RBSE Solutions Class 10 Science Chapter 10 Electricity Current

I. Multiple Choice Questions

Question 1: If a 5 V battery is providing 2 A current in a conductor then what is the resistance of conductor?
(a) 3 Ohm
(b) 2.5 Ohm
(c) 10 Ohm
(d) 2 Ohm
Answer: (b) 2.5 Ohm
In simple words: To find resistance, divide the voltage (5V) by the current (2A). This calculation gives 2.5 Ohm.

🎯 Exam Tip: Remember Ohm's Law: Resistance (R) = Voltage (V) / Current (I). Always state the formula before calculation.

Question 2: Resistivity depends on which of the following?
(a) Length of conductor
(b) Transverse section of conductor
(c) Material of conductor
(d) None of these
Answer: (c) Material of conductor
In simple words: Resistivity is a fixed property of a material, like copper or iron, and does not change with its size or shape. Different materials have different resistivities.

🎯 Exam Tip: Distinguish between resistance (which depends on length and cross-section) and resistivity (which is an intrinsic material property).

Question 3: Volt is the unit of which of the following?
(a) Current
(b) Potential difference
(c) Charge
(d) Work
Answer: (b) Potential difference
In simple words: A Volt measures the "push" or energy per unit charge that makes current flow in a circuit, which is known as potential difference.

🎯 Exam Tip: Clearly know the SI units for fundamental electrical quantities: Ampere for current, Volt for potential difference, Coulomb for charge, and Ohm for resistance.

Question 5: What is the frequency of AC in India?
(a) 45 Hz
(b) 50 Hz
(c) 55 Hz
(d) 60 Hz
Answer: (b) 50 Hz
In simple words: In India, the alternating current (AC) changes direction 50 times every second. This rate of change is called its frequency.

🎯 Exam Tip: Be aware of standard electrical frequencies in different regions; 50 Hz is common in many parts of the world, while 60 Hz is used elsewhere like North America.

Question 6: What happens in each resistor when resistor of different values in parallel combination are connected to a source of electricity?
(a) Values of current and potential difference are different
(b) Values of current and potential difference are same
(c) Values of current are different but value of potential difference is same
(d) Value of current is same but values of potential difference are different
Answer: (c) Values of current are different but value of potential difference is same
In simple words: When resistors are connected side-by-side (in parallel), all of them get the same voltage from the power source. However, the current that flows through each one will be different, depending on its individual resistance.

🎯 Exam Tip: For parallel circuits, the voltage across each component is the same, but the current divides. For series circuits, the current is the same, but the voltage divides.

Question 7: If 2 Coulomb charge flows for 0.5 second in a circuit then what is the value of electric current?
(a) 1 A
(b) 4 A
(c) 1.5 A
(d) 10 A
Answer: (b) 4 A
In simple words: Electric current is found by dividing the amount of charge by the time it takes to flow. So, 2 Coulombs divided by 0.5 seconds gives 4 Amperes.

🎯 Exam Tip: Remember the definition of current: Current (I) = Charge (Q) / Time (t). Be careful with units and decimal values in calculations.

Electricity Current Very Short Answer Type Questions

Question 9: What is the unit of specific resistance or resistivity?
Answer: Ohm metre
In simple words: Resistivity is measured in "Ohm-meter", which describes how strongly a material resists electric current.

🎯 Exam Tip: Ensure you write "Ohm metre" (or "Ohm-meter") and not just "Ohm" (which is the unit of resistance). Using the symbol \( \Omega \text{m} \) is also correct.

Question 10: Define electric current.
Answer: Electric current is the flow of electric charge. This flow is usually carried by electrons moving through a conductor.
In simple words: Electric current is simply the movement of electric charges.

🎯 Exam Tip: A precise definition should include "flow of electric charge per unit time," but for a very short answer, "flow of electric charge" is often accepted.

Question 11: What is electric potential?
Answer: Electric potential, also known as electric field potential or electrostatic potential, is the amount of potential energy that a single unit of positive electric charge would have if placed at any point in space. It represents the work done per unit charge.
In simple words: Electric potential tells us how much energy each unit of charge has at a certain point. It's like how much "push" a charge can get from that spot.

🎯 Exam Tip: Focus on the idea of "energy per unit charge" or "work done per unit charge" when defining electric potential.

Question 12: How will you define 1 Ω resistance?
Answer: One Ohm (\( \Omega \)) of resistance is defined as the resistance of a conductor through which 1 Ampere (A) of electric current flows when a potential difference of 1 Volt (V) is applied across its two ends. This definition is based on Ohm's Law.
In simple words: If you push with 1 Volt and get 1 Ampere of current, then the resistance is 1 Ohm.

🎯 Exam Tip: Always relate the definition of 1 Ohm to the units of voltage (Volt) and current (Ampere) via Ohm's law.

Question 13: How does resistance depend on transverse section?
Answer: Resistance depends inversely on the area of the transverse section. This means that if the cross-sectional area of a conductor increases, its resistance decreases, and vice-versa. A thicker wire has more space for electrons to flow, reducing resistance.
In simple words: A thicker wire has less resistance, while a thinner wire has more resistance.

🎯 Exam Tip: Remember that resistance is directly proportional to length and inversely proportional to the cross-sectional area.

Question 15: What is electric power?
Answer: Electric power is the rate at which work is done by an electric current, or the rate at which electrical energy is transferred in an electric circuit per unit time. It is measured in Watts.
In simple words: Electric power is how fast electrical energy is used or produced.

🎯 Exam Tip: State clearly that power is "work done per unit time" or "energy consumed per unit time."

Question 16: 100W - 220V is written on an electric bulb. What does it mean?
Answer: This rating means that the electric bulb consumes 100 Watts of electrical power when it is operated at a potential difference of 220 Volts. The "220V" indicates the maximum voltage for the bulb to function properly and safely.
In simple words: It means the bulb uses 100 Watts of power when it gets 220 Volts of electricity.

🎯 Exam Tip: Always explain both the power rating (W) and the voltage rating (V) in terms of ideal operating conditions and power consumption.

Question 17: Which type of combination of resistors is seen in household wiring?
Answer: Parallel combination of resistors is used in household wiring. This ensures that each appliance receives the full supply voltage and can be operated independently.
In simple words: Household wiring uses parallel connections so all appliances get the same voltage and can be turned on or off separately.

🎯 Exam Tip: Be ready to explain the advantages of parallel connection (independent operation, full voltage) over series connection for household circuits.

Electricity Current Short Answer Type Questions

Question 18: What is the difference between series and parallel combinations of resistors?
Answer:

In simple words: In series, parts are connected one after another, so the total resistance goes up and the current is the same everywhere. In parallel, parts are connected side-by-side, so the total resistance goes down and the voltage is the same across each part.

🎯 Exam Tip: Clearly state the key characteristics for each combination: current and voltage distribution, and how total resistance changes.

Question 20: Two resistance wires are made up of same material and have same length. If ratio of areas of cross section is 2:11 then what is the ratio of their resistance?
Answer: We know that resistance \( R = \rho \times \frac{L}{A} \), where \( \rho \) is resistivity, \( L \) is length, and \( A \) is area of cross-section.
Since both wires are made of the same material and have the same length, \( \rho \) and \( L \) are constant.
So, resistance is inversely proportional to the area of cross-section: \( R \propto \frac{1}{A} \).
Therefore, the ratio of resistances will be the inverse of the ratio of areas.
Given the ratio of areas \( \frac{A_1}{A_2} = \frac{2}{11} \).
This means the ratio of resistances \( \frac{R_1}{R_2} = \frac{A_2}{A_1} = \frac{11}{2} \).
The ratio of their resistance is 11:2.
In simple words: Because the wires are the same length and material, resistance is opposite to area. If the areas are in a 2:11 ratio, the resistances will be in an 11:2 ratio.

🎯 Exam Tip: Always remember the formula for resistance and identify constant factors to simplify the relationship (e.g., if length and resistivity are constant, \( R \propto 1/A \)).

Question 21: Define electric potential and potential difference.
Answer: An electric potential is the amount of work needed to move a unit positive charge from a reference point to a specific point in an electric field. This is also called electric field potential.
Potential Difference: The potential difference between two points in an electric circuit is the work done to move a unit charge from one point to another. It drives the current in a circuit.
In simple words: Electric potential is the energy a charge has at a certain spot. Potential difference is the energy difference between two spots, which pushes the current.

🎯 Exam Tip: Differentiate between electric potential (energy at a point) and potential difference (energy change between two points), both being work per unit charge.

Question 22: What is the difference between AC generator and DC generator?
Answer:

In simple words: An AC generator makes current that keeps changing direction, using slip rings. A DC generator makes current flow in only one direction, using split rings to achieve this.

🎯 Exam Tip: The main difference lies in the type of current produced (AC vs DC) and the component used to collect/rectify it (slip rings vs split rings/commutator).

