around a straight conductor carrying current is given by......
(a) the right hand thumb rule
(b) Fleming's left hand rule
(c) Fleming's right hand rule
(d) none of these
Answer: (a) the right hand thumb rule
In simple words: The right-hand thumb rule helps determine the direction of the magnetic field circling a straight current-carrying wire.

🎯 Exam Tip: Distinguish clearly between the right-hand thumb rule (for magnetic field around a current-carrying wire) and Fleming's rules (for force and induced current).

 

Question 4.The resistance of a wire is 100 Ω. If it carries a current of 1A for 10 seconds, the heat produced will be..........
(a) 1000 J
(b) 10 J
(c) 0.1 J
(d) 10000 J
Answer: (a) 1000 J
In simple words: Using Joule's law (H = I²Rt), the heat produced is (1A)² × 100 Ω × 10s = 1000 Joules.

🎯 Exam Tip: For numerical problems involving heat, always use H = I²Rt. Ensure units are consistent (Amperes, Ohms, seconds, Joules).

 

Question 5.If 220 V potential difference is applied across an electric bulb, a current of 0.45 A flows in the bulb. What must be the power of the bulb?
(a) 99 W
(b) 70 W
(c) 45 W
(d) 22 W
Answer: (a) 99 W
In simple words: The power of the bulb is calculated by multiplying the voltage (220 V) by the current (0.45 A), which gives 99 Watts.

🎯 Exam Tip: Use the formula P = VI for calculating power. Always double-check calculations and units.

 

Question 6.Electromagnetic induction means
(a) charging of an electric conductor.
(b) production of magnetic field due to a current flowing through a coil.
(c) generation of a current in a coil due to relative motion between the coil and the magnet.
(d) motion of the coil around the axle in an electric motor.
Answer: (c) generation of a current in a coil due to relative motion between the coil and the magnet.
In simple words: Electromagnetic induction is the process where relative movement between a coil and a magnet creates an electric current in the coil.

🎯 Exam Tip: This question tests the core definition of electromagnetic induction. Focus on the 'relative motion' aspect leading to 'induced current'.

 

Question 7.Write the correct option by observing the figures.
ℹ️ चित्र व्याख्या (Diagram Explanation): इसमें दो परिपथ दिखाए गए हैं, 'A' और 'B'। दोनों में एक सीधा चालक तार है जो एक कार्डबोर्ड से होकर गुजरता है, जिसके चारों ओर चुंबकीय क्षेत्र की रेखाएँ संकेंद्रित वृत्तों के रूप में दर्शाई गई हैं। परिपथ 'A' में, प्रतिरोध 'R' और एक एमीटर (A) वोल्टेज स्रोत (V) से श्रृंखला में जुड़े हैं। परिपथ 'B' में, दो प्रतिरोध (R1 और R2) श्रृंखला में जुड़े हैं, और एक एमीटर (A) वोल्टेज स्रोत (V) के साथ श्रृंखला में जुड़ा है, लेकिन कुल प्रतिरोध का मान परिपथ A से भिन्न है। चुंबकीय क्षेत्र की तीव्रता तार में प्रवाहित धारा पर निर्भर करती है।
(a) Magnetic field in A is stronger.
(b) Magnetic field in B is stronger.
(c) Magnetic fields in A and B are same.
(d) Magnetic fields in A and B are weaker.
Answer: (b) Magnetic field in B is stronger.
In simple words: The magnetic field in circuit B is stronger because its parallel resistance is lower than in A, leading to a higher current.

🎯 Exam Tip: Magnetic field strength around a current-carrying conductor is directly proportional to the current. Analyze the circuit to determine which configuration allows more current (e.g., lower total resistance). If circuit B has parallel resistors (as implied by the explanation's "parallel combination"), its equivalent resistance would be lower than circuit A (which shows a single resistor). Hence, current in B would be higher, leading to a stronger magnetic field. If the diagram showed two resistors in series for A and one resistor for B, then the explanation would be different. Given the explanation, circuit B implies lower resistance, therefore higher current.

[Explanation : The resistance in circuit B is less (parallel combination) than that in A. Hence, the current in B is more than that in A. Therefore, the magnetic field in B is stronger than that in A.]

 

Question 8.Observe the following diagram and choose the correct alternative:
ℹ️ चित्र व्याख्या (Diagram Explanation): ये चित्र दो विद्युत परिपथ दिखाते हैं, जिन्हें A और B से दर्शाया गया है, प्रत्येक में एक सीधा चालक तार एक कार्डबोर्ड से होकर गुजरता है। तार के चारों ओर चुंबकीय क्षेत्र की रेखाएँ वृत्ताकार पैटर्न में दर्शाई गई हैं। दोनों परिपथों में एक एमीटर (A) और एक वोल्टेज स्रोत (5V) जुड़े हैं। परिपथ A में, एक श्रृंखला में दो प्रतिरोध R1 और R2 जुड़े हैं। परिपथ B में, वही दो प्रतिरोध R1 और R2 एक समानांतर संयोजन में जुड़े हैं। चुंबकीय क्षेत्र की तीव्रता परिपथ में प्रवाहित धारा पर निर्भर करती है, जो प्रतिरोध के संयोजन पर निर्भर करती है।
(a) The intensity of magnetic field in A is larger than in B.
(b) The intensity of magnetic field in B is less than in A.
(c) The intensity of magnetic field in A and B is same.
(d) The intensity of magnetic field in A is less than in B.
Answer: (d) The intensity of magnetic field in A is less than in B.
In simple words: Since circuit B has resistors in parallel, its total resistance is less than circuit A (which has resistors in series), meaning more current flows in B, resulting in a stronger magnetic field.

🎯 Exam Tip: This question tests understanding of series vs. parallel resistance and its impact on current and magnetic field strength. Remember that parallel combinations have lower equivalent resistance than series combinations (for the same individual resistors), leading to higher current and thus stronger magnetic fields if voltage is constant.

State Whether The Following Statements Are True Or False. (If A Statement Is False, Correct It And Rewrite It.) :

 

Question 1.Electric power = I²R.
Answer:True.
In simple words: The statement correctly defines electric power as the product of the square of current and resistance.

🎯 Exam Tip: This is one of the fundamental formulas for electric power. Make sure you can recall P = I²R, P = VI, and P = V²/R.

 

Question 2.Magnetic poles exist in pairs.
Answer:True.
In simple words: Magnetic poles always occur in pairs, a north pole and a south pole; they cannot be isolated.

🎯 Exam Tip: This is a key property of magnets - monopoles do not exist. Always remember magnets have both a North and a South pole.

 

Question 3.Electromagnetism was discovered by Oersted.
Answer:True.
In simple words: Oersted was the first to demonstrate that electric currents produce magnetic fields, leading to the discovery of electromagnetism.

🎯 Exam Tip: Associate Oersted with the discovery of the magnetic effect of electric current, a cornerstone of electromagnetism.

 

Question 4.Magnetic field increases as we go away from a magnet.
Answer:False. (Magnetic field decreases as we go away from a magnet.)
In simple words: The strength of a magnetic field actually gets weaker as you move further away from the magnet.

🎯 Exam Tip: Remember the inverse square law relationship in many physical phenomena; magnetic field strength generally diminishes with distance.

 

Question 5.Magnetic lines of force cross each other.
Answer:False. (Magnetic lines of force do not cross each other.)
In simple words: Magnetic field lines never intersect because if they did, it would mean two directions of magnetic field at one point, which is impossible.

🎯 Exam Tip: This is a crucial property of magnetic field lines. Emphasize that if they crossed, a compass needle would point in two directions at once, which is physically impossible.

 

Question 6.Electric generator is used to generate current.
Answer:True.
In simple words: An electric generator serves the purpose of producing electric current.

🎯 Exam Tip: Focus on the primary function of an electric generator, which is to convert mechanical energy into electrical energy (current).

 

Question 7.An electric motor converts mechanical energy into electric energy.
Answer:False. (An electric motor converts electric energy into mechanical energy.)
In simple words: An electric motor actually converts electrical energy into mechanical energy, not the other way around.

🎯 Exam Tip: It is crucial to distinguish the energy conversion in an electric motor (electrical to mechanical) from that in an electric generator (mechanical to electrical).

 

Question 8.In India, the frequency of AC is 50 Hz.
Answer:True.
In simple words: The standard frequency for alternating current in India is 50 Hertz.

🎯 Exam Tip: This is a factual detail about power supply standards in India. Ensure you remember the 50 Hz frequency for AC.

 

Question 9.The electricity meter in the domestic electric circuit measures electrical energy consumption in kilowatt-hours.
Answer:True.
In simple words: Home electricity meters track power usage using kilowatt-hours, commonly known as 'units'.

🎯 Exam Tip: Understand that electricity meters measure energy (kW.h), not power (kW), and relate it to 'units' on your electricity bill.

 

Question 10.Electric generator converts mechanical energy into electric energy.
Answer:True.
In simple words: An electric generator transforms mechanical motion into electrical energy.

🎯 Exam Tip: Reinforce the energy transformation for a generator: mechanical input produces electrical output.

 

Question 11.Split rings are used in a DC generator and in an electric motor.
Answer:True.
In simple words: Split rings, also known as commutators, are essential components in both DC generators and electric motors to ensure unidirectional current flow or torque.

🎯 Exam Tip: Recognize the critical role of split rings (commutators) in both DC motors and DC generators for maintaining consistent direction of current/torque.

 

Question 12.Electromagnetic induction was discovered by Coulomb.
Answer:False. (Electromagnetic induction was discovered by Faraday and independently by Henry.)
In simple words: Coulomb discovered the force between electric charges, while electromagnetic induction was discovered by Faraday and Henry.

🎯 Exam Tip: Do not confuse scientists and their discoveries. Coulomb is known for Coulomb's Law of electrostatics, not electromagnetic induction.

 

Question 13.Faraday found that electricity could produce rotational motion.
Answer:True.
In simple words: Faraday's experiments demonstrated how electric currents interact with magnetic fields to create forces that can cause rotational motion.

🎯 Exam Tip: Connect Faraday not just to induction, but also to the fundamental principle behind electric motors (force on a current-carrying conductor in a magnetic field).

Tell The Odd One Out. Give Proper Explanation:

 

Question 1.Find the odd one out and justify it. Fuse wire, M.C.B., rubber gloves, generator.
Answer:Generator. It converts mechanical energy into electric energy. All others are related to safety measures to avoid mishap due to electricity.
In simple words: A generator produces electricity, while fuse wire, MCB, and rubber gloves are all safety devices against electrical hazards.

🎯 Exam Tip: Grouping items based on their primary function (production vs. safety) is a common pattern in "odd one out" questions.

Match The Columns:

 

Column I	Column II
1. The right hand thumb rule	a. The direction of the force on a current-carrying conductor placed in a magnetic field.
2. Fleming's right hand rule	b. The direction of the magnetic field around a straight conductor carrying a current.
3. Fleming's left hand rule	c. The direction of induced current in a conductor.

Answer:(1) The right hand thumb rule – The direction of the magnetic field around a straight conductor carrying a current. (2) Fleming's right hand rule – The direction of induced current in a conductor. (3) Fleming's left hand rule – The direction of the force on a current-carrying conductor placed in a magnetic field.
In simple words: The right-hand thumb rule shows magnetic field direction, Fleming's right-hand rule shows induced current direction, and Fleming's left-hand rule shows force direction on a current-carrying conductor.