Electricity Current Numerical Questions

Question 24: Find the number of joule in 1 kilo watt hour.
Answer: One kilowatt-hour (kWh) is a unit of energy. To convert it to Joules, we use the definitions of power and time:
1 Kilowatt Hour = 1000 Watts \( \times \) 1 Hour
1 Kilowatt Hour = 1000 Watts \( \times \) (60 minutes \( \times \) 60 seconds)
1 Kilowatt Hour = 1000 Watts \( \times \) 3600 seconds
Since 1 Watt = 1 Joule/second:
1 Kilowatt Hour = 1000 (Joule/second) \( \times \) 3600 seconds
1 Kilowatt Hour = 3,600,000 Joules
1 Kilowatt Hour = \( 3.6 \times 10^6 \) Joules.
This unit is commonly used for electricity billing.
In simple words: One kilowatt-hour is equal to 3.6 million Joules of energy.

🎯 Exam Tip: Remember the conversion factors: 1 kilowatt = 1000 watts, 1 hour = 3600 seconds, and 1 watt = 1 joule/second. Perform multiplication carefully.

Question 25: Write Joule's law of heating.
Answer: Joule's law of heating states that when an electric current flows through a resistor, the heat (H) produced in the conductor is directly proportional to the square of the current (I), the resistance (R) of the conductor, and the time (t) for which the current flows.
The formula for the heat produced is:
\( H = I^2 R t \)
Here, \( I \) is the current, \( R \) is the resistance, and \( t \) is the time. This law explains why wires get hot when electricity passes through them.
In simple words: Joule's law says that how much heat a wire makes depends on how strong the current is (squared), how much it resists, and for how long the current flows.

🎯 Exam Tip: State the direct proportionalities clearly and remember the formula \( H = I^2 R t \). Make sure to mention the constant factors for proportionality.

Question 26: Make a diagram of experimental verification of Ohm's.
Answer: The experimental setup for Ohm's law verification involves a circuit diagram and a V-I graph. The circuit shows a battery, a key, an ammeter in series, a resistor, a voltmeter in parallel across the resistor, and a rheostat for varying current. The graph shows a linear relationship between potential difference (V) and current (I), confirming Ohm's law.

In simple words: The diagram shows how to set up the experiment with a battery, switch, a changing resistor, and meters. The graph plots the voltage against the current, which should be a straight line for Ohm's law.

🎯 Exam Tip: When drawing circuit diagrams, ensure all components are correctly symbolized and connected (ammeter in series, voltmeter in parallel). For graphs, label axes correctly and show a linear relationship for Ohm's law.

Question 28: What are the main components of an AC generator and how does it work?
Answer: An AC generator has several main components:

• Field Magnet: This is a very strong magnet that creates a magnetic field, as shown by the North (N) and South (S) poles in the figure.
• Armature: This is a rectangular coil (ABCD) made of insulated copper wire wound around a soft iron core. It rotates within the magnetic field.
• Slip Rings: These are two metal rings (S1 and S2) that are connected to the ends (A and D) of the coil. They rotate along with the coil and provide contact for the brushes.
• Brushes: These are usually made of carbon. One end of each brush touches a slip ring, and the other end connects to the external circuit.

Working of AC Generator:
When the armature coil rotates in the magnetic field, the magnetic flux passing through it changes continuously. This change in magnetic flux induces an electric current in the coil. As the coil rotates, its plane alternately becomes perpendicular and parallel to the magnetic field.
In the first half of the rotation, if the coil rotates clockwise, the current flows in one direction in the external circuit (e.g., from B1 to B2). In the next half of the rotation, the direction of the induced current reverses (e.g., from B2 to B1). This continuous change in the direction of the current after every half cycle means the generator produces alternating current (AC).

In simple words: An AC generator uses a spinning coil between magnets to make electricity. As the coil turns, it creates current that keeps switching direction. Slip rings and brushes collect this changing current.

🎯 Exam Tip: When describing the working, emphasize the change in magnetic flux as the cause of induced current and how slip rings ensure AC output. Make sure your diagram clearly labels all parts.

Question 29: With the help of suitable diagram derive equivalent resistance of resistors in parallel combination.
Answer: Let us consider a circuit with three resistors, \( R_1, R_2, \) and \( R_3 \), connected in parallel, as shown in the diagram below. A battery provides a potential difference \( V \) across this parallel combination.
In a parallel circuit, the potential difference (voltage) across each resistor is the same and equal to the supply voltage \( V \). However, the total current \( I \) from the battery splits among the three resistors. Let the currents flowing through \( R_1, R_2, \) and \( R_3 \) be \( I_1, I_2, \) and \( I_3 \) respectively.
According to Kirchhoff's Current Law, the total current is the sum of the individual currents:
\( I = I_1 + I_2 + I_3 \)
According to Ohm's Law, for each resistor:
\( I_1 = \frac{V}{R_1} \)
\( I_2 = \frac{V}{R_2} \)
\( I_3 = \frac{V}{R_3} \)
Substituting these into the total current equation:
\( I = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} \)
\( I = V \left( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right) \)
If \( R_p \) is the equivalent resistance of the parallel combination, then the total current can also be written as:
\( I = \frac{V}{R_p} \)
Comparing the two expressions for \( I \):
\( \frac{V}{R_p} = V \left( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right) \)
Dividing both sides by \( V \), we get the formula for equivalent resistance in a parallel combination:
\( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \)
This shows that the reciprocal of the equivalent resistance in a parallel combination is equal to the sum of the reciprocals of the individual resistances. This means adding more resistors in parallel actually reduces the overall resistance.

In simple words: When resistors are connected side-by-side, the total resistance becomes smaller. You find the total by adding the inverse of each resistance and then taking the inverse of that sum.

🎯 Exam Tip: Clearly state that voltage is constant and current adds up in parallel circuits. Emphasize taking the reciprocal for each resistance and for the final equivalent resistance.

Question 30: Find the maximum and minimum resistance by combination of three resistors of 10, 20 and 30.
Answer:
To find the maximum resistance, the resistors should be connected in series combination.
For series combination, the equivalent resistance \( R_s \) is the sum of individual resistances:
\( R_s = R_1 + R_2 + R_3 \)
\( R_s = 10 \Omega + 20 \Omega + 30 \Omega \)
\( R_s = 60 \Omega \)
To find the minimum resistance, the resistors should be connected in parallel combination.
For parallel combination, the reciprocal of the equivalent resistance \( R_p \) is the sum of the reciprocals of individual resistances:
\( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \)
\( \frac{1}{R_p} = \frac{1}{10 \Omega} + \frac{1}{20 \Omega} + \frac{1}{30 \Omega} \)
To add these fractions, find a common denominator, which is 60:
\( \frac{1}{R_p} = \frac{6}{60 \Omega} + \frac{3}{60 \Omega} + \frac{2}{60 \Omega} \)
\( \frac{1}{R_p} = \frac{6 + 3 + 2}{60 \Omega} \)
\( \frac{1}{R_p} = \frac{11}{60 \Omega} \)
Now, take the reciprocal to find \( R_p \):
\( R_p = \frac{60}{11} \Omega \)
\( R_p \approx 5.45 \Omega \).
The maximum resistance is \( 60 \Omega \) and the minimum resistance is approximately \( 5.45 \Omega \). Connecting resistors in parallel always gives a total resistance less than the smallest individual resistor.
In simple words: For the most resistance, put them in a line (series) to get 60 Ohm. For the least resistance, put them side-by-side (parallel) to get about 5.45 Ohm.

🎯 Exam Tip: Clearly state whether you are calculating for series or parallel. For parallel combinations, remember to find a common denominator for the reciprocals and then invert the final sum to get \( R_p \).

Question 32: Find the equivalent resistance between A and B in following circuits.
(a)

Question 3: If a wire of resistance R is melted and recast into half of its length, the new resistance of wire will be
(a) R/4
(b) R/2
(c) R
(d) 2R
Answer: (a) R/4
In simple words: When a wire is melted and reshaped to half its original length, its new resistance becomes one-fourth of the first resistance. This is because both length and area change.

🎯 Exam Tip: Remember that melting and recasting a wire changes both its length and cross-sectional area, affecting its resistance significantly. For these problems, volume remains constant.

Question 4: If \( R_1 \) and \( R_2 \) are the resistance of filaments of a 400 W and 200 W lamp designed to operate at same voltage then:
(a) \( R_1 = 2R_2 \)
(b) \( R_2 = 2R_1 \)
(c) \( R_2 = 4R_1 \)
(d) \( R_1 = R_2 \)
Answer: (b) \( R_2 = 2R_1 \)
In simple words: For bulbs operating at the same voltage, the one with less power (like the 200 W bulb) will have more resistance. The 200 W bulb has double the resistance of the 400 W bulb.

🎯 Exam Tip: Use the formula \( P = \frac{V^2}{R} \) to relate power, voltage, and resistance. Since V is constant, P is inversely proportional to R.

Question 5: An electric bulb is rated 220 V -100 W. If it is operated at 110 V then power consumed by it will be
(a) 100 W
(b) 50 W
(c) 25 W
(d) 400 W
Answer: (c) 25 W
In simple words: A bulb designed for 220 V will use much less power if you connect it to only 110 V. The power used drops by a lot when the voltage is cut in half.

🎯 Exam Tip: First calculate the bulb's resistance using its rated power and voltage (\( R = V^2/P \)). Then, use this resistance with the new voltage to find the new power consumption.

Question 7: A current of 2 A passes through a conductor and produces 80 joules of heat in 10 seconds. The resistance of the conductor is
(a) 0.5 Ω
(b) 2 Ω
(c) 4 Ω
(d) 20 Ω
Answer: (b) 2 Ω
In simple words: To find the resistance, we can use Joule's heating law, which connects heat, current, resistance, and time. If a certain amount of heat is made by a current flowing for some time, the resistance helps us figure out the missing part.

🎯 Exam Tip: Use Joule's law of heating, \( H = I^2Rt \), where H is heat, I is current, R is resistance, and t is time. Rearrange the formula to solve for R.

Question 8: A house is fitted with 10 tubes each of 40 W. If all tubes are lighted for 10 hours and if the cost of one unit of electricity energy is Rs. 2.50 the total cost of electricity consumption is
(a) Rs. 100
(b) Rs. 20
(c) Rs. 25
(d) Rs. 10
Answer: (d) Rs. 10
In simple words: We first find the total power used by all the tubes, then multiply it by the hours they are on to get total energy in kilowatt-hours. Then, we multiply this by the cost per unit to find the final bill.

🎯 Exam Tip: Remember that 1 unit of electricity is 1 kilowatt-hour (kWh). Always convert total power to kilowatts before multiplying by hours.

Question 9: Ohm's law is valid only when
(a) Temperature increases
(b) Temperature decreases
(c) Current remains constant
(d) Temperature remains constant.
Answer: (d) Temperature remains constant.
In simple words: Ohm's law says that current and voltage are directly linked, but this rule only works if the temperature of the material does not change. If the temperature changes, the resistance also changes, and the law won't hold true anymore.

🎯 Exam Tip: The condition "temperature remains constant" is crucial for Ohm's law to be strictly applicable. It is often referred to as a "constant physical conditions" clause.

Question 11: Seven identical lamps of resistance 220 Ω each are connected to a 220 V line as shown in figure. Then reading of ammeter will be
(a) \( \frac{1}{10} \) A
(b) \( \frac{2}{5} \) A
(c) \( \frac{3}{10} \) A
(d) None of these
Answer: (d) None of these
In simple words: Since all seven identical lamps are connected in series, their total resistance is the sum of their individual resistances. Then, we use Ohm's law to find the total current in the circuit. If the diagram showed a parallel connection, the current would be much higher.

🎯 Exam Tip: Always identify whether components are in series or parallel before calculating total resistance and current. Series components add up resistance, parallel components add up current in a different way.

Question 12: Number of kilowatt-hours \( = \frac { \text{Volt} \times \text{ampere} \times \text{hours} }{1000} \)

🎯 Exam Tip: To calculate electricity consumption in kilowatt-hours (units), remember to divide the product of voltage, current, and time (in hours) by 1000 to convert from watt-hours to kilowatt-hours.

Question 13: A man has five resistors each of 1/5 Ω. What is the maximum resistance he can obtain by connecting them?
(a) 1 Ω
(b) 5 Ω
(c) \( \frac{1}{2} \) Ω
(d) \( \frac{2}{5} \) Ω
Answer: (a) 1 Ω
In simple words: To get the highest possible resistance from several resistors, you should connect them one after another in a straight line. This is called a series connection, where you simply add up all the individual resistance values.

🎯 Exam Tip: Maximum resistance is achieved by connecting resistors in series. Minimum resistance is achieved by connecting resistors in parallel.

Question 14: 1 kilowatt hour (kWh) is equal to
(a) \( 3.6 \times 10^6 \) MJ
(b) \( 3.6 \times 10^5 \) MJ
(c) \( 3.6 \times 10^2 \) MJ
(d) 3.6 MJ
Answer: (d) 3.6 MJ
In simple words: A kilowatt-hour is a unit of energy, often used for electricity bills. It is equivalent to a very large amount of energy measured in joules, or in this case, megajoules (MJ). This conversion is useful for understanding energy costs.

🎯 Exam Tip: The conversion factor is \( 1 \text{ kWh} = 3.6 \times 10^6 \text{ J} \). Since 1 MJ = \( 10^6 \) J, then 1 kWh = 3.6 MJ.

Question 15: Kilowatt hour is unit of
(a) Energy
(b) Power
(c) Impulse
(d) Force
Answer: (a) Energy
In simple words: The term "kilowatt hour" tells you how much work can be done or how much heat can be produced by electricity over a period of time. It measures the total amount of energy used, not how quickly it's used.

🎯 Exam Tip: Power is the rate at which energy is used or produced (e.g., Watts or kilowatts). Energy is power multiplied by time (e.g., Watt-hours or kilowatt-hours).

Question 16: Conventionally the direction of the current is taken as
(a) from negative charge to positive charge
(b) from positive charge to negative charge
(c) from positive terminal to negative terminal
(d) from negative terminal to positive terminal
Answer: (c) from positive terminal to negative terminal
In simple words: Even though we know electrons (which are negative) move to carry current, people agreed a long time ago to say that current flows from the positive end of a battery to the negative end. This is called conventional current.

🎯 Exam Tip: Distinguish between conventional current (direction of positive charge flow) and electron flow (actual movement of electrons).

Question 17: The resistance of a conductor is reduced to half its initial value. In doing so the heating effects in the conductor will become
(A) half
(b) double
(c) one fourth
(d) four times
Answer: (a) half
In simple words: When the resistance of a wire is cut in half, the amount of heat it produces also becomes half. This is assuming the current remains the same. If resistance decreases, less energy is converted into heat.

🎯 Exam Tip: Recall Joule's law, \( H = I^2Rt \). If I and t are constant, H is directly proportional to R. So, if R is halved, H is also halved.

Question 18: Materials which allow larger current to flow through them are called
(a) Insulators
(b) Conductors
(c) Semiconductors
(d) Alloy
Answer: (b) Conductors
In simple words: Conductors are special materials like metals that let electricity pass through them very easily. They have many free electrons that can move around.

🎯 Exam Tip: Conductors have low resistance, while insulators have very high resistance. Semiconductors have resistance between these two.

Question 19: The resistivity of a wire
(a) varies with its length
(b) varies with its mass
(c) varies with its cross section
(d) is independent of length, cross section and mass of wire
Answer: (d) is independent of length, cross section and mass of wire
In simple words: Resistivity is a basic property of the material itself, not how long or thick the wire is. It only changes if the material itself changes, or if its temperature changes.

🎯 Exam Tip: Resistivity (ρ) is an intrinsic property of a material, depending only on its nature and temperature, not its dimensions. Resistance (R) depends on resistivity, length (L), and cross-sectional area (A) as \( R = \rho \frac{L}{A} \).

Question 20: What sets electron into motion in an electric circuit?
(a) Battery/cell
(b) Resistor
(c) Switch
(d) Conductor
Answer: (a) Battery/cell
In simple words: A battery or cell acts like a pump for electricity. It creates a difference in electric pressure, called potential difference, which pushes the electrons around the circuit, making the current flow.

🎯 Exam Tip: The battery provides the electromotive force (EMF) or potential difference, which is the "push" required to move charges and establish a current in a circuit.

Question 21: An electric geyser has rating 2000 W, 220 V on it. What is the minimum setting of fuse wire that may be required for use with this geyser?
(a) 5 A
(b) 10 A
(c) 15 A
(d) 20 A
Answer: (b) 10 A
In simple words: A fuse protects electrical items by breaking the circuit if too much current flows. To choose the right fuse for a geyser, you need to know how much current it normally uses. The fuse rating should be slightly higher than this normal current.

🎯 Exam Tip: Calculate the maximum current the appliance draws using \( P = VI \) (Current \( I = P/V \)). The fuse rating should be just above this calculated current to allow normal operation while still providing protection.

Question 22: Graphs between electric current and potential difference across two conductors A and B are shown in the figure. Which of the following conductor has more resistance?
(a) B
(b) A
(c) Both have equal resistance
(d) Can't be predicted
Answer: (b) A
In simple words: In a V-I graph, the steeper the line, the lower the resistance. Here, line A is less steep than line B, meaning for the same current, A requires a higher voltage, so A has more resistance. It means the conductor 'A' resists the flow of current more.

🎯 Exam Tip: On a V-I graph (Voltage on x-axis, Current on y-axis), the slope (\( I/V \)) represents the conductance, and its inverse (\( V/I \)) represents the resistance. A smaller slope means higher resistance.