🎯 Exam Tip: Clearly memorize the application of each rule: Right-Hand Thumb Rule (magnetic field from current), Fleming's Left-Hand Rule (force on current in magnetic field), and Fleming's Right-Hand Rule (induced current in moving conductor in magnetic field).

Name The Following:

 

Question 1.The negatively charged particle considered as a free particle moving in a metallic conductor.
Answer:Electron.
In simple words: The free-moving negatively charged particle in metals that conducts electricity is called an electron.

🎯 Exam Tip: This is a basic definition. Always remember that electrons are the charge carriers in metallic conductors.

 

Question 2.The quantity expressed in ampere.
Answer:Electric current.
In simple words: Electric current is the quantity measured in amperes.

🎯 Exam Tip: Ampere is the SI unit for electric current. Know the units for all fundamental electrical quantities.

 

Question 3.The quantity expressed in ohm.
Answer:Electric resistance.
In simple words: Ohm is the unit for measuring electric resistance.

🎯 Exam Tip: Ohm is the SI unit for resistance. Relate it to Ohm's Law (V=IR).

 

Question 4.The quantity expressed in volt.
Answer:Electric potential.
In simple words: Volt is the unit used to express electric potential or potential difference.

🎯 Exam Tip: Volt is the SI unit for potential difference. Understand that it represents the energy per unit charge.

 

Question 5.The quantity expressed in joule.
Answer:Work (and energy).
In simple words: Joule is the standard unit for measuring both work and energy.

🎯 Exam Tip: Joule is the SI unit for energy and work. Recognize its use across different forms of energy.

 

Question 6.The quantity expressed in watt.
Answer:Power.
In simple words: Watt is the unit used to express power.

🎯 Exam Tip: Watt is the SI unit for power. Remember it's the rate at which work is done or energy is transferred.

 

Question 7.The quantity expressed in kilowatt-hour.
Answer:Electric energy.
In simple words: Kilowatt-hour is a commercial unit used to measure electric energy consumption.

🎯 Exam Tip: Kilowatt-hour (kW.h) is a practical unit for energy, commonly seen on electricity bills, and distinct from the SI unit joule.

 

Question 8.A component used to control the current.
Answer:Resistor.
In simple words: A resistor is an electrical component that limits or controls the flow of electric current in a circuit.

🎯 Exam Tip: Resistors are fundamental circuit components. Understand their function in controlling current according to Ohm's Law.

 

Question 9.An instrument used to measure electric current.
Answer:Ammeter
In simple words: An ammeter is a device specifically designed to measure the magnitude of electric current.

🎯 Exam Tip: Differentiate ammeter (measures current) from voltmeter (measures voltage) and galvanometer (detects current).

 

Question 10.An instrument used to measure electric potential difference.
Answer:Voltmeter.
In simple words: A voltmeter is an instrument used to measure the electric potential difference, or voltage, between two points in a circuit.

🎯 Exam Tip: Remember that a voltmeter is always connected in parallel across the points where potential difference is to be measured.

 

Question 11.The ratio of the work done to the quantity of charge transferred.
Answer:Electric potential difference.
In simple words: Electric potential difference is defined as the work done per unit charge moved between two points.

🎯 Exam Tip: This is the definition of electric potential difference (V = W/Q). It's a key concept in electricity.

 

Question 12.An alloy of Ni, Cr, Mn and Fe.
Answer:Nichrome.
In simple words: Nichrome is an alloy primarily composed of nickel, chromium, manganese, and iron.

🎯 Exam Tip: Nichrome is important for its high resistance and high melting point, making it suitable for heating elements. Remember its main constituents.

 

Question 13.The Sl unit of resistance.
Answer:The ohm.
In simple words: The standard international unit for resistance is the ohm.

🎯 Exam Tip: Ensure consistent use of SI units in all calculations and definitions.

 

Question 14.A metal used to make the filament of an electric bulb.
Answer:Tungsten.
In simple words: Tungsten is the metal chosen for electric bulb filaments due to its extremely high melting point.

🎯 Exam Tip: Tungsten's high melting point and strength at high temperatures make it ideal for light bulb filaments.

 

Question 15.An alloy used to prepare a coil of high resistance for use in electric appliances such as an electric heater.
Answer:Nichrome.
In simple words: Nichrome is an alloy with high resistance, commonly used in heating elements like those in electric heaters.

🎯 Exam Tip: Remember Nichrome's specific properties (high resistivity, resistance to oxidation at high temperatures) that make it suitable for heating coils.

 

Question 16.Constituents of the alloy used to make a fuse wire.
Answer:Lead and tin.
In simple words: Fuse wire is typically made from an alloy of lead and tin.

🎯 Exam Tip: Fuse wires need a low melting point to protect circuits; lead-tin alloy is chosen for this property.

 

Question 17.The unit same as the watt-second.
Answer:The joule.
In simple words: A watt-second is equivalent to a joule, both representing a unit of energy.

🎯 Exam Tip: Watt-second is directly equivalent to Joule (Power × time = Energy). This reinforces the definition of power.

 

Question 18.A unit for intensity of magnetic field.
Answer:The oersted.
In simple words: Oersted is a unit used to quantify the intensity of a magnetic field.

🎯 Exam Tip: Reiterate that Oersted is a CGS unit for magnetic field intensity, distinct from the SI unit (Tesla or A/m).

 

Question 19.The scientist in whose honour the Sl unit of power is named.
Answer:James Watt
In simple words: The SI unit of power, the watt, is named after the Scottish inventor James Watt.

🎯 Exam Tip: Link the unit 'Watt' to the scientist James Watt, who made significant contributions to the steam engine, though not directly electricity.

 

Question 20.A device that converts electric energy into mechanical energy.
Answer:Electric motor.
In simple words: An electric motor is a device that changes electrical energy into mechanical movement.

🎯 Exam Tip: This is the definition of an electric motor. Clearly differentiate it from a generator.

 

Question 21.A device that converts mechanical energy into electric energy.
Answer:Electric generator.
In simple words: An electric generator is a machine that transforms mechanical energy into electrical energy.

🎯 Exam Tip: This is the definition of an electric generator. Clearly differentiate it from a motor.

Answer The Following Questions In One Sentence Each :

Answer The Following Questions In One Sentence Each :

Question 1. What is the production of magnetism by an electric current called?
Answer: The production of magnetism by an electric current is called electromagnetism.
In simple words: When an electric current flows, it creates magnetism, a phenomenon known as electromagnetism.

🎯 Exam Tip: This is a fundamental definition in electromagnetism and is often asked directly.

Question 2. Is magnetic field a scalar or a vector?
Answer: Magnetic field is a vector.
In simple words: A magnetic field has both magnitude and direction, making it a vector quantity.

🎯 Exam Tip: Understanding whether a quantity is scalar or vector is crucial for solving problems involving directions and magnitudes.

Question 3. In India, what is the time interval in which AC changes direction?
Answer: In India, AC changes direction every \( \frac{1}{100} \) s.
In simple words: In India, the alternating current reverses its direction 100 times per second, meaning it changes direction every hundredth of a second.

🎯 Exam Tip: This relates to the frequency of AC supply, which is 50 Hz, meaning two changes in direction per cycle.

Question 4. What is the periodic time of AC in India?
Answer: In India, the periodic time of AC is \( \frac{1}{50} \) s.
In simple words: The time taken for one complete cycle of alternating current in India is one-fiftieth of a second.

🎯 Exam Tip: Periodic time is the inverse of frequency (T=1/f). Knowing the standard frequency in India (50 Hz) helps quickly determine this value.

Answer The Following Questions:

Question 1. Define electric power.
Answer: Electric power is the electric work done per unit time.
OR
Electric power is the rate at which electric energy is used.
In simple words: Electric power is how quickly electrical energy is used or work is done by electricity.

🎯 Exam Tip: This definition is fundamental to understanding electrical circuits and energy consumption. Remember both forms of the definition.

Question 2. State the formula for electric power. Hence, obtain its Sl unit.
Answer: Electric power (P) = \( \frac{\text{electric work (W) done or electric energy used}}{\text{time (t)}} \)
The SI unit of work is the joule and that of time is the second. Hence, the SI unit of power is the joule per second. It is given the special name: the watt (W). One watt equals one joule per second.
W = VIt = \( \text{I}^2\text{Rt} \) = \( \frac{\text{V}^2}{\text{R}} \) t
\( \implies \) P = W/t = VI = \( \text{I}^2\text{R} \) = \( \text{V}^2\text{/R} \).
Here, V is the potential difference applied across an electrical appliance, R is the resistance of the appliance and I is the current through the appliance.
[Note The SI unit of power, the watt, is named in honour of James Watt (1736-1819), British instrument maker and engineer. ]
In simple words: Electric power is calculated as the work done or energy used divided by time (P=W/t). Its standard unit is the watt, which is one joule per second.

🎯 Exam Tip: Memorize the different formulas for power (P=VI, P=\( \text{I}^2\text{R} \), P=\( \text{V}^2\text{/R} \)) as they are frequently used in numerical problems. The SI unit of watt is also crucial.

Question 3. What is the commercial unit of electric energy? Obtain the relation between this unit and the Sl unit of energy.
Answer: The commercial unit of electric energy is the kilowatt-hour (kW-h) and the SI unit of energy is the joule (J).
1 kW-h = \( 10^3 \) \( \frac{\text{J}}{\text{s}} \) \( \times \) 3600 s
= \( 3.6 \times 10^6 \) J
[Note: The kilowatt-hour is often called simply the unit. (See the energy bill, i.e., the electricity bill.)]
In simple words: The commercial unit for electric energy is the kilowatt-hour (unit), while the standard SI unit is the joule. One kilowatt-hour is equivalent to 3.6 million joules.

🎯 Exam Tip: Distinguish between commercial (kW-h) and SI (Joule) units of energy. The conversion factor (1 kW-h = \( 3.6 \times 10^6 \) J) is very important for numerical problems.

Question 4. What is one kilowatt.hour?
Answer: One kilowatt-hour is the electric energy used in one hour by an electrical appliance of power one kilowatt. It is equal to \( 3.6 \times 10^6 \) J.
In simple words: One kilowatt-hour is the amount of electrical energy consumed by a 1-kilowatt appliance operating for one hour, equivalent to 3.6 million joules.

🎯 Exam Tip: This definition clearly explains what a "unit" on an electricity bill represents and its relation to power and time.

Question 5. What is heating effect of electric current? What is its origin?
Answer: The production of heat in a resistance due to the electric current flowing through it when it is connected in an electrical circuit, is called the heating effect of electric current.
When a potential difference is applied across a metallic conductor, free electrons in the conductor move from the end at the lower potential to the end at the higher potential giving rise to electric current. These electrons collide with the atoms and positive ions and transfer some kinetic energy to them. This energy is converted into heat. Hence, the temperature of the conductor begins to rise i.e., the conductor becomes hot. This is the origin of the heating effect of electric current.
In simple words: The heating effect of electric current is the generation of heat when electricity flows through a resistor. It occurs because moving electrons collide with atoms in the conductor, transferring kinetic energy which becomes heat.

🎯 Exam Tip: Understand the mechanism behind the heating effect-it's due to electron-atom collisions. This principle is applied in many everyday appliances.