Question 23: Two wires A and B of same metal are connected in parallel. Wire A has length and radius V and wire B has a length '2l' and radius '2r'. Then the ratio of total resistance of parallel combination and the resistance of wire A

🎯 Exam Tip: When analyzing resistance problems, always consider how length, area, and material affect resistance. For parallel combinations, remember the formula for equivalent resistance.

Question 24: Correct formula for Joules law of heating is
(a) \( H = I^2RT \)
(b) \( H = IVT \)
(c) \( H = \frac{V^2}{R} \)
(d) All of these
Answer: (d) All of these
In simple words: Joule's law explains how electrical energy is changed into heat. There are several ways to write this law using different electrical measurements, all of which are correct ways to describe the heat produced. For example, \( H = IVT \) connects heat with current, voltage, and time.

🎯 Exam Tip: Be familiar with all three common forms of Joule's law of heating: \( H = I^2Rt \), \( H = VIt \), and \( H = \frac{V^2}{R}t \). All three are derived from Ohm's law and the definition of electrical power.

Question 25: In household electric circuit different appliances are connected in parallel to each other because
(a) The appliances work at same voltage
(b) The appliances can be operated independent of each other
(c) Even if a component of a electric circuit fails other can work efficiently
(d) All of the above
Answer: (d) All of the above
In simple words: Connecting household devices in parallel ensures that each device gets the full power from the main supply and can be turned on or off without affecting other devices. This setup makes sure that if one device breaks, others keep working.

🎯 Exam Tip: Understanding the advantages of parallel connection (constant voltage, independent operation, system reliability) is key to explaining household wiring choices.

Question 26: The condition necessary for electromagnetic induction is that
(a) there must be a relative motion between the coil of wire and galvanometer.
(b) there must be a relative motion between the coil of wire and a magnet.
(c) there must be a relative motion between the galvanometer and a magnet
(d) all of the above.
Answer: (b) there must be a relative motion between the coil of wire and a magnet.
In simple words: For electricity to be created by magnetism, either the wire coil or the magnet must be moving relative to each other. This movement changes the magnetic field passing through the coil, which then makes current flow.

🎯 Exam Tip: Electromagnetic induction requires a change in magnetic flux through a coil. This change is most commonly achieved by relative motion between a coil and a magnet or by changing the current in a nearby coil.

Electricity Current Very Short Answer Type Questions

Question 1: What is the S.I. unit of charge?
Answer: The S.I. unit of charge is Coulomb (C). It is a fundamental unit in electricity.
In simple words: The standard way to measure electric charge is in Coulombs.

🎯 Exam Tip: Always specify the full name of the unit (Coulomb) along with its symbol (C) for clarity.

Question 3: What is the S.I. unit of electric potential?
Answer: The S.I. unit of electric potential is Volt (V). It represents the energy per unit charge.
In simple words: Electric potential is measured in Volts.

🎯 Exam Tip: Electric potential is often called voltage and is measured in Volts. Remember that 1 Volt is equal to 1 Joule per Coulomb.

Question 4: Give the unit of electric resistance.
Answer: The S.I. unit of electric resistance is Ohm (Ω). It indicates how much a material opposes current flow.
In simple words: Electric resistance is measured using the unit called Ohm.

🎯 Exam Tip: Use the Greek letter omega (Ω) as the symbol for Ohm. The concept of resistance is central to Ohm's law.

Question 5: How is resistance, volt and current (I) related?
Answer: Resistance (\( R \)), potential difference (Volt, \( V \)), and current (\( I \)) are related by Ohm's Law: \( R = \frac{V}{I} \). This means resistance is the ratio of voltage to current.
In simple words: Resistance tells you how much voltage is needed to push a certain amount of current. We can find it by dividing the voltage by the current.

🎯 Exam Tip: Ohm's law, \( V = IR \), is fundamental. Be ready to rearrange it to find any of the three quantities if the other two are known.

Question 6: How is 1 ohm related to ampere and volt?
Answer: 1 Ohm is defined as the resistance through which 1 Ampere of current flows when a potential difference of 1 Volt is applied across it. So, \( 1 \text{ ohm} = \frac { 1 \text{ Volt} }{ 1 \text{ ampere} } \).
In simple words: If you have 1 Volt of push and get 1 Ampere of flow, then the resistance is 1 Ohm.

🎯 Exam Tip: This is the direct definition of the unit Ohm based on Ohm's law. Make sure to state it clearly, linking all three units.

Question 7: What constitutes the current?
Answer: The flow of free electrons constitutes the electric current in a conductor. These tiny particles move from one point to another, carrying electric charge.
In simple words: Current is made of tiny free electrons moving.

🎯 Exam Tip: Although conventional current is defined by the flow of positive charge, the actual charge carriers in metallic conductors are negatively charged electrons.

Question 8: When the given resistances are connected in series, which physical quantity does not change.
Answer: When resistances are connected in series, the current (\( I \)) flowing through each resistor remains the same. The current has only one path to follow.
In simple words: In a series circuit, the electricity (current) flowing through each part is the same.

🎯 Exam Tip: In a series circuit, current is constant, but potential difference (voltage) adds up. In a parallel circuit, voltage is constant, but current adds up.

Question 10: What is a voltmeter?
Answer: A voltmeter is a measuring instrument used to measure the potential difference (voltage) between two points in an electric circuit. It is always connected in parallel.
In simple words: A voltmeter is a tool that checks the "push" of electricity between two points in a circuit.

🎯 Exam Tip: Voltmeters have very high resistance so they draw minimal current when connected in parallel, accurately measuring the voltage across a component without altering the circuit significantly.

Question 11 Give the unit of power.
Answer: The S.I. unit of power is Watt (W). Power measures how fast energy is used or produced.
In simple words: Power is measured in Watts.

🎯 Exam Tip: Remember that 1 Watt is equal to 1 Joule per second (\( 1 \text{ W} = 1 \text{ J/s} \)). Power is the rate of doing work or transferring energy.

Question 12: How is power related to current and voltage?
Answer: Electric power (\( P \)) is related to current (\( I \)) and voltage (\( V \)) by the formula \( P = V \times I \). This means power is the product of potential difference and current.
In simple words: Power is found by multiplying the voltage (push) by the current (flow of electricity).

🎯 Exam Tip: This is the fundamental formula for electrical power. You can derive other forms using Ohm's law, such as \( P = I^2R \) or \( P = \frac{V^2}{R} \).

Question 13: How is ammeter and voltmeter connected in a circuit?
Answer: An ammeter is connected in series with the component to measure the current flowing through it. A voltmeter is connected in parallel across the component to measure the potential difference.
In simple words: An ammeter goes in line with the circuit (series) to measure current, and a voltmeter goes across the circuit (parallel) to measure voltage.

🎯 Exam Tip: Ammeters have very low resistance, so they don't significantly impede current when in series. Voltmeters have very high resistance, so they don't divert much current when in parallel.

Question 14: What happens to the resistance when length of conductor is doubled without affecting the thickness of conductor?
Answer: If the length of a conductor is doubled while its thickness (cross-sectional area) remains unchanged, its resistance will also double. This is because resistance is directly proportional to length.
In simple words: Making a wire twice as long also makes its resistance twice as big.

🎯 Exam Tip: Remember the formula for resistance \( R = \rho \frac{L}{A} \). If \( L \) doubles and \( A \) stays constant, \( R \) doubles.

Question 15: What is the S.I. unit of resistivity?
Answer: The S.I. unit of resistivity is Ohm-meter (Ωm). Resistivity is a measure of a material's inherent ability to resist electric current.
In simple words: Resistivity is measured in Ohm-meters

🎯 Exam Tip: Do not confuse resistance (Ohm, Ω) with resistivity (Ohm-meter, Ωm). Resistivity is a material property, while resistance depends on the material's dimensions.

Question 17: The combined resistance in the given circuit is 5 Ω. What is the value of R?
Answer: The two resistors, 10 Ω and R, are connected in parallel. The formula for equivalent resistance (\( R_{eq} \)) in parallel is: \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \) We are given \( R_{eq} = 5 \Omega \) and \( R_1 = 10 \Omega \). \( \frac{1}{5} = \frac{1}{10} + \frac{1}{R} \) Now, we solve for R: \( \frac{1}{R} = \frac{1}{5} - \frac{1}{10} \) \( \frac{1}{R} = \frac{2}{10} - \frac{1}{10} \) \( \frac{1}{R} = \frac{1}{10} \)
\( \implies R = 10 \Omega \) So, the value of R is \( 10 \Omega \).
In simple words: When resistors are in parallel, their combined resistance is less than the smallest individual resistance. We use a special formula for parallel resistors to work backwards and find the unknown resistance 'R'.

🎯 Exam Tip: For parallel combinations, remember that the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances. Ensure to calculate the common denominator correctly when adding fractions.

Question 18: A wire of resistance 4 Ω is bent in the form of a closed circle. What is the resistance between the two points at the ends of any diameter of the circle?
Answer: When a wire of total resistance \( 4 \Omega \) is bent into a closed circle, and we want to find the resistance between two points at the ends of a diameter, it means the circuit is split into two equal semicircular parts. Each semicircular part will have half the total resistance, so each will be \( \frac{4 \Omega}{2} = 2 \Omega \). These two \( 2 \Omega \) parts are then connected in parallel between the two points. The equivalent resistance (\( R_{eq} \)) for two resistors in parallel is: \( \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \) Here, \( R_1 = 2 \Omega \) and \( R_2 = 2 \Omega \). \( \frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{2} \) \( \frac{1}{R_{eq}} = \frac{2}{2} \) \( \frac{1}{R_{eq}} = 1 \)
\( \implies R_{eq} = 1 \Omega \) So, the resistance between the two points at the ends of any diameter is \( 1 \Omega \).
In simple words: When a wire forms a full circle, picking two points on opposite sides of the circle makes two paths for electricity, each with half the wire's total resistance. These two paths work side-by-side, which is a parallel connection.

🎯 Exam Tip: Visualize the paths for current flow. When a circular wire is cut by a diameter, the two semi-circular paths are effectively in parallel, each having half the original wire's resistance.

Question 19: Define magnetic field.
Answer: A magnetic field is the region or space around a magnet or a current-carrying conductor where its magnetic influence can be felt. Within this area, a magnetic force can attract or repel other magnetic substances or moving charges. We can think of it as an invisible area where magnetic forces are present.
In simple words: A magnetic field is the area around a magnet where its push or pull force can be felt.

🎯 Exam Tip: When defining magnetic field, emphasize that it is a region where magnetic force is experienced. Mentioning both magnets and current-carrying conductors as sources is good.

Question 20: On which factors do the magnetic field due to current carrying conductor depend?
Answer: The magnitude of the magnetic field produced at a given point due to a current-carrying conductor depends on the following factors:
1. It increases as the current through the wire increases. More current means a stronger magnetic field.
2. It decreases as the distance from the wire increases. The further you are from the wire, the weaker the magnetic field.
In simple words: The strength of a magnetic field around a wire depends on how much electricity flows through the wire and how far away you are from it.

🎯 Exam Tip: Remember these two key direct and inverse relationships. A simple analogy is heat from a fire: hotter fire (more current), less heat as you move away (more distance).

Question 21: Name and state the rule used to determine the direction of field in a current carrying straight conductor.
Answer: The rule used to determine the direction of the magnetic field around a current-carrying straight conductor is the Right-Hand Thumb Rule. According to this rule, if you hold the current-carrying straight conductor in your right hand with your thumb pointing in the direction of the current, then the direction in which your fingers curl around the conductor gives the direction of the magnetic field lines. These field lines form concentric circles around the wire.
In simple words: We use the Right-Hand Thumb Rule. Point your right thumb where electricity flows, and your fingers will show the direction of the magnetic field.

🎯 Exam Tip: Clearly state the rule and explain how to apply it (thumb for current, fingers for field). A simple sketch of a wire with circular field lines and an arrow for current and field can score extra points.

Question 22: Draw magnetic field lines due to current through circular loop.
Answer: When current flows through a circular loop, the magnetic field lines produced are concentric circles near the wire at each point, just like for a straight wire. However, as you move towards the center of the loop, the field lines become straighter and stronger, appearing almost parallel to each other at the very center, creating a strong and uniform magnetic field.
In simple words: Around a circle with electricity, the magnetic field lines look like small circles near the wire, and then become straighter lines as they get closer to the center of the big circle.

🎯 Exam Tip: For circular loops, magnetic field lines are concentric circles near the wire edges and nearly straight lines through the center, pointing perpendicular to the plane of the loop.

Question 23: What is an electromagnet?
Answer: An electromagnet is a temporary magnet created when electric current flows through a coil of wire, usually wrapped around a soft iron core. The magnetic field only exists as long as the current is flowing. This allows its magnetic strength to be controlled by changing the current.
In simple words: An electromagnet is a magnet that only works when electricity is running through it.

🎯 Exam Tip: Highlight that an electromagnet is *temporary* and its strength can be *controlled*. This is its key distinction from permanent magnets.

Question 24: Name some devices that use current carrying conductors and magnetic field.
Answer: Many devices utilize the interaction between current-carrying conductors and magnetic fields. Some common examples include electric motors (which convert electrical energy to mechanical energy), electric generators (which convert mechanical energy to electrical energy), loud speakers (which use magnetic fields to convert electrical signals into sound), and various measuring instruments (like galvanometers and ammeters). These devices depend on the force experienced by a current in a magnetic field.
In simple words: Electric motors, electric generators, loudspeakers, and measuring tools all use electricity flowing through wires along with magnetic fields to work.

🎯 Exam Tip: Think about applications where electricity causes motion (motors, speakers) or motion causes electricity (generators) – these are often based on electromagnetic principles.

Question 25: Define electromagnetic induction.
Answer: Electromagnetic induction is the phenomenon where an electric current is produced (or induced) in a conductor or a coil when it is exposed to a changing magnetic field. This change can happen either by moving the conductor through a magnetic field, moving the magnet near the conductor, or by changing the strength of the magnetic field itself. It is the core principle behind electric generators.
In simple words: Electromagnetic induction is when electricity is made in a wire by a changing magnetic field near it.

🎯 Exam Tip: The key phrase is "changing magnetic field" or "change in magnetic flux." Simply having a conductor in a magnetic field is not enough; there must be relative motion or a change in field strength.

Question 26: What is electric generator?
Answer: An electric generator is a device that converts mechanical energy into electrical energy. It works on the principle of electromagnetic induction, where the rotation of a coil in a magnetic field generates an electric current. This mechanical energy often comes from sources like steam, water, or wind turbines.
In simple words: An electric generator is a machine that turns movement energy into electrical energy.

🎯 Exam Tip: Clearly state the energy conversion (mechanical to electrical) and the underlying principle (electromagnetic induction) for a complete definition.

Question 27: What is the difference between AC generator and DC generator?
Answer: The main difference between an AC generator and a DC generator lies in the type of current they produce and their commutator (slip ring) arrangement.
An AC generator produces alternating current, meaning the direction of the current changes periodically after equal intervals of time. It uses slip rings.
A DC generator produces direct current, meaning the current flows in only one direction. It uses a split-ring commutator (also called a rectifier) to ensure unidirectional current.
In simple words: AC generators make electricity that keeps changing direction, while DC generators make electricity that flows in only one steady direction.

🎯 Exam Tip: Focus on the type of current produced (AC vs. DC) and the mechanism that achieves this (slip rings vs. split-ring commutator).

Electricity Current Short Answer Type Questions

Question 1: Give two points of difference between ammeter and voltmeter and show their arrangement in an electric circuit.
Answer:

In simple words: An ammeter measures how much electricity is flowing and is put right into the path of the flow. A voltmeter measures the electric "pressure" across a part and is connected around that part

🎯 Exam Tip: Remember the connection rules: Ammeter in series (low resistance), Voltmeter in parallel (high resistance). This is a common practical application question.

Question 2: Give two points of difference between resistance and resistivity.
Answer:

In simple words: Resistance is how much a specific wire or component blocks electricity, while resistivity is a basic number that tells how much the material itself blocks electricity, no matter its shape or size.

🎯 Exam Tip: The key difference is that resistance is for a specific object, while resistivity is for the material itself. Remember the formula \( R = \rho \frac{L}{A} \) to understand their relationship.

Question 3: Give the differences between series and parallel connections of resistors.
Answer:

In simple words: In series, electricity flows through each part one by one, making the total resistance higher. In parallel, electricity splits and flows through many paths at once, which makes the total resistance lower.

🎯 Exam Tip: Clearly remember the key characteristics of series (current same, voltage divides, resistances add) and parallel (voltage same, current divides, reciprocals of resistances add). Drawing simple diagrams helps illustrate these differences.

Question 4: Why are decorative lights always connected in parallel?

🎯 Exam Tip: For decorative lights, parallel connection is preferred because if one bulb fuses, the others can continue to glow. Also, each bulb receives the full voltage, ensuring uniform brightness.

Question 5. What is purely resistive electric circuit?
Answer: A purely resistive electric circuit is a circuit where only resistors are connected to a power source. All the energy from the source is turned into heat by these resistors.
In simple words: In this circuit, all the electricity becomes heat because of the resistors.

🎯 Exam Tip: Remember that "purely resistive" means there are no other components like capacitors or inductors to store energy, so all electrical energy is converted into heat.

Question 6. Give the difference in unit of electric energy and commercial unit of electric energy.
Answer: The standard unit for electric energy is the Joule, which can also be expressed as a watt-hour. However, for commercial use, like for electricity bills, a larger unit called the kilowatt-hour is used. This helps in measuring bigger amounts of energy.
In simple words: Electric energy is measured in Joules or watt-hours. For household bills, we use kilowatt-hours because it's a bigger unit.