Question 6. Statement 1: Electric current (flow of electrons) creates heat in a resistor. Statement 2: Heat in the resistor is created according to the law of energy conservation. Explain Statement 1 with the help of Statement 2.
Answer: (1) When electrons flow through a resistor (during flow of electric current) electrons possess kinetic energy.
(2) During the flow of electrons there is a decrease in the kinetic energy of the electrons due to collisions with atoms, ions and molecules.
(3) According to the law of conservation of energy, this decrease in the kinetic energy of the electrons gets converted into heat.
In simple words: As electrons move through a resistor, their kinetic energy decreases due to collisions. By the law of energy conservation, this lost kinetic energy transforms into heat, warming the resistor.

🎯 Exam Tip: This question connects the microscopic behavior of electrons to the macroscopic phenomenon of heat production, emphasizing the law of conservation of energy. It's a key conceptual understanding.

Question 7. State Joule's law about heating effect of electric current.
Answer: Joule's law about heating effect of electric current: The quantity of heat produced in a conductor when a current flows through it is directly proportional to (1) the square of the current (2) the resistance of the conductor (3) the time for which the current flows.
In simple words: Joule's law states that the heat generated in a conductor is directly proportional to the square of the current, the resistance of the conductor, and the duration the current flows.

🎯 Exam Tip: The mathematical form \( \text{H} = \text{I}^2\text{Rt} \) is crucial. Remember the direct proportionality with \( \text{I}^2 \), R, and t. This is a fundamental law for electrical heating.

Question 8. Obtain the mathematical expression for the heat generated in a metallic conductor by electric current (Joule's law).
Answer: If V is the potential difference applied across a metallic conductor of resistance R, the current through the conductor, given by Ohm's law, is
I = V/R ......(1)
The charge passing through the conductor in time t when the current I flows in the conductor is
Q = It.......(2)
The work done in this process is W = VQ .....(3)
From Eqs. (1), (2) and (3), we have,
W = (IR) (It) = \( \text{I}^2\text{Rt} \) = VIt
\( \implies \) W = \( \text{V} (\frac{\text{V}}{\text{R}}) \text{t} = \frac{\text{V}^2}{\text{R}} \) t
This work is converted into heat.
When I is expressed in ampere, R in ohm, t in second and V in volt, W is expressed in joule. In that case,
W = \( \text{I}^2\text{Rt} \) = VIt = \( \frac{\text{V}^2}{\text{R}} \) t (in joule)
Usually heat energy (H) is expressed in calorie. Using the relation 4.18 J = 1 cal, we have
H= \( \text{W} = \frac{\text{I}^2\text{Rt}}{4.18} \) (in cal)
\( \implies \frac{\text{VIt}}{4.18} \) (in cal) = \( \frac{\text{V}^2\text{t}}{4.18\text{R}} \) (in cal)
This is the required equation.
In simple words: Joule's law for heat generated (H) can be derived using Ohm's law (I=V/R) and the definition of work done (W=VQ). This leads to \( \text{H} = \text{I}^2\text{Rt} \), or VIt, or \( \text{V}^2\text{t/R} \), expressing the conversion of electrical work into heat.

🎯 Exam Tip: Be able to derive Joule's law from first principles (Ohm's law, work, charge). Also, remember the conversion factor between joules and calories (4.18 J = 1 cal).

Question 9. Two dissimilar bulbs are connected in series. Which bulb will be brighter? (Hint: Consider the resistance of each bulb.)
Answer: The bulb of higher resistance will be brighter, assuming that the filaments of the two bulbs have the same length and the same area of cross section, but are made of metals with different resistivities.
[Explanation Heat produced (H) in time t = \( \text{I}^2\text{Rt} \), where I is the current through a conductor and R is the resistance of the conductor. In a series combination, the current through each conductor is the same.
\( \implies \) H \( \propto \) R for a given t. Hence, the bulb with higher R will become more hot and hence emit more light energy per second. Here it is assumed that the filaments of the two bulbs have the same length and the same area of cross section, but are made of metals with different resistivities.]
In simple words: When two dissimilar bulbs are connected in series, the bulb with higher resistance will glow brighter because, in a series circuit, the same current flows through both, and heat produced (and thus brightness) is directly proportional to resistance (\( \text{H} = \text{I}^2\text{Rt} \)).

🎯 Exam Tip: For series circuits, current (I) is constant. Therefore, use \( \text{H} = \text{I}^2\text{Rt} \) to compare heating/brightness. For parallel circuits, voltage (V) is constant, so \( \text{H} = \text{V}^2\text{t/R} \) would be more appropriate.

Question 10. Name any six domestic appliances whose working is based on the heating effect of electric current. (OR) State applications of heating effect of electric current.
Answer: Domestic appliances whose working is based on the heating effect of electric current:
1. Electric heater
2. Electric iron
3. Electric oven
4. Electric toaster
5. Electric kettle
6. Electric geyser
7. Fuse.
Some other applications of heating effect of electric current:
1. Electric bulb
2. Electric furnace
3. In industry for soldering, welding, cutting, drilling
4. In surgery for cutting tissues with a finely heated platinum wire.
In simple words: Many home appliances like heaters, irons, ovens, toasters, kettles, and geysers use the heating effect of electric current to function, converting electrical energy into heat.

🎯 Exam Tip: Be ready to list multiple examples of appliances that utilize the heating effect of electric current. Understanding this practical application reinforces the theoretical concept.

Question 11. Explain the application of heating effect of electric current in an electric bulb.
Answer: In an electric bulb, there is a filament of metal such as tungsten having high melting point. When an electric current is passed through the filament, it becomes hot and emits light. The bulbs are usually filled with chemically inactive gases such as nitrogen and argon to prevent oxidation of the filament and hence prolong their life.
In simple words: An electric bulb works by heating a high-melting-point tungsten filament with electric current until it glows and emits light. Inactive gases inside prevent the filament from oxidizing and extend its lifespan.

🎯 Exam Tip: Focus on the properties of tungsten (high melting point, high resistance) and the role of inert gases (prevent oxidation, prolong life) as key points for this explanation.

Question 12. Why is tungsten used to make solenoid type coil in an electric bulb?
Answer: Tungsten is used to make solenoid type coil in an electric bulb for the following reasons:
1. Tungsten has high resistance and high melting point (nearly 3422° C).
2. Using current, it can be heated to high temperature so that it emits more light.
In simple words: Tungsten is chosen for electric bulb filaments because its high resistance allows it to heat up significantly, and its extremely high melting point prevents it from melting at the high temperatures required to emit bright light.

🎯 Exam Tip: The two key properties of tungsten are its high melting point and high resistance. Link these properties directly to the function of a bulb (heating to emit light without melting).

Question 13. Explain the application of heating effect of electric current in an electric iron.
Answer: In an electric iron, a coil of high resistance is held between mica sheets and placed inside a heavy metal block provided with a handle made of an insulator such as plastic. When an electric current is passed through the coil, it becomes hot. Mica is a good conductor of heat. Hence, heat produced in the coil is transferred to the metal block which can then be used for ironing clothes.
Mica is a bad conductor of electricity. Hence, there is no electrical contact between the coil and the metal block. Therefore, the person using the iron does not get an electric shock even if he or she happens to touch it by chance.
In simple words: An electric iron uses a high-resistance coil, insulated by mica, to generate heat when current flows. Mica transfers this heat to the metal soleplate for ironing while electrically isolating the user from the current, preventing shocks.

🎯 Exam Tip: Highlight the roles of the high-resistance coil (heat generation), mica (heat conductor, electrical insulator), and the metal block (heat transfer) in the working of an electric iron.

Question 14. Take any electricity bill of your home. In the bill there is one table which shows the units consumed by you for the last eleven months. Find the average consumption of electricity in your home for each season (i.e., summer, winter and rainy season). Are they the same? Why?
Answer: The units consumed, on an average, in a home are different for each season.
The energy requirement depends very much on the temperature of the surroundings. For example, a refrigerator, electric fans, an air conditioner, etc. are used more in summer than in winter or rainy season. On the contrary, an electric heater, geyser, etc., are used more in winter than in summer. Hence, there is variation in the average consumption of electricity from season to season.
In simple words: Electricity consumption varies by season because appliances like ACs and fans are used more in summer, while heaters and geysers are used more in winter, directly impacting the overall energy demand based on ambient temperature.

🎯 Exam Tip: This question assesses real-world understanding of energy consumption. The key idea is that energy usage is driven by external environmental factors (temperature) and the seasonal need for specific appliances.

Question 15. Name the types of wires or cables used in the electric power supply provided by the State Electricity Board for houses and factories.
Answer: The wires or cables used in the electric power supply provided by the State Electricity Board are of three types:
1. phase wire (or live wire, the wire that carries an electric current)
2. Neutral wire
3. The earth wire.
In simple words: The three types of wires used for electricity supply to homes and factories are the live (phase) wire, the neutral wire, and the earth wire.

🎯 Exam Tip: Knowing the three types of wires and their basic function (live-carries current, neutral-return path, earth-safety) is important for understanding household wiring.

Question 16. In a domestic electric supply in India, what is the potential difference between the live wire and the neutral wire?
Answer: In a domestic electric supply in India, the potential difference between the live wire and the neutral wire is 220 V.
[Note: AC is used in domestic electric supply.]
In simple words: In India, the standard voltage supplied to homes between the live and neutral wires is 220 Volts.

🎯 Exam Tip: Remember the standard domestic voltage in India (220 V) and that it's an AC supply.

Question 17. Name the type of wire to which the main fuse is connected.
Answer: The main fuse is connected to the live wire (phase wire).
In simple words: The main fuse, a safety device, is always connected to the live wire to break the circuit and protect appliances from overcurrent.

🎯 Exam Tip: The main fuse must be connected to the live wire to effectively disconnect the appliance from the high potential side in case of overcurrent, ensuring safety.

Question 18. What does the electricity meter measure?
Answer: The electricity meter measures electric energy consumption. It is expressed in 'units', where 1 unit means 1 kilowatt-hour ( = \( 3.6 \times 10^6 \) joules).
In simple words: An electricity meter measures the total electrical energy consumed, typically expressed in kilowatt-hours (units), which indicates how much energy has been used over time.

🎯 Exam Tip: Clearly state that the meter measures "electric energy consumption" and associate "units" with "kilowatt-hour" and its Joule equivalent.

Question 19. Is the electric potential difference across each appliance (in a domestic electric circuit) the same?
Answer: Yes, the electric potential difference across each appliance (in a domestic electric circuit) is the same.
In simple words: In a typical home electrical setup, all appliances receive the same voltage because they are connected in parallel.

🎯 Exam Tip: This points to the parallel connection of appliances in domestic circuits, which ensures each appliance receives the full supply voltage.

Question 20. Name the types of wire across which an electric appliance is connected.
Answer: An electric appliance is connected across the live wire (phase wire) and the neutral wire.
In simple words: An electrical appliance is connected between the live wire (which carries current) and the neutral wire (which provides the return path) to complete the circuit.

🎯 Exam Tip: Appliances need a potential difference to operate, which is established between the live and neutral wires.

Question 21. Electrical appliances are connected in parallel. What are the advantages of this arrangement?
Answer: In the parallel arrangement of electric appliances, the applied potential difference is the same in each case. Further, even if one of the appliances does not work or is removed for repairing, the other appliances can still be used.
In simple words: Parallel connections for appliances ensure they all get the same voltage and can operate independently, meaning if one appliance fails, others continue to work.

🎯 Exam Tip: The two main advantages of parallel connections are uniform voltage across all appliances and independent operation (if one fails, others remain functional). These are high-scoring points.

Question 22. In a domestic electric supply, if two bulbs are connected in series instead of parallel, what will happen if the filament of one of the bulbs breaks?
Answer: In a domestic electric supply, if two bulbs are connected in series instead of parallel, if the filament of one of the bulbs breaks, there will be no current through the other bulb as well even if the circuit is switched on. Hence the good bulb will also not glow.
In simple words: If bulbs were connected in series, a broken filament in one bulb would break the entire circuit, stopping current flow to all other bulbs, causing them to not glow either.

🎯 Exam Tip: This illustrates a key disadvantage of series connections: a single point of failure (like a broken filament) disrupts the entire circuit. Contrast this with the robustness of parallel connections.

Question 23. What is overloading? When does it occur? What does it cause? How can overloading be avoided?
Answer: A flow of large amount of current in a circuit, beyond the permissible value of current, is called overloading.
It occurs when many electrical appliances of high power rating, such as a geyser, a heater, an oven, a motor, etc., are switched on simultaneously. This causes fire.
Overloading can be avoided by not connecting many electrical appliances of high power rating in the same circuit.
In simple words: Overloading happens when too much current flows through a circuit by connecting too many high-power devices, which can cause overheating and fire. It's avoided by distributing appliances across different circuits or using fewer high-power devices at once.

🎯 Exam Tip: Define overloading, explain its cause (simultaneous use of high-power appliances), its effect (fire), and the primary prevention method (avoid connecting too many high-power appliances to one circuit).

Question 24. Explain the application of heating effect of electric current in a fuse.
Answer: A fuse protects electrical circuits and appliances by stopping the flow of electric current when it exceeds a specified value. For this, it is connected in series with the appliance (or circuit) to be protected. A fuse is a piece of wire made of an alloy of low melting point (e.g. an alloy of lead and tin). If a current larger than the specified value flows through the fuse, its temperature increases enough to melt it. Hence, the circuit breaks and the appliance is protected from damage.
[Note: The fuse wire is usually enclosed in a cartridge of an insulator such as glass or porcelain provided with metal caps. The current rating (such as 1 A, 2 A) may be printed on the cartridge.]
In simple words: A fuse protects circuits by using a low-melting-point wire connected in series. When current exceeds a safe limit, the wire heats up, melts, and breaks the circuit, preventing damage to appliances.

🎯 Exam Tip: The key elements are: series connection, low melting point alloy, and the mechanism of melting to break the circuit and protect devices. Relate this to the heating effect of current.

Question 25. State the conclusions that can be drawn from Oersted's experiment.
Answer: Conclusions that can be drawn from Oersted's experiment:
1. An electric current produces a magnetic field around it. The moving charge in the conducting wire is a source of magnetic field.
2. The direction of the magnetic field produced by the current is the direction in which the north pole of the magnetic needle is deflected. Hence, from the experimental observations we can conclude that at any point near the current-carrying conductor, the magnetic field is perpendicular to (i) the length of the conductor and (ii) the line joining the conductor and the given point.
In simple words: Oersted's experiment concluded that an electric current creates a magnetic field around it, and the direction of this field is perpendicular to both the current flow and the line connecting the conductor to the point of observation.

🎯 Exam Tip: Remember the two main conclusions: electricity produces magnetism, and the magnetic field direction is perpendicular to the current. This is the foundation of electromagnetism.

Question 26. What is the effect on the magnetic needle in Oersted's experiment, when (1) a current is passed through the wire (2) the current through the wire is increased (3) the current through the wire is stopped (4) the current through the wire is reversed (5) the distance between the magnetic needle and the wire is increased, keeping the current through the wire constant?
Answer: In Oersted's experiment, when there is no current in the wire, the magnetic needle is at rest along the north-south direction.
(1) When a current is passed through the wire, the needle is deflected.
(2) When the current through the wire is increased, the deflection of the needle increases.
(3) When the current through the wire is stopped, the needle comes to rest in its original position along the north-south direction.
(4) When the current through the wire is reversed, the needle is deflected in the direction opposite to that in the first case.
(5) When the distance between the magnetic needle and the wire is increased, keeping the current through the wire constant, the deflection of the needle becomes less.
In simple words: In Oersted's experiment, current deflects the needle; increasing current increases deflection. Stopping current returns the needle to rest, reversing current reverses deflection, and increasing distance reduces deflection.

🎯 Exam Tip: This question tests understanding of the direct observation and relationship between current, magnetic field strength, and direction, as demonstrated by Oersted. Know how each change affects the compass needle.

Question 27. State the factors on which the magnitude of the magnetic field due to a current-carrying conductor depends and how it depends.
Answer: The magnetic field at a point due to a current-carrying conductor depends on the current through the conductor and the distance of the point from the conductor.
1. The magnitude of the magnetic field produced at a given point is directly proportional to the magnitude of the current passing through the conductor.
2. The magnitude of the magnetic field produced by a given current in the conductor decreases as the distance from the conductor increases.
[Note If the direction of the current is reversed, the direction of the magnetic field is also reversed.]
In simple words: The strength of the magnetic field created by a current-carrying wire depends on two factors: it increases with the amount of current and decreases as you move further away from the wire.

🎯 Exam Tip: Remember the two inverse-square law type relationships: magnetic field strength is directly proportional to current (I) and inversely proportional to the distance (r) from the conductor. This is fundamental to Biot-Savart Law and Ampere's Law.

Question 28. State the right hand thumb rule.
Answer: Imagine that you have held a current-carrying straight conductor in your right hand in such a way that your thumb points in the direction of the current. Then turn your fingers around the conductor. The direction of the fingers in the direction of the magnetic lines of force produced by the current.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दाहिने हाथ के अंगूठे के नियम को दर्शाता है। इसमें एक सीधे धारावाही चालक को दाहिने हाथ से इस तरह पकड़ा गया है कि अंगूठा धारा की दिशा में इंगित कर रहा है। मुड़ी हुई उंगलियां चालक के चारों ओर चुंबकीय क्षेत्र रेखाओं की दिशा को दर्शाती हैं, जो एक संकेंद्रित वृत्त बनाती हैं।
In simple words: The right hand thumb rule states that if you point your right thumb in the direction of current in a straight wire, your curled fingers will show the direction of the magnetic field lines around the wire.

🎯 Exam Tip: This rule is crucial for determining the direction of the magnetic field around a straight current-carrying conductor. Practice visualizing it for different current directions.

Question 29. With a neat labelled diagram, describe the pattern of magnetic lines of force due to a current through a circular loop. Also explain how the magnetic field depends on the number of turns (n) in the loop.
Answer: The pattern of magnetic lines of force due to a current through a circular loop is shown in Figure
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक वृत्ताकार चालक लूप से प्रवाहित धारा के कारण उत्पन्न चुंबकीय क्षेत्र को दर्शाता है। चुंबकीय क्षेत्र रेखाएँ लूप के पास संकेंद्रित वृत्त बनाती हैं और लूप के केंद्र में लगभग सीधी रेखाएँ प्रतीत होती हैं। N और S ध्रुव चुंबकीय क्षेत्र की दिशा को इंगित करते हैं।
1. It is seen that every point of the loop forms a centre of a large number of concentric magnetic lines of force forming a series. The circles are small near the wire and become large as we move away from the wire. At the centre of the loop, the arcs of these circles appear as straight lines because of very large radius of the circle.
2. The magnetic field produced by a current-carrying wire at a given point is directly proportional to the current through the wire. If the loop has n turns, the field produced is n times that produced by a single turn (assuming that all the turns have practically the same radius and are in the same plane). The reason is the current in each turn has the same direction and the field due to each turn contributes equally to the total field.
In simple words: For a circular current loop, magnetic field lines form concentric circles near the wire and appear straight at the center. The magnetic field strength is directly proportional to the current and, for a coil, increases N times if it has N turns.

🎯 Exam Tip: Understand that the magnetic field pattern resembles that of a bar magnet. The proportionality to 'n' (number of turns) is important for solenoids and coils.

Question 30. Write Fleming's left hand rule.
Answer: Fleming's left hand rule: The left hand thumb, index finger, and the middle finger are stretched so as to be perpendicular to each other. If the index finger is in the direction of the magnetic field, and the middle finger points in the direction of the current, then the direction of the thumb in the direction of the force on the conductor.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फ्लेमिंग के बाएं हाथ के नियम को दर्शाता है। इसमें बाएं हाथ के अंगूठे, तर्जनी उंगली (index finger) और मध्यमा उंगली (middle finger) को एक-दूसरे के लंबवत फैलाया गया है। अंगूठा चालक पर लगने वाले बल की दिशा, तर्जनी उंगली चुंबकीय क्षेत्र की दिशा, और मध्यमा उंगली विद्युत धारा की दिशा को इंगित करती है।
[Note: A magnetic field exerts a force on a current-carrying conductor. Electric current is the time rate of flow of electric charge. Thus, a magnetic field exerts a force on a moving charge. This property is used to accelerate charged particles such as protons, deuterons and alpha particles, as well as electrons, to very high energies. A machine used for this purpose is called a charged particle accelerator. It may be linear or circular in design and very big in size. Such high energy particles are used to study the structure of matter. ]
In simple words: Fleming's left-hand rule helps find the direction of force on a current-carrying conductor in a magnetic field: thumb points to force, forefinger to magnetic field, and middle finger to current, all perpendicular to each other.

🎯 Exam Tip: Fleming's left-hand rule is crucial for motors. Remember "F-B-I" (Force-Thumb, Field-Forefinger, Current-Middle finger) to recall the directions for easy application.

Question 31. What is electric motor?
Answer: A device that converts electric energy into mechanical energy is called an electric motor.
In simple words: An electric motor is a device that changes electrical energy into useful mechanical motion or energy.

🎯 Exam Tip: The core function of an electric motor is energy conversion: electrical to mechanical. This is a fundamental definition.

Question 32. State the principle on which the working of an electric motor is based.
Answer: An electric motor works on the principle that a current-carrying conductor placed in a magnetic field experiences a force. In this case, the forces acting on different parts of the coil of the motor produce the rotational motion of the coil.
In simple words: Electric motors operate on the principle that a current-carrying wire experiences a force when placed in a magnetic field, causing the motor's coil to rotate.

🎯 Exam Tip: The principle is the magnetic force on a current-carrying conductor. Link this to the rotational motion in a motor.

Question 33. State the uses/applications of an electric motor.
Answer: Uses/applications of an electric motor:
1. In domestic appliances such as a mixer, a blender, a refrigerator and washing machine.
2. In an electric fan, a hair dryer, a record player, a tape recorder and a blower.
3. In an electric car, a rolling mill, an electric crane, an electric lift, a pump, a computer and an electric train.
In simple words: Electric motors are widely used in daily life, powering everything from kitchen appliances like mixers and washing machines to electric fans, cars, cranes, and even computers.

🎯 Exam Tip: Be able to provide multiple examples of electric motor applications, demonstrating a broad understanding of their impact in technology and daily life.