🎯 Exam Tip: Distinguish between the fundamental unit (Joule) and practical units for everyday use (watt-hour, kilowatt-hour). Knowing the conversion (1 kWh = 3.6 x 10⁶ J) is also crucial.

Question 7. Define one watt hour and one kilowatt hour.
Answer: One watt-hour is a measure of energy. It means the amount of energy used when a device consumes 1 watt of power for 1 hour. Similarly, one kilowatt-hour (kWh) is the energy consumed when a device uses 1000 watts of power for 1 hour.
In simple words: A watt-hour is using 1 watt for one hour. A kilowatt-hour is using 1000 watts for one hour.

🎯 Exam Tip: Focus on understanding that both are units of energy, with kilowatt-hour being 1000 times larger and commonly used for billing.

Question 8. Give two disadvantages of the heating effect of current.
Answer: When electric current flows through a wire, it creates heat. Two disadvantages of this heating effect are:
1. It can damage circuits and other materials by causing high temperatures or melting them.
2. This heat often cannot be used and might require extra cooling systems, like fans in computers, which adds to the device's complexity and energy use.
In simple words: The heat from electricity can break circuits and melt things. Sometimes, this heat is wasted and needs fans to cool things down.

🎯 Exam Tip: Think about common household appliances or devices where unwanted heat is a problem, such as computers overheating or wires melting due to short circuits.

Question 9. Give two advantages of heating effect of electric current.
Answer: The heating effect of electric current has many useful applications. Two advantages include:
1. It is used in devices like electric room heaters, where electrical energy is intentionally converted into heat to warm spaces.
2. This principle also works in electric kettles and water heaters, quickly changing electricity into heat for boiling water or heating water.
In simple words: The heating effect of electricity is good for things like room heaters and water heaters because it makes heat.

🎯 Exam Tip: When listing advantages, try to give specific examples of devices that utilize the heating effect for a clear answer

Question 11. What do the following symbols represent in the electric circuit?
(a)
(b)
(c)
(d)
(e)
(f)
Answer: The symbols in electric circuits help us draw them easily. Here are what these common symbols represent:
(a) An electric cell, which is a single power source.
(b) A voltmeter, used to measure potential difference across two points.
(c) A plug key (or switch), used to switch the circuit on or off.
(d) A resistor, which limits the flow of current.
(e) A variable resistance or rheostat, which allows us to change the resistance in the circuit.
(f) A bulb, which glows when current passes through it.
In simple words: These are circuit symbols. They show things like batteries, meters for voltage, switches, parts that slow electricity down, parts that change resistance, and lights.

🎯 Exam Tip: Practice drawing and identifying these common circuit symbols, as they are fundamental for understanding and representing electrical circuits.

Question 12. What is potential difference? Explain and give its unit with definition.
Answer: Potential difference, also known as voltage, is the work done to move a unit of electric charge from one point to another in an electric circuit. We can write it as \( V = \frac{W}{Q} \), where \( W \) is the work done and \( Q \) is the charge. The standard (S.I.) unit for potential difference is the Volt (\( V \)). One volt is defined as the potential difference between two points if 1 Joule of work is done to move 1 Coulomb of charge between them. This means \( 1 \, V = \frac{1 \, J}{1 \, C} \).
In simple words: Potential difference is how much energy it takes to move electricity from one spot to another. Its unit is the Volt. If you use 1 Joule of energy to move 1 Coulomb of electricity, that's 1 Volt.

🎯 Exam Tip: Understand that potential difference is about the "energy per charge" and is often compared to water pressure driving water through a pipe.

Question 13. A circuit diagram is given below. Calculate (a) Current through each resistor. (b) The total current in the circuit. (c) The total effective resistance of the circuit.
Answer: We are given a circuit with a 10V battery and three resistors (2Ω, 5Ω, 10Ω) connected in parallel.
(a) To find the current through each resistor, we use Ohm's Law (\( I = \frac{V}{R} \)). Since they are in parallel, the voltage \( V \) across each resistor is the same as the battery voltage, 10V.
\( I_1 = \frac{V}{R_1} = \frac{10\,V}{2\,\Omega} = 5\,A \)
\( I_2 = \frac{V}{R_2} = \frac{10\,V}{5\,\Omega} = 2\,A \)
\( I_3 = \frac{V}{R_3} = \frac{10\,V}{10\,\Omega} = 1\,A \)
(b) The total current in a parallel circuit is the sum of the currents through each branch.
\( I_{total} = I_1 + I_2 + I_3 = 5\,A + 2\,A + 1\,A = 8\,A \)
(c) For resistors connected in parallel, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances.
\( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{2\,\Omega} + \frac{1}{5\,\Omega} + \frac{1}{10\,\Omega} \)
To add these fractions, we find a common denominator, which is 10.
\( \frac{1}{R_p} = \frac{5}{10\,\Omega} + \frac{2}{10\,\Omega} + \frac{1}{10\,\Omega} = \frac{5+2+1}{10\,\Omega} = \frac{8}{10\,\Omega} = \frac{4}{5\,\Omega} \)
\( R_p = \frac{5}{4}\,\Omega = 1.25\,\Omega \)
In simple words: In this parallel circuit, we calculate current for each resistor by dividing the voltage by its resistance. Then, we add these currents to get the total current. To find the total resistance, we add the "upside-down" values of each resistance and then flip the final answer.

🎯 Exam Tip: Remembr that in parallel circuits, voltage across each component is the same, but current divides. The total resistance is always less than the smallest individual resistance.

Question 14. You have three resistors, each with a resistance of 3 ohms. Determine all the possible equivalent resistance values you can achieve by combining them.
Answer: To find all possible equivalent resistance values using three 3Ω resistors, we can combine them in different ways:
(i) **All three resistors in series:**
When resistors are connected in series, their total resistance is the sum of individual resistances.
\( R_{series} = R_1 + R_2 + R_3 = 3\,\Omega + 3\,\Omega + 3\,\Omega = 9\,\Omega \)
(ii) **All three resistors in parallel:**
When resistors are connected in parallel, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances.
\( \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{3\,\Omega} + \frac{1}{3\,\Omega} + \frac{1}{3\,\Omega} = \frac{3}{3\,\Omega} = \frac{1}{1\,\Omega} \)
\( R_{parallel} = 1\,\Omega \)
(iii) **Two resistors in parallel, and this combination in series with the third resistor:**
First, calculate the equivalent resistance of two 3Ω resistors in parallel.
\( \frac{1}{R_{p1}} = \frac{1}{3\,\Omega} + \frac{1}{3\,\Omega} = \frac{2}{3\,\Omega} \)
\( R_{p1} = \frac{3}{2}\,\Omega = 1.5\,\Omega \)
Next, this \( 1.5\,\Omega \) equivalent resistance is connected in series with the third 3Ω resistor.
\( R_{total} = R_{p1} + R_3 = 1.5\,\Omega + 3\,\Omega = 4.5\,\Omega \)
(iv) **Two resistors in series, and this combination in parallel with the third resistor:**
First, calculate the equivalent resistance of two 3Ω resistors in series.
\( R_{s1} = R_1 + R_2 = 3\,\Omega + 3\,\Omega = 6\,\Omega \)
Next, this \( 6\,\Omega \) equivalent resistance is connected in parallel with the third 3Ω resistor.
\( \frac{1}{R_{total}} = \frac{1}{R_{s1}} + \frac{1}{R_3} = \frac{1}{6\,\Omega} + \frac{1}{3\,\Omega} = \frac{1}{6\,\Omega} + \frac{2}{6\,\Omega} = \frac{3}{6\,\Omega} = \frac{1}{2\,\Omega} \)
\( R_{total} = 2\,\Omega \)
In simple words: We can combine three 3-ohm resistors in four main ways. Putting them all in a row (series) gives 9 ohms. Stacking them side-by-side (parallel) gives 1 ohm. If two are stacked and one is in a row with them, it gives 4.5 ohms. If two are in a row and one is stacked with them, it gives 2 ohms. These are all the possible resistance values.

🎯 Exam Tip: When solving combination problems, always break down the circuit into smaller series or parallel groups and calculate their equivalent resistance step-by-step.

Question 15. A battery of 10 V is connected in a circuit with three resistors of 3 Ω, 4 Ω, and 6 Ω connected in series. How much current flows through each resistor?
Answer: To find the current flowing through each resistor in a series circuit, we first need to calculate the total equivalent resistance of the circuit.
When resistors are connected in series, the total resistance (\( R_s \)) is the sum of their individual resistances:
\( R_s = R_1 + R_2 + R_3 = 3\,\Omega + 4\,\Omega + 6\,\Omega = 13\,\Omega \)
Now, using Ohm's Law (\( I = \frac{V}{R} \)), we can find the total current flowing from the 10 V battery:
\( I = \frac{V_{total}}{R_s} = \frac{10\,V}{13\,\Omega} \approx 0.769\,A \)
In a series circuit, the current is the same through every component. Therefore, the current flowing through each resistor (3 Ω, 4 Ω, and 6 Ω) is approximately \( 0.769\,A \).
In simple words: First, we add all the resistor values together because they are in a series. Then, we use Ohm's Law (Voltage divided by Resistance) to find the total current. Since it's a series circuit, this same current flows through every single resistor.