Question 34. (i) Which principle is explained in this figure?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक धारावाही चालक को एक चुंबकीय क्षेत्र (N और S ध्रुवों के बीच) में दर्शा रहा है। चालक से प्रवाहित धारा (I) और चुंबकीय क्षेत्र की दिशा दिखाई गई है, जो यह बताता है कि चालक पर एक बल कार्य करेगा जिससे यह गति करेगा।
Answer: (i) A force is exerted on a current-carrying conductor in the presence of a magnetic field.
(ii) Which rule is used to find out the direction of force in this principle?
Answer: (ii) Fleming's left hand rule is used.
(iii) In which machine is this principle used? Draw a diagram showing the working of that machine.
Answer: (iii) Electric motor.
In simple words: This figure illustrates the principle that a current-carrying conductor experiences a force in a magnetic field. This force's direction is found using Fleming's left-hand rule, and this principle is fundamental to the operation of an electric motor.

🎯 Exam Tip: Connect the visual representation (diagram) to the underlying principle (force on a current-carrying conductor in a magnetic field), the rule for direction (Fleming's left-hand rule), and its primary application (electric motor). This shows a comprehensive understanding.

 

**Question 35.** Observe the following diagram and answer the questions.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक इलेक्ट्रिक मोटर की संरचना को दर्शाता है। इसमें एक आयताकार कुंडली (ABCD) को शक्तिशाली चुंबक (N और S ध्रुव) के ध्रुव खंडों के बीच रखा गया है। कुंडली के सिरे स्प्लिट रिंग्स (X और Y) से जुड़े होते हैं, जो कार्बन ब्रश (E और F) के संपर्क में रहते हैं और एक धुरी (Axle) पर लगे होते हैं। (a) Construction of which equipment does the following diagram show? (b) On which principle does this equipment work? (c) According to which law does the coil ABCD rotate? (d) Write the law in your own words. (e) Where is this equipment used?
Answer:(a) Given diagram shows the construction of an electric motor. (b) An electric motor works on the principle that a current-carrying conductor placed in a magnetic field experiences a force. In this case, the forces acting on different parts of the coil of the motor produce the rotational motion of the coil. (c) The rotation of the coil is based on Fleming's left hand rule. (d) Fleming's left hand rule: The left hand thumb, index finger, and the middle finger are stretched so as to be perpendicular to each other. If the index finger is in the direction of the magnetic field, and the middle finger points in the direction of the current, then the direction of the thumb in the direction of the force on the conductor. (e) Uses / applications of an electric motor:
(1) In domestic appliances such as a mixer, a blender, a refrigerator and washing machine.
(2) In an electric fan, a hair dryer, a record player, a tape recorder and a blower.
(3) In an electric car, a rolling mill, an electric crane, an electric lift, a pump, a computer and an electric train.In simple words: This question describes an electric motor, how it's built, the principle it uses (force on a current-carrying conductor in a magnetic field), the rule for determining force direction (Fleming's left-hand rule), and its various real-world applications.

🎯 Exam Tip: Understanding the construction, working principle, and applications of electric motors is crucial. Be prepared to explain Fleming's left-hand rule clearly and list multiple uses.

 

**Question 36.** Study the following principle and answer the questions. A force is excreted on a current-carrying conductor placed in a magnetic field. The direction of this force depends on both the direction of the current and the direction of the magnetic field. This force is maximum when the direction of the current is perpendicular to the direction of the magnetic field. (a) By which law can we determine the direction of the force excreted on the current-carrying conductor? (b) In which electrical equipment is this principle used? (c) Draw a diagram representing the construction of this equipment. (d) Write the working of this equipment in brief.
Answer:(a) Fleming's left hand rule. (b) Electric motor.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक इलेक्ट्रिक मोटर की संरचना को दर्शाता है। इसमें एक आयताकार कुंडली (ABCD) को शक्तिशाली चुंबक (N और S ध्रुव) के ध्रुव खंडों के बीच रखा गया है। कुंडली के सिरे स्प्लिट रिंग्स (X और Y) से जुड़े होते हैं, जो कार्बन ब्रश (E और F) के संपर्क में रहते हैं और एक धुरी (Axle) पर लगे होते हैं। यह एक धारावाही चालक पर चुंबकीय क्षेत्र में लगने वाले बल को दर्शाता है। (d) Working:
1. When the circuit is completed with a plug key or switch, the current flows in the direction E \( \implies \) A\( \implies \)B\( \implies \)C\( \implies \) D \( \implies \) F. As the magnetic field is directed from the north pole to the south pole, the force on AB is downward and that on CD is upward by Fleming's left hand rule. Hence, AB moves downward and CD upward. These forces are equal in magnitude and opposite in direction. Therefore, as observed from the side AD, the loop ABCD and the axle start rotating in anticlockwise direction.
2. After half a rotation, X and Y come in contact with brushes F and E respectively and the current flows in the direction EDCBAF. Hence the force on CD is downward and that on AB is upward. Therefore, the loop and the axle continue to rotate in the anticlockwise direction.
3. After every half rotation, the current in the loop is reversed and the loop and the axle continue to rotate in anti clockwise direction. When the current is switched off, the loop stops rotating after some time.In simple words: This question explains the principle of force on a current-carrying conductor in a magnetic field, identifies Fleming's left-hand rule for direction, and describes the construction and working of an electric motor based on this principle.

🎯 Exam Tip: Memorize Fleming's left-hand rule and its application in an electric motor. Practice drawing and labeling the diagram of an electric motor, and be able to explain its working steps clearly.

 

**Question 37.** What is a galvanometer used for? Explain in brief the working of a galvanometer.
Answer:Galvanometer is a sensitive device used to detect the presence of current in a circuit as well as to determine the direction of the current in the circuit. With suitable modification, it can be used to measure charge, current and voltage. Its working is based on the same principle as that of an electric motor. Here, a coil is pivoted (or suspended) between the pole pieces of a magnet and a pointer is connected to the coil. As the coil rotates when a current is passed through it, the pointer also rotates. The rotation of the coil and hence the deflection of the coil is proportional to the current. The pointer deflects on both sides of the Central zero mark depending on' the direction of the current.In simple words: A galvanometer is a device for detecting and measuring small electric currents and voltages, working on the principle that current causes a coil to rotate in a magnetic field, deflecting a pointer.

🎯 Exam Tip: Know the primary function of a galvanometer and its basic operating principle. Differentiate it from an ammeter or voltmeter by its sensitivity and ability to detect direction.

 

**Question 38.** Take a coil AB having 10-15 turns. Connect the two ends of the coil to the galvanometer as shown in Figure. Take a strong bar magnet.
(1) Move the north pole of the magnet towards the end B of the coil.
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में एक कुंडली (coil AB) को एक गैल्वेनोमीटर (G) से जोड़ा गया है। एक छड़ चुंबक (जिसके N और S ध्रुव चिह्नित हैं) को कुंडली के पास दिखाया गया है, जो विद्युत चुम्बकीय प्रेरण प्रयोग की व्यवस्था को दर्शाता है। Observe the deflection of the pointer in the galvanometer. Note the direction of the deflection (i.e. right or left).
(2) Now repeat this with the south pole of the magnet towards the end B of the coil. Again observe the deflection. Note its direction.
(3) What will happen if instead of the magnet, the coil is moved?
(4) If both the coil and the magnet are kept stationary, do you observe any deflection?
(5) Compare the direction of the deflection when the north pole of the magnet is moved towards the end B of the coil with that when the end B of the coil is moved away from the north pole of the magnet.
(6) What conclusions do you draw from the observations?
Answer:Observations: The two deflections, in parts (1) and (2) of the experiment, are in the opposite directions. (3) If instead of the magnet, the coil is moved towards the stationary magnet, the deflection of the pointer in the galvanometer is observed in one direction, while if the coil is moved away from the magnet, the deflection is observed in the opposite direction. The effect of moving the north pole of the magnet towards the coil and the effect of moving the coil towards the north pole of the magnet are the same. (4) If both the coil and the magnet are kept stationary, no deflection is observed. (5) The two deflections are in opposite directions. (6) Whenever there is relative motion of the coil and the magnet, electric potential difference is induced in the circuit which gives rise to, i.e., induces, an electric current in the circuit causing the deflection of the pointer in the galvanometer. The direction of the current and hence that of the deflection of the pointer in the galvanometer depends on which pole of the magnet faces the coil as well as the direction of relative motion.In simple words: This experiment demonstrates electromagnetic induction, showing that current is induced in a coil only when there's relative motion between the coil and a magnet, and the direction of the induced current depends on the direction of relative motion and the magnet's pole.

🎯 Exam Tip: This question tests your understanding of Faraday's law of electromagnetic induction. Focus on explaining the role of relative motion and how it influences the induced current's direction and magnitude.

 

**Question 39.** Take two coils of about 50 turns. Insert them over a nonconducting cylindrical roll as shown in Figure.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र विद्युत चुम्बकीय प्रेरण के एक प्रयोग को दर्शाता है जिसमें दो कुंडलियाँ (Coil 1 और Coil 2) एक अचालक बेलनाकार रोल पर रखी गई हैं। Coil 1 एक बैटरी और प्लग कुंजी (K) से जुड़ी है, जबकि Coil 2 एक गैल्वेनोमीटर (G) से जुड़ी है। यह यह देखने के लिए है कि कैसे एक कुंडली में धारा बदलने से दूसरी कुंडली में प्रेरण होता है। (A thick paper roll can be used.) Connect coil 1 to a battery with a plug key K. Connect coil 2 to a galvanometer G.
(1) Plug the key and observe the deflection in the galvanometer.
(2) Unplug the key and again observe the deflection. Note your observations. What conclusions do you draw from these observations?
Answer:Observations :
1. When the key is plugged, the galvanometer shows a momentary deflection. When the current in coil 1 becomes steady, the galvanometer shows zero deflection, i.e., its pointer returns to the zero mark at the centre of the scale.
2. When the key is unplugged, the galvanometer shows a momentary deflection in the opposite direction relative to that in part (1) of the experiment. When the current in coil 1 becomes zero as the circuit is broken on unplugging the key, the galvanometer shows zero deflection, i.e., its pointer returns to the zero mark at the centre of the scale. Conclusions: As the current in coil 1 changes, the magnetic field associated with the current changes. This induces an electric potential difference in coil 2 which gives rise to an electric current and hence the deflection of the galvanometer. The direction of the induced current and hence that of the deflection of the pointer in the galvanometer depends on whether the current through coil 1 increases or decreases with time. When there is a steady current in coil 1, there is no change in the associated magnetic field and hence no production of induced potential difference in coil 2. In that case there is no current in coil 2 and hence the galvanometer shows zero deflection.In simple words: This experiment demonstrates mutual induction, where a changing current in one coil induces a momentary current in a nearby second coil, as observed by a galvanometer deflection, but only when the current is changing, not steady.

🎯 Exam Tip: This question illustrates mutual induction. Focus on explaining that induced current only occurs during a change in magnetic flux (i.e., when current in the primary coil starts or stops, not when it's steady).

 

**Question 40.** What is electromagnetic induction? Who discovered it?
Answer:The process by which a changing magnetic field in a conductor induces a current in another conductor is called electromagnetic induction. A current can be induced in a conductor either by moving it in a magnetic field or by changing the magnetic field around the conductor. Electromagnetic induction was discovered by Michael Faraday in 1831 and independently by Joseph Henry in 1830.In simple words: Electromagnetic induction is the process of generating an electric current in a conductor by changing the magnetic field around it, discovered by Michael Faraday and Joseph Henry.

🎯 Exam Tip: Clearly define electromagnetic induction and remember the names of its discoverers. This is a fundamental concept in electricity and magnetism.