🎯 Exam Tip: Always remember the key characteristic of series circuits: current is uniform throughout, while voltage divides across components.

Question 16. Two electric bulbs are rated 60 W, 220 V and 20 W, 220 V, are connected in parallel to a 220 V supply. Calculate the total electric current in the circuit.
Answer: We have two electric bulbs connected in parallel to a 220 V power supply. The first bulb is rated 60 W at 220 V, and the second bulb is rated 20 W at 220 V. We need to find the total current flowing in the circuit.
For **Bulb 1**:
Power \( P_1 = 60\,W \)
Voltage \( V = 220\,V \)
Using the formula \( P = V \times I \), we can find the current \( I_1 \):
\( I_1 = \frac{P_1}{V} = \frac{60\,W}{220\,V} \approx 0.2727\,A \)
For **Bulb 2**:
Power \( P_2 = 20\,W \)
Voltage \( V = 220\,V \)
Using the formula \( P = V \times I \), we can find the current \( I_2 \):
\( I_2 = \frac{P_2}{V} = \frac{20\,W}{220\,V} \approx 0.0909\,A \)
Since the bulbs are connected in parallel, the total current (\( I_{total} \)) is the sum of the currents through each bulb:
\( I_{total} = I_1 + I_2 = 0.2727\,A + 0.0909\,A = 0.3636\,A \)
The total electric current in the circuit is approximately 0.36 A.
In simple words: First, we find the current for each light bulb by dividing its power by the voltage. Since the bulbs are connected side-by-side (parallel), the total current is simply the sum of the currents from each bulb.

🎯 Exam Tip: In parallel circuits, remember that voltage is constant across components, so you can calculate individual currents and then sum them for the total circuit current.

Question 18. What is the heating effect of electric current? Find the expression for calculating 'Heat'.
Answer: The heating effect of electric current occurs when current flows through a conductor, causing its temperature to rise and heat to be released. This phenomenon is also known as Joule heating.
To find the expression for the heat produced (\( H \)), consider a current \( I \) flowing through a resistor with resistance \( R \). Let \( V \) be the potential difference across the resistor and \( t \) be the time for which a charge \( Q \) flows through it.
1. **Work done by charge:** The work done (\( W \)) to move a charge \( Q \) across a potential difference \( V \) is given by:
\( W = V \times Q \)
2. **Relation between work and heat:** According to the law of conservation of energy, the electrical work done is converted into heat energy. So, \( H = W \).
\( H = V \times Q \)
3. **Current and charge:** Electric current (\( I \)) is the rate of flow of charge, so \( I = \frac{Q}{t} \), which means \( Q = I \times t \).
4. **Substituting Q:** Substitute \( Q = I \times t \) into the equation for \( H \):
\( H = V \times (I \times t) = VIt \)
5. **Using Ohm's Law:** According to Ohm's Law, \( V = IR \). Substitute this into the equation for \( H \):
\( H = (IR) \times It = I^2 Rt \)
Therefore, the expression for the heat produced is \( H = I^2 Rt \). This formula shows that heat generated is directly proportional to the square of current, resistance, and time.
In simple words: When electricity moves through a wire, it makes the wire hot. This is called the heating effect. To calculate this heat, we use a formula: Heat equals the square of the current, multiplied by the resistance, multiplied by the time.

🎯 Exam Tip: The \( H = I^2 Rt \) formula is crucial. Remember that the heat generated depends significantly on the current (squared), so even small increases in current can lead to much larger heating effects.

Question 19. Two wires of equal lengths, one of copper and the other of manganin (an alloy) have the same thickness. Which one can be used for: (i) electrical transmission lines, (ii) electrical heating devices? Why?
Answer: We have two wires, one made of copper and one of manganin, both of the same length and thickness. We need to decide which material is better for electrical transmission lines and which for heating devices, and why.
(i) **For electrical transmission lines:**
**Copper** would be used for electrical transmission lines. Transmission lines need to carry current with minimum energy loss, which means they should have very low electrical resistance. Materials with low resistivity are good conductors. From general knowledge (and the table provided on the next page which shows copper's low resistivity), copper is an excellent conductor. Silver has slightly lower resistivity, but copper is more practical due to cost.
(ii) **For electrical heating devices:**
**Manganin** would be used for electrical heating devices. Heating devices work by converting electrical energy into heat. This requires a material with high electrical resistance, which means high resistivity. Manganin is an alloy designed to have high resistivity and can withstand high temperatures without melting easily, making it suitable for generating heat.
In simple words: For carrying electricity over long distances, we need materials that let electricity flow very easily, like copper, to avoid losing energy. For things that make heat, like heaters, we need materials that resist electricity a lot, like manganin, so they get hot when electricity flows through them.

🎯 Exam Tip: Remember that low resistivity means a good conductor (for transmission), while high resistivity means a good heater (for heating elements).

Question 21. Study the following circuit and answer the questions that follows: (a) State the type of combination of the two resistors in the circuit. (b) How much current is flowing through (i) 10 Ω and (ii) 15 Ω resistor? (c) What is the ammeter reading?
Answer: We are given an electric circuit with a 3 V battery and two resistors, 10 Ω and 15 Ω, connected in parallel. An ammeter is also in the circuit.
(a) The two resistors are connected in a **parallel combination**. This means they are connected across the same two points, so the voltage across each of them is the same as the battery voltage.
(b) To find the current flowing through each resistor, we use Ohm's Law (\( I = \frac{V}{R} \)). Since the resistors are in parallel, the voltage \( V \) across each resistor is 3 V.
(i) Current through the **10 Ω resistor**:
\( I_{10\Omega} = \frac{V}{R_{10\Omega}} = \frac{3\,V}{10\,\Omega} = 0.3\,A \)
(ii) Current through the **15 Ω resistor**:
\( I_{15\Omega} = \frac{V}{R_{15\Omega}} = \frac{3\,V}{15\,\Omega} = 0.2\,A \)
(c) The ammeter measures the total current flowing out of the battery. In a parallel circuit, the total current is the sum of the currents in each branch.
Ammeter reading = Total current = \( I_{10\Omega} + I_{15\Omega} = 0.3\,A + 0.2\,A = 0.5\,A \)
In simple words: (a) The resistors are connected side-by-side (parallel). (b) For the 10-ohm resistor, 0.3 amps of current flow. For the 15-ohm resistor, 0.2 amps flow. We get these by dividing the battery voltage by each resistor's value. (c) The ammeter shows the total current, which is 0.5 amps (0.3 + 0.2).

🎯 Exam Tip: When analyzing parallel circuits, always remember that voltage is the same across all branches, making current calculation for each branch straightforward using Ohm's Law.

Question 23. Calculate the area of cross section of a wire of length 1.0 m, its resistance is 23 Ω and the resistivity of the material of the wire is 1.84 x 10^-6 Ωm.
Answer: We need to calculate the cross-sectional area of a wire given its length, resistance, and the resistivity of its material.
Given:
Length of wire \( l = 1.0\,m \)
Resistance of wire \( R = 23\,\Omega \)
Resistivity of the material \( \rho = 1.84 \times 10^{-6}\,\Omega m \)
The formula relating resistance, resistivity, length, and cross-sectional area is:
\( R = \frac{\rho l}{A} \)
To find the area \( A \), we can rearrange the formula:
\( A = \frac{\rho l}{R} \)
Now, substitute the given values:
\( A = \frac{(1.84 \times 10^{-6}\,\Omega m) \times (1.0\,m)}{23\,\Omega} \)
\( A = \frac{1.84 \times 10^{-6}}{23}\,m^2 \)
\( A = 0.08 \times 10^{-6}\,m^2 = 8.0 \times 10^{-8}\,m^2 \)
The area of the cross-section of the wire is \( 8.0 \times 10^{-8}\,m^2 \).
In simple words: We know how much a wire resists electricity, how long it is, and how easily its material conducts electricity. To find how thick the wire is (its cross-sectional area), we use a formula: Area equals resistivity times length, divided by resistance.

🎯 Exam Tip: Ensure all units are consistent (S.I. units) before plugging values into the formula to avoid errors in calculation.

Question 24. Draw a schematic diagram of an electric circuit consisting of a battery of two cells each of 1.5 V, 5 Ω, 10 Ω and 15 Ω resistor and a plug key, all connected in series.
Answer: We need to draw a circuit diagram. It should show a battery made of two 1.5 V cells connected in series (so a total of 3 V), along with three resistors (5 Ω, 10 Ω, and 15 Ω) and a plug key, all connected in a single series circuit.