 

**Question 41.** State Faraday's law of induction.
Answer:Whenever the number of magnetic lines of force passing through a coil changes, a current is induced in the coil.In simple words: Faraday's law states that an electric current is induced in a coil if the magnetic field lines passing through it change.

🎯 Exam Tip: This is a core law. Ensure your statement is precise, emphasizing the "change" in magnetic flux as the condition for induction.

 

**Question 42.** State Fleming's right hand rule.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र फ्लेमिंग के दाहिने हाथ के नियम को दर्शाता है, जहाँ अंगूठा चालक की गति की दिशा (Motion of the conductor), तर्जनी उंगली चुंबकीय क्षेत्र की दिशा (Direction of the magnetic field), और मध्यमा उंगली प्रेरित धारा की दिशा (Direction of the induced current) को इंगित करती है। तीनों उंगलियां एक-दूसरे के लंबवत फैली हुई हैं। Stretch the thumb, the index finger and the middle finger of the right hand in such a way that they are perpendicular to each other. In this position, the thumb indicates the direction of the motion of the conductor, the index finger the direction of the magnetic field, and the middle finger shows the direction of the induced current.In simple words: Fleming's right-hand rule helps determine the direction of induced current when a conductor moves in a magnetic field: thumb for motion, forefinger for field, and middle finger for induced current.

🎯 Exam Tip: Clearly differentiate Fleming's right-hand rule (for induced current/generators) from the left-hand rule (for force/motors). Practice illustrating how each finger represents a specific direction.

 

**Question 43.** Observe the following figure. If the current in the coil A is changed, will some current be induced in the coil B? Explain.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में दो कुंडलियाँ (A और B) एक-दूसरे के पास रखी गई हैं। कुंडली A एक बैटरी और कुंजी से जुड़ी है जो उसमें धारा प्रवाहित करती है, जबकि कुंडली B एक गैल्वेनोमीटर (G) से जुड़ी है। यह व्यवस्था चुंबकीय प्रेरण के सिद्धांत को समझाने के लिए है, जहाँ कुंडली A में धारा के परिवर्तन से कुंडली B में एक धारा प्रेरित होती है। If the current in the coil A is changed, there will be some current induced in the coil B. Explanation: When the current in the coil A is changed, the magnetic field associated with the current changes. This induces potential difference in the coil B. This gives rise to (i.e., induces) a current in the coil B. The greater the rate at which the current in the coil A is changed with respect to time, the greater is the current induced in the coil B as can be seen from the deflection of the pointer in the galvanometer. This phenomenon is known as electromagnetic induction.In simple words: Yes, changing the current in coil A will induce current in coil B due to electromagnetic induction, because the changing current in coil A creates a changing magnetic field that links with coil B.

🎯 Exam Tip: This question is another example of mutual induction. Focus on the concept that a *changing* magnetic field is necessary to induce current. A steady current produces a steady magnetic field, which does not induce current in another coil.

 

**Question 44.** What is a direct current (DC)?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक DC परिपथ को दर्शाता है जिसमें एक बैटरी, प्लग कुंजी और एक प्रतिरोधक (R) जुड़ा हुआ है। यह DC धारा के स्थिर प्रवाह के लिए एक सरल सेटअप है। A nonoscillatory current that flows only in one direction is called a direct current (DC). It can change in magnitude, but its direction remains the same.In simple words: Direct current (DC) is an electric current that flows in only one constant direction, although its magnitude can vary.

🎯 Exam Tip: Define DC clearly by its unidirectional flow. Understand that while its direction is constant, its magnitude can still fluctuate, unlike alternating current.

 

**Question 45.** What is an alternating current (AC)?
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक AC जनरेटर परिपथ को दर्शाता है जिसमें एक AC जनरेटर, प्लग कुंजी और एक प्रतिरोधक (R) जुड़ा हुआ है। इसके नीचे एक ग्राफ दिखाया गया है जो समय के साथ AC धारा के परिमाण और दिशा में आवधिक परिवर्तन को प्रदर्शित करता है, जिसमें DC धारा की तुलना भी की गई है। A current that changes in magnitude and direction after equal intervals of time is called an alternating current (AC).In simple words: Alternating current (AC) is an electric current that periodically reverses its direction and continuously changes its magnitude over time.

🎯 Exam Tip: The key characteristics of AC are its periodic change in both magnitude and direction. Be ready to explain why this makes it suitable for long-distance power transmission.

 

**Question 45.** What is the value of frequency of AC in India?
Answer:In India, the value of frequency of AC is 50 hertz.In simple words: In India, the alternating current completes 50 cycles of changing direction every second, meaning its frequency is 50 hertz.

🎯 Exam Tip: Remember the standard frequency of AC in India (50 Hz). This is a common general knowledge question related to electricity.

 

**Question 46.** What is the periodic time of AC in India?
Answer:In India, the periodic time of AC is 0.02 s \(( = \frac{1}{50} \text{ s})\)In simple words: The periodic time for AC in India is 0.02 seconds, meaning it takes 0.02 seconds for one complete cycle of current direction change.

🎯 Exam Tip: Relate periodic time to frequency (T = 1/f). If the frequency is 50 Hz, the periodic time is 1/50 = 0.02 seconds.

 

**Question 47.** State one advantage of AC over DC.
Answer:One advantage of AC over DC is that electric power can be transmitted over long distances without much loss of energy.In simple words: AC is preferred over DC for long-distance power transmission because it can be easily stepped up or down using transformers, minimizing energy loss during transmission.

🎯 Exam Tip: The ability to easily step up/down voltage with transformers is the primary reason AC is used for long-distance power transmission, leading to reduced energy loss.

 

**Question 48.** Name two appliances/devices in which a direct current is used.
Answer:A direct current is used in a portable electric torch and radio.In simple words: Devices like flashlights and radios typically use direct current, often supplied by batteries.

🎯 Exam Tip: Think of battery-powered devices as examples of DC usage. Common examples include torches, mobile phones, and small electronic gadgets.

 

**Question 49.** Name two appliances/devices in which an alternating current is used. (OR) State any two uses of an AC generator.
Answer:An alternating current is used in an electric heater and a refrigerator.In simple words: Appliances commonly found in homes, such as electric heaters and refrigerators, operate using alternating current.

🎯 Exam Tip: Most large household appliances and industrial equipment use AC. Examples like electric irons, washing machines, and air conditioners are good to remember.

 

**Question 50.** What is
(1) an electric generator
(2) an AC generator
(3) a DC generator?
Answer:(1) A device which converts mechanical energy into electric energy is called an electric generator. (2) A generator which converts mechanical energy into electric energy in the form of an alternating current (AC) is called an AC generator. (3) A generator which converts mechanical energy into electric energy in the form of a direct current (DC) is called a DC generator.In simple words: An electric generator converts mechanical energy into electrical energy; an AC generator produces alternating current, while a DC generator produces direct current.

🎯 Exam Tip: Clearly define each type of generator based on its energy conversion and the type of current it produces (AC or DC).

 

**Question 51.** State the principle on which the working of an electric generator is based.
Answer:The working of an electric generator is based on the principle of electromagnetic induction. When the coil of an electric generator rotates in a magnetic field, a current is induced in the coil. This induced current then flows in the circuit connected to the coil.In simple words: An electric generator works on electromagnetic induction, where rotating a coil in a magnetic field induces an electric current.

🎯 Exam Tip: Electromagnetic induction is the core principle. Emphasize the conversion of mechanical energy to electrical energy via this induction.

 

**Question 52.** Show graphically variation of AC with time. Explain the nature of the graph.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह ग्राफ समय के साथ AC धारा के परिवर्तन को दर्शाता है। इसमें धारा का मान शून्य से अधिकतम होकर वापस शून्य पर आता है, फिर विपरीत दिशा में अधिकतम होकर फिर शून्य पर लौटता है, इस प्रकार एक पूर्ण चक्र पूरा करता है, जो AC की आवधिक और दोलनी प्रकृति को स्पष्ट करता है। In this case, the frequency of the alternating current (AC) produced is 50 Hz. The coil completes 50 rotations every second. The time for one rotation of the coil is \(\frac{1}{50}\) second. It is called the periodic time or simply the period of AC. Positive current means the current flows in one direction and negative current means the current flows in the opposite direction in the external circuit. Here, the maximum value of AC is 5 A.In simple words: The graph of AC current against time shows a sinusoidal wave, indicating that the current's magnitude and direction continuously change periodically, with positive values representing one direction and negative values the opposite.

🎯 Exam Tip: Be able to draw and interpret the sinusoidal waveform of AC. Understand that the positive and negative halves represent current flowing in opposite directions.

 

**Question 53.** Observe the figure and answer the following questions.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक सरल जनरेटर (generator) की योजनाबद्ध व्यवस्था को दर्शाता है, जहाँ एक आयताकार कुंडली (armature coil) को दो शक्तिशाली चुम्बकों (S-poles) के बीच एक चुंबकीय क्षेत्र में घुमाया जा रहा है, जिससे विद्युत धारा प्रेरित होती है, जैसा कि एक बल्ब के जलने से संकेत मिलता है। (a) Identify the machine shown in the figure. (b) Write a use of this machine. (c) How transformation of energy takes place in this machine.
Answer:(a) The instrument shown in the figure is generator. (b) This machine is used to generate electricity. (c) The generator generates electricity through following transformation: Mechanical Energy \( \implies \) Electrical EnergyIn simple words: The figure displays a generator, a device that produces electricity by converting mechanical energy into electrical energy through electromagnetic induction.

🎯 Exam Tip: Recognize the basic components and function of a generator from its diagram. Clearly state the energy transformation involved (mechanical to electrical).

 

**Question 54.** Observe the following figure. Which bulb will fuse?
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दो बल्बों (Bulb A और Bulb B) को एक विद्युत परिपथ में दर्शाता है, जो एक वोल्टेज स्रोत से जुड़े हैं। Bulb A को 50W-220V और Bulb B को 100W-220V के रूप में मूल्यांकित किया गया है, लेकिन वे 440V के स्रोत से जुड़े हैं। यह यह आकलन करने के लिए है कि कौन सा बल्ब अधिक वोल्टेज के कारण फ्यूज होगा।
Answer: Bulb A.In simple words: Bulb A will fuse because, despite having a higher resistance, its filament is typically thinner and designed for a lower power output, making it more vulnerable to damage from the significantly increased power (four times its rated power) when connected to a much higher voltage.

🎯 Exam Tip: To determine which bulb fuses under overvoltage, calculate their actual power dissipation and compare it to their rated power. Generally, the bulb with the lower power rating and thinner filament will be more susceptible to fusing under severe overload conditions.

 

Give Scientific Reasons:

 

**Question 1.** In an electric iron, the coil of high resistance is kept between mica sheets.
Answer:
1. Mica is a bad conductor of electricity and good conductor of heat.
2. In an electric iron, the coil of high resistance is kept between mica sheets so that there is no electrical contact between the coil and the heavy metal block of the iron though there is heat transfer. This protects the user from getting an electric shock.In simple words: Mica sheets are used in electric irons to insulate the high-resistance heating coil electrically from the outer metal body while efficiently transferring the generated heat to the iron's base, preventing electric shocks to the user.

🎯 Exam Tip: Understand the dual properties of mica (electrical insulator, good heat conductor) that make it ideal for heating appliances where heat transfer is needed without electrical contact.

 

**Question 2.** The material used for fuse has low melting point. OR A fuse should be made of a material of low melting point.
Answer:
1. A fuse is used to protect a circuit and the appliances connected in the circuit by stopping the flow of an excessive electric current. For this, a fuse is connected in series in the circuit.
2. When the current in the circuit passes through the fuse, its temperature increases. When the current exceeds the specified value, the fuse must melt to break the circuit. For this, the material used for a fuse has low melting point.In simple words: A fuse is made of a low melting point material so that it can quickly melt and break the circuit when an excessive current flows through it, thereby protecting appliances from damage.

🎯 Exam Tip: The low melting point is critical for the fuse's protective function. Explain the chain reaction: excess current \( \implies \) increased temperature \( \implies \) melting \( \implies \) circuit break \( \implies \) protection.

 

Distinguish Between The Following:

 

**Question 1.** Direct current and Alternating current.
Answer:Direct current:
1. Direct current flows only in one direction.
2. It cannot be used for large scale of electricity for household purpose. Alternating current:
1. Alternating current reverses its direction periodically with time.
2. It is used in household electrical appliances such as an electric heater, an electric iron, a refrigerator, etc.In simple words: Direct current (DC) flows in a single direction, while alternating current (AC) periodically reverses its direction; AC is widely used for household power and long-distance transmission, whereas DC is common in battery-powered devices.

🎯 Exam Tip: For distinction questions, use a clear point-by-point comparison. Focus on direction of flow, ease of transmission, and common applications for both AC and DC.

 

**Question 2.** Electric motor and Electric generator.
Answer:Electric motor:
1. A battery is used in an electric motor to pass a current through the coil.
2. In this case, a current-carrying coil is set in rotation due to the magnetic field.
3. Split rings are used in an electric motor.
4. In this case, electric energy is converted into mechanical energy. Electric generator:
1. A battery is not used in an electric generator.
2. In this case, a potential difference and hence a current is produced when the coil is set into rotation in the magnetic field by an external agent.
3. Rings used in an AC generator are not split.
4. In this case, mechanical energy is converted into electric energy.In simple words: An electric motor converts electrical energy into mechanical energy by using current to create motion, while an electric generator converts mechanical energy into electrical energy by inducing current through rotational motion in a magnetic field.

🎯 Exam Tip: The core distinction lies in energy conversion (motor: electrical to mechanical; generator: mechanical to electrical) and the role of current (motor: current causes motion; generator: motion causes current). Note the ring types (split for DC motor/generator, slip for AC generator).

 

Question 1.An electric bulb is connected to a source of 250 volts. The current passing through it is 0.27 A. What is the power of the bulb?
Answer:Solution: Data: P = 100 W, I = 3 A, R = ?, P = I\(^2\)R P = VI = 250 V × 0.27 A = 67.5 W The power of the bulb = 67.5 W.
In simple words: To find the power of the bulb, multiply the given voltage (250 V) by the current (0.27 A) flowing through it. This calculation directly gives the power consumed by the bulb.

🎯 Exam Tip: Remember the formula P = VI for calculating power when voltage and current are known. Ensure units are consistent (Volts, Amperes, Watts).

 

Question 2.If a bulb of 60 W is connected across a source of 220 V, find the current drawn by it.
Answer:Solution: Data: P = 60 W, V = 220 V, I = ? P = VI
\( \implies \) I = \(\frac{P}{V}\) = \(\frac{60 W}{220 V}\) = \(\frac{3}{11}\) A = 0.2727 A The current drawn by the bulb = \(\frac{3}{11}\) A = 0.2727 A
In simple words: To find the current drawn by the bulb, divide its power (60 W) by the voltage (220 V) of the source it's connected to. This gives the rate of charge flow through the bulb.

🎯 Exam Tip: Practice rearranging the power formula (P=VI) to solve for unknown variables like current or voltage. Accuracy in calculation is key for numerical problems.

 

Question 3.A bulb of 40 W is connected across a source of 220 V. Find the resistance of the bulb.
Answer:Solution: Data: P = 40 W, V = 220 V, R = ? P = VI = V (\(\frac{V}{R}\)) = \(\frac{V^2}{R}\)
\( \implies \) R = \(\frac{V^2}{P}\) = \(\frac{(220 V)^2}{40 W}\) = \(\frac{220 \times 220}{40}\) \(\Omega\) = \(\frac{220 \times 220}{2 \times 20}\) \(\Omega\) = 40 × 110 \(\Omega\) = 1210 \(\Omega\) The resistance of the bulb = 1210 \(\Omega\)
In simple words: To find the resistance of the bulb, square the voltage (220 V) and divide it by the power (40 W). This calculation applies Ohm's law in conjunction with the power formula.

🎯 Exam Tip: Remember the three common power formulas (P=VI, P=I\(^2\)R, P=V\(^2\)/R) and choose the appropriate one based on the given parameters. Show all steps clearly.

 

Question 4.If the current passing through a bulb is 0.2 A and the power of the bulb is 20 W, find the voltage applied across the bulb.
Answer:Solution: Data: I = 0.2 A, P = 20 W, V = ? P = VI
\( \implies \) V = \(\frac{P}{I}\) = \(\frac{20W}{0.2A}\) = 100 V The voltage across the bulb = 100 V.
In simple words: To find the voltage across the bulb, divide its power (20 W) by the current (0.2 A) flowing through it. This directly applies the power formula to solve for voltage.

🎯 Exam Tip: Always state the formula used before substituting values. Double-check calculations, especially with decimals, to avoid errors.

 

Question 5.Two tungsten bulbs of power 50 W and 60 W work on 220 V potential difference. If they are connected in parallel, how much current will flow in the main conductor?
Answer:Solution: Data: P\( _1 \) = 50 W, P\( _2 \) = 60 W, V = 220 V, I = ? P=VI
\( \implies \) I\( _1 \) = \(\frac{P_1}{V}\) and I\( _2 \) = \(\frac{P_2}{V}\) Current in the main conductor, I = I\( _1 \) + I\( _2 \) ..........(parallel combination) = \(\frac{P_1}{V}\) + \(\frac{P_2}{V}\) = \(\frac{P_1 + P_2}{V}\) = \(\frac{50 W + 60 V}{220V}\) = \(\frac{110}{220}\) A = \(\frac{1}{2}\) A = 0.5 A
In simple words: First, calculate the current for each bulb using P=VI. Since the bulbs are in parallel, the total current in the main conductor is the sum of the individual currents.

🎯 Exam Tip: For parallel connections, remember that voltage remains constant across components, and total current is the sum of individual currents. Clearly identify the type of connection (series/parallel) before solving.

 

Question 6.An electric iron rated 750 W is operated for 2 hours/day. How much energy is consumed by the electric iron for 30 days?
Answer:Solution: Data: P = 750 W, t = 2 \(\frac{hours}{day}\) for 30 days The energy consumed = Pt = 750 × 2 × 30 = 1500 × 30 = 45000 W-h = 45 kW-h The energy consumed by the electric iron for 30 days = 45 kW-h.
In simple words: To find the total energy consumed, multiply the power of the iron (750 W) by its daily operating hours (2 hours) and then by the number of days (30). Convert the final answer to kilowatt-hours for standard energy units.

🎯 Exam Tip: When calculating energy consumption, ensure the time is in hours and power in kilowatts if the final answer needs to be in kilowatt-hours (units). Pay attention to unit conversions.

 

Question 7.If a TV of rating 100 W operates for 6 hours per day, find the number of units consumed in a leap year.
Answer:Solution: Data: P = 100 W, t = 6 \(\frac{hours}{day}\) × 366 days = 2196 hours 1 unit = 1 kW-h = 1000 W-h N = \(\frac{Pt}{1000 W-h/unit}\) = \(\frac{100 W \times 2196 hours}{1000 W-h/unit}\) = 219.6 units 219.6 units are consumed in a leap year.
In simple words: Calculate the total operating hours in a leap year (366 days * 6 hours/day). Then, multiply this total time by the TV's power (100 W) to get energy in W-h, and divide by 1000 to convert to kilowatt-hours (units).

🎯 Exam Tip: For leap years, remember to use 366 days in your calculations. Energy consumption is typically measured in kilowatt-hours (units), so ensure the final conversion is done correctly.

 

Question 8.An electric appliance of rating 300 W is used 5 hours per day in the month of March. Find the number of units consumed.
Answer:Solution: Data: P = 300 W, t = 5 \(\frac{hour}{day}\) × 31 days = 155 hours, 1 unit = 1 kW-h = 1000 W-h N = \(\frac{Pt}{1000 W-h/unit}\) = \(\frac{300 W \times 155 hours}{1000 W-h/unit}\) = 46.5 units 46.5 units are consumed in the month of March.
In simple words: Multiply the appliance's power (300 W) by the daily operating hours (5) and the number of days in March (31) to get total energy in W-h. Then, divide by 1000 to find the units consumed in kW-h.

🎯 Exam Tip: Be mindful of the number of days in the specified month. Ensure consistent units (Watts to kilowatts, hours for time) for accurate energy unit calculation.

 

Question 9.A washing machine rated 300 W operates one hour/day. If the cost of a unit is Rs. 3.00, find the cost of the energy to operate the washing machine for the month of March.
Answer:Solution: Data: P = 300 W, Rs. 3.00 per unit, t = 1 \(\frac{hour}{day}\) × 31 days = 31 hours, 1 unit = 1 kW-h = 1000 W-h, cost of the energy = ? N = \(\frac{Pt}{1 kWh/unit}\) = \(\frac{300 W \times 31 hours}{1000 W-h/unit}\) Cost = 9.3 units × Rs. 3.00 per unit = Rs. 27.9. The cost of the energy to operate the washing machine for the month of March = Rs. 27.9.
In simple words: First, calculate the total energy consumed by multiplying the washing machine's power (300 W) by its daily operating hours (1 hour) and the days in March (31), then convert to units (kW-h). Finally, multiply the total units by the cost per unit to get the total cost.

🎯 Exam Tip: When calculating electricity costs, convert total energy consumed into kilowatt-hours (units) first, then multiply by the cost per unit. Carefully note the number of days in the specific month.

 

Question 10.Find the heat produced in joule if a current of 0.1 A is passed through a coil of resistance 50 \(\Omega\) for two minutes. Keeping other conditions the same if the length of the wire is reduced to the original length (by cutting 4 the wire), what will be the heat produced?
Answer:Solution: Data: I = 0.1 A, R = 50 \(\Omega\), t = 2 minutes = 2 × 60 s = 120 s, H = ? H = I\(^2\)Rt = (0.1A)\(^2\) × 50 \(\Omega\) × 120 s = 0.01 × 50 × 120 J = 60 J Heat produced = 60 joules. In the second case, the resistance of the wire will be \(\frac{50\Omega}{4}\) Hence, the heat produced = \(\frac{60J}{4}\) = 15 J.
In simple words: Use Joule's law (H=I\(^2\)Rt) to find the heat produced with the initial resistance and time. If the wire length is quartered, the resistance also becomes one-fourth (assuming uniform wire), so the heat produced will also be one-fourth of the original, assuming current remains constant.

🎯 Exam Tip: Remember Joule's law (H=I\(^2\)Rt). When wire length is reduced, resistance changes proportionally, affecting the heat produced. Ensure time is always in seconds for Joule's law calculations.

 

Question 11.Calculate the heat produced in calorie when a current of 0.1 A is passed through a wire of resistance 41.8 \(\Omega\) for 10 minutes.
Answer:Solution: Data: I = 0.1 A, R = 41.8 \(\Omega\), t = 10minutes = 10 × 60 s = 600 s, H = ? H = \(\frac{I^2Rt}{4.18}\) cal = \(\frac{(0.1)^2 \times 41.8 \times 600}{4.18}\) cal = \(\frac{0.01 \times 41.8 \times 600}{4.18}\) cal = 0.01 × 10 × 600 cal = 60 cal Heat produced = 60 calories.
In simple words: First, convert time to seconds. Then, calculate heat in joules using Joule's law (H=I\(^2\)Rt). Finally, convert joules to calories by dividing by 4.18 (since 1 calorie ≈ 4.18 Joules).

🎯 Exam Tip: Be careful with unit conversions, especially between joules and calories. Always convert time to seconds before applying Joule's law for heat calculations.

 

Question 12.A potential difference of 250 volts is applied across a resistance of 1000 \(\Omega\) in an electric iron. Find (1) the current (2) the heat produced in joule in 12 seconds. Keeping other conditions the same, if the length of the wire in the iron is reduced to half the original length (by cutting the wire), what will be the current and heat produced?
Answer:Solution: Data: V = 250 V, R = 1000 \(\Omega\), t = 12 s, I = ? H = ? (1) V = IR
\( \implies \) I = \(\frac{V}{R}\) = \(\frac{250V}{1000\Omega}\) = 0.25 A The current through the resistance = 0.25 A. (2) H = I\(^2\)RT = (0.25 A)\(^2\) × 1000 \(\Omega\) × 12 s = (\(\frac{1}{4}\) × \(\frac{1}{4}\) × 1000) × 12 J = \(\frac{1}{16}\) × 1000 × 12 J = 62.5 x 12 J = 750 J The heat energy produced in the resistance in 12 seconds = 750 joules. On cutting the wire, the resistance of the wire will become half the initial resistance (R' = R/2 = 1000/2 = 500 \(\Omega\)). Hence, the current will become double the initial current as I = V/R and V is the same in both the cases. New current I' = V/R' = 250V / 500\(\Omega\) = 0.5 A. New heat produced H' = Vlt = 250 V × 0.5 A × 12 s = 250 × 6 J = 1500 J.
In simple words: First, use Ohm's law (V=IR) to find the current. Then, use Joule's law (H=I\(^2\)Rt or H=VIt) to calculate heat. If the wire length is halved, resistance halves, and current doubles (for constant voltage). Recalculate heat with the new current and resistance.

🎯 Exam Tip: When a conductor's length changes, its resistance changes proportionally (R \(\propto\) L). This will affect current (if voltage is constant) and subsequently the heat produced by Joule's law. Show both initial and final calculations clearly.

 

Question 13.A potential difference of 100 V is applied across a resistor of resistance 50 \(\Omega\) for 6 minutes and 58 seconds. Find the heat produced in (i) joule (ii) calorie.
Answer:Solution: Data: V = 100 V, R = 50 \(\Omega\), t = 6 minutes and 58 seconds = (6 × 60 + 58) s = (360 + 58) = 418 s, H = ? (i) H = \(\frac{V^2t}{R}\) = \(\frac{(100 V)^2 \times 418 s}{50 \Omega}\) = \(\frac{100 \times 100 \times 418}{50}\) J = 200 × 418 J = 83600 J Heat generated = 83600 joules. (ii) H = \(\frac{V^2t}{4.18 R}\) cal = \(\frac{(100)^2 \times 418}{4.18 \times 50}\) cal = \(\frac{100 \times 100 \times 418}{4.18 \times 50}\) cal = 2 × 10\(^4\) cal Heat produced = 2 × 10\(^4\) calories.
In simple words: First, convert the total time into seconds. Then, use the formula H=\(\frac{V^2t}{R}\) to calculate the heat produced in joules. To convert to calories, divide the joule value by 4.18.

🎯 Exam Tip: Always convert time to seconds when using Joule's law formulas. Be careful with calculations involving powers of ten and the conversion factor between joules and calories.

 

Numerical Problems For Practice:

 

Question 1.When the voltage applied across a bulb is 200 V, the current passing through the bulb is 0.1 A. Find the power of the bulb.
Answer:20 W
In simple words: To find the power, simply multiply the given voltage (200 V) by the current (0.1 A). Power is the product of voltage and current.

🎯 Exam Tip: This is a direct application of the power formula P = VI. Ensure units are correct (Volts, Amperes, Watts).

 

Question 2.A bulb of 100 W is connected across a source of 200 V. Find the current drawn by it.
Answer:0.5 A
In simple words: To find the current, divide the bulb's power (100 W) by the voltage of the source (200 V). Current is power divided by voltage.

🎯 Exam Tip: Practice rearranging the power formula (P=VI) to solve for current. Always write down the formula before substituting values.

 

Question 3.A bulb of 60 W is connected across a source of 240 V. Find the resistance of the bulb.
Answer:960 \(\Omega\)
In simple words: To find the resistance, square the voltage (240 V) and divide it by the power of the bulb (60 W). This uses the relation between power, voltage, and resistance.

🎯 Exam Tip: Utilize the formula R = V\(^2\)/P. Accuracy in squaring the voltage and division is crucial.

 

Question 4.If the current passing through a bulb is 0.15 A and the power of the bulb is 30 W, find the voltage applied across the bulb.
Answer:200 V
In simple words: To find the voltage, divide the power of the bulb (30 W) by the current (0.15 A) flowing through it. Voltage is power divided by current.

🎯 Exam Tip: Another direct application of P=VI. Remember to isolate V (V=P/I) for solving.

 

Question 5.An electric appliance of rating 800 W is used 4 hours per day in the month of December. Find the number of units consumed.
Answer:99.2 units
In simple words: Multiply the appliance's power (800 W) by the daily operating hours (4) and the number of days in December (31). Then, divide the total energy in W-h by 1000 to convert it into kilowatt-hours, which are "units".

🎯 Exam Tip: Ensure you use the correct number of days for the given month (December has 31 days). Convert W-h to kW-h (divide by 1000) for "units" consumed.

 

Question 6.An electric appliance rated 400 W is used 5 hours per day in the month of June. If the cost of a unit is Rs. 3.00, find the energy bill for June.
Answer:Rs. 180
In simple words: Calculate total energy in W-h by multiplying power (400 W), daily hours (5), and days in June (30). Convert this to units (kW-h) by dividing by 1000. Finally, multiply the total units by the cost per unit (Rs. 3.00) to get the bill.

🎯 Exam Tip: Pay attention to the specific month (June has 30 days). The final step involves multiplying the total units by the given cost per unit for the energy bill.

 

Question 7.An electric bulb rated 60 W is used 10 hours per day for 20 days. If the cost of a unit is Rs. 3.00, find the energy bill.
Answer:Rs. 36
In simple words: Multiply the bulb's power (60 W) by its daily operating hours (10) and the number of days (20) to find total energy in W-h. Convert this to units (kW-h) and then multiply by the cost per unit (Rs. 3.00) to get the energy bill.

🎯 Exam Tip: Total operating time (daily hours x number of days) is a critical component for calculating energy consumption. Remember the conversion to kilowatt-hours before calculating the cost.

 

Question 8.Two electric bulbs rated 60 W and 40 W respectively are used 5 hours per day for 20 days. If the cost of a unit is Rs. 4.00, find the cost of the energy used.
Answer:Rs. 40
In simple words: Calculate the total power of both bulbs (60 W + 40 W). Then, multiply this combined power by daily operating hours (5) and the number of days (20) to get total energy in W-h. Convert to units (kW-h) and multiply by the cost per unit (Rs. 4.00).

🎯 Exam Tip: When multiple appliances are used, sum their powers if they operate simultaneously for the same duration. The process then follows standard energy consumption and cost calculation.

 

Question 9.Find the heat produced in joule if a current of 0.1 A is passed through a coil of resistance 25 \(\Omega\) for one minute.
Answer:15 J
In simple words: Convert the time (one minute) to seconds. Then, use Joule's law (H=I\(^2\)Rt) by multiplying the square of the current (0.1 A), the resistance (25 \(\Omega\)), and the time in seconds.

🎯 Exam Tip: Time for Joule's law must always be in seconds. Square the current value carefully before multiplying by resistance and time.

 

Question 10.Calculate the heat produced in calorie when a current of 0.1 A is passed through a wire of resistance 41.8 \(\Omega\) for 5 minutes.
Answer:30 calories
In simple words: Convert time (5 minutes) to seconds. Calculate heat in joules using H=I\(^2\)Rt. Then, divide the result by 4.18 to convert it into calories.

🎯 Exam Tip: Remember the conversion factor: 1 calorie = 4.18 Joules. Ensure accurate unit conversions and calculations.

 

Question 11.Calculate the heat produced in calorie when a current of 0.2 A is passed through a wire of resistance 41.8 \(\Omega\) for 10 minutes.
Answer:240 calories
In simple words: Convert time (10 minutes) to seconds. Use H=I\(^2\)Rt to find heat in joules, then divide by 4.18 to get calories.

🎯 Exam Tip: Pay close attention to the current (0.2 A) and the duration (10 minutes) to avoid calculation errors. The constant 4.18 for J to cal conversion is essential.

 

Question 12.Find the heat produced in calorie when a current of 0.2 A is passed through a wire of resistance 20.9 \(\Omega\) for 10 minutes.
Answer:120 calories
In simple words: Convert time (10 minutes) to seconds. Apply H=I\(^2\)Rt to calculate heat in joules. Finally, divide by 4.18 to express the heat in calories.

🎯 Exam Tip: Double-check the input values for current, resistance, and time. Ensure the conversion from minutes to seconds and joules to calories is performed correctly.

 

Question 13.A potential difference of 100 V is applied across a wire of resistance 50 \(\Omega\) for one minute. Find the heat produced in joule.
Answer:1.2 × 10\(^4\) joules
In simple words: Convert time (one minute) to seconds. Use the formula H=\(\frac{V^2t}{R}\) by squaring the voltage (100 V), multiplying by time (60 s), and dividing by resistance (50 \(\Omega\)) to get heat in joules.

🎯 Exam Tip: When voltage, resistance, and time are given, using H=\(\frac{V^2t}{R}\) is the most direct method. Remember to convert minutes to seconds.

 

Question 14.A potential difference of 100 V is applied across a wire for two minutes. If the current through the wire is 0.1 A, find the heat produced in joule.
Answer:1200 joules
In simple words: Convert time (two minutes) to seconds. Use the formula H=VIt by multiplying the voltage (100 V), current (0.1 A), and time in seconds to find the heat produced in joules.

🎯 Exam Tip: This problem involves V, I, and t, so H=VIt is the most suitable formula. Always ensure time is in seconds.

 

Question 15.A potential difference of 100 V is applied across a wire for 6 minutes and 58 seconds. If the current through the wire is 0.1 A, find the heat produced in calorie.
Answer:1000 calories
In simple words: Convert the total time (6 minutes 58 seconds) to seconds. Calculate heat in joules using H=VIt. Then, divide the joule value by 4.18 to convert it into calories.

🎯 Exam Tip: Convert mixed time units (minutes and seconds) into total seconds accurately. Always remember to divide by 4.18 to convert joules to calories.