🎯 Exam Tip: Pay close attention to standard symbols for each component (battery, resistor, key) and ensure the connections reflect a series circuit accurately.

Question 25. A lamp rated 100 W at 220 V is connected to the mains electric supply. What current is drawn from the supply line if the voltage is 220 V?
Answer: An electric lamp is rated at 100 W and 220 V. It is connected to a 220 V main power supply. We need to calculate the current it draws from the supply line.
Given:
Power of the lamp \( P = 100\,W \)
Voltage of the supply \( V = 220\,V \)
We can use the formula relating power, voltage, and current:
\( P = V \times I \)
To find the current \( I \), rearrange the formula:
\( I = \frac{P}{V} \)
Substitute the given values:
\( I = \frac{100\,W}{220\,V} \)
\( I = \frac{10}{22}\,A = \frac{5}{11}\,A \)
\( I \approx 0.45\,A \)
The current drawn from the supply line is approximately 0.45 A.
In simple words: To find how much electricity (current) the lamp uses, we divide its power (watts) by the voltage of the electricity supply.

🎯 Exam Tip: Always use the correct power formula (\( P = VI \), \( P = I^2R \), or \( P = \frac{V^2}{R} \)) based on the information provided in the problem. Here, \( P \) and \( V \) are given, making \( P = VI \) the most direct.

Question 26. Ten bulbs are connected in a series circuit to a power supply line and ten identical bulbs are connected in a parallel circuit to an identical power supply line. (a) Which circuit would have the highest voltage across each bulb? (b) In which circuit would the bulb be brighter? (c) If one bulb blows out, in which circuit will other bulb stop glowing?
Answer: We are comparing two circuits, both with ten identical bulbs and connected to the same power supply. One circuit has bulbs in series, and the other has bulbs in parallel.
(a) **Which circuit would have the highest voltage across each bulb?**
The **parallel circuit** would have the highest voltage across each bulb. In a parallel circuit, each component receives the full voltage of the power supply. In a series circuit, the total voltage is divided among all the bulbs.
(b) **In which circuit would the bulb be brighter?**
The bulbs in the **parallel circuit** would be brighter. Since each bulb in parallel receives the full supply voltage, more current flows through it (compared to a series bulb), making it glow more brightly.
(c) **If one bulb blows out, in which circuit will other bulb stop glowing?**
If one bulb blows out, all other bulbs in the **series circuit** will stop glowing. This is because a series circuit is a single path for current, and if that path is broken (by a blown bulb), no current can flow through any other component. In a parallel circuit, other paths remain, so other bulbs would continue to glow.
In simple words: In a parallel circuit, each bulb gets the full power and stays bright, and if one breaks, the others still work. In a series circuit, power is shared, so bulbs are dimmer, and if one breaks, all go out.

🎯 Exam Tip: Clearly state the characteristics of series and parallel circuits (voltage, current, brightness, and effect of a break) to answer comparison questions effectively.

Question 27. In what ways can the magnitude of the induced current be increased?
Answer: Induced current is created when a conductor moves through a magnetic field, or when the magnetic field around a coil changes. To increase the amount of this induced current, we can do the following:
1. **Increase the number of coils of the wire:** More turns in the coil mean the wire cuts through more magnetic field lines, leading to a larger induced voltage and thus more current.
2. **Increase the power of the magnet (magnetic field strength):** Using a stronger magnet creates a more intense magnetic field, which also results in a greater change in magnetic flux, leading to a larger induced current. We can also increase the speed of relative motion between the coil and the magnet.
In simple words: To make more electricity flow when a magnet and wire are used (induced current), you can use more loops of wire or a stronger magnet.

🎯 Exam Tip: Remember Faraday's law of electromagnetic induction, which links the magnitude of induced current to the rate of change of magnetic flux. Increasing turns, magnetic field strength, or speed all contribute to this change.

Question 29. How will the magnetic field produced in a current carrying circular coil change if we: (i) increase the value of current? (ii) increase the distance from the coil? (iii) increase the number of turns of coil?
Answer: We are looking at how the magnetic field around a circular coil carrying current changes based on certain factors.
(i) **If we increase the value of current:**
The magnetic field produced in the circular coil will **increase**. A stronger current creates a more intense magnetic field.
(ii) **If we increase the distance from the coil:**
The magnetic field will **decrease** as you move further away from the coil. The strength of a magnetic field diminishes with distance from its source.
(iii) **If we increase the number of turns of coil:**
The magnetic field produced will **increase**. More turns in the coil mean that each turn contributes to the magnetic field, effectively making the coil a stronger electromagnet.
In simple words: The magnetic field around a wire coil gets stronger if you send more electricity through it or add more loops to the coil. It gets weaker if you move further away from the coil.

🎯 Exam Tip: For circular coils, the magnetic field strength is directly proportional to both the current and the number of turns, and inversely related to the distance from the coil (at its center).

Electricity Current Long Answer Type Questions

Question 1. State Ohm's law. Draw a graph between voltage and current for a metallic conductor. Draw a circuit diagram of a circuit which consists of battery, ammeter, voltmeter, resistor, rheostat and a key.
Answer: Ohm's Law states that for a metallic conductor, the potential difference (\( V \)) across its ends is directly proportional to the current (\( I \)) flowing through it, provided its temperature and other physical conditions remain constant.
Mathematically, this can be written as:
\( V \propto I \)
\( \implies \) \( V = IR \)
Where \( R \) is the constant of proportionality called resistance.
**Graph between Voltage (V) and Current (I):**
For a metallic conductor obeying Ohm's Law, a graph of potential difference (\( V \)) versus current (\( I \)) is a straight line passing through the origin. The slope of this V-I graph represents the resistance (\( R \)) of the conductor.

**Circuit Diagram:**
A circuit diagram showing a battery, ammeter, voltmeter, resistor, rheostat, and a key, all connected properly for an experiment. The battery provides the power, the ammeter measures current in series, the voltmeter measures voltage in parallel across the resistor, the rheostat changes resistance, and the key controls the circuit.

In simple words: Ohm's law says that electricity flowing through a metal wire is directly related to the push (voltage) given to it, as long as the wire's temperature stays the same. The graph of voltage against current is a straight line. The circuit diagram shows all the parts needed to test this law, like a battery to give power, meters to measure electricity and push, and a way to change how much the wire resists.

🎯 Exam Tip: When drawing circuit diagrams, ensure all components are correctly symbolized and connected. Ammeters are always in series, and voltmeters are always in parallel with the component they measure.

Question 2. (a) A torch bulb is rated 2.5 V and 750 mA. Calculate (i) its power, (ii) its resistance, (iii) the energy consumed if this bulb is lighted for 4 hours. (b) Two identical resistors, each of resistance 2 Ohm, are connected (i) in series and (ii) in parallel, to a battery of 12 volts. Calculate the ratio of power consumed in two cases.
Answer: (a) We are given a torch bulb rated 2.5 V and 750 mA. We need to calculate its power, resistance, and energy consumed.
Given:
Voltage \( V = 2.5\,V \)
Current \( I = 750\,mA = 0.75\,A \)
Time \( t = 4\,hours = 4 \times 3600\,s = 14400\,s \)
(i) **Power (\( P \)):**
\( P = V \times I = 2.5\,V \times 0.75\,A = 1.875\,W \)
(ii) **Resistance (\( R \)):**
Using Ohm's Law \( V = IR \), so \( R = \frac{V}{I} \):
\( R = \frac{2.5\,V}{0.75\,A} = \frac{250}{75}\,\Omega = \frac{10}{3}\,\Omega \approx 3.33\,\Omega \)
(iii) **Energy consumed (\( E \)) if lighted for 4 hours:**
\( E = P \times t = 1.875\,W \times 4\,h = 7.5\,Wh \)
In Joules: \( E = P \times t = 1.875\,W \times 14400\,s = 27000\,J \)
This bulb uses 7.5 watt-hours or 27000 Joules of energy.

(b) Two identical resistors, each with a resistance of 2 Ω, are connected to a 12 V battery. We need to calculate the ratio of power consumed when they are connected in series versus in parallel.
Given:
Resistance of each resistor \( R_1 = R_2 = 2\,\Omega \)
Battery voltage \( V = 12\,V \)

(i) **When connected in series:**
Total resistance \( R_s = R_1 + R_2 = 2\,\Omega + 2\,\Omega = 4\,\Omega \)
Current in series circuit \( I_s = \frac{V}{R_s} = \frac{12\,V}{4\,\Omega} = 3\,A \)
Power consumed in series \( P_s = V \times I_s = 12\,V \times 3\,A = 36\,W \)

(ii) **When connected in parallel:**
Equivalent resistance \( R_p \):
\( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2\,\Omega} + \frac{1}{2\,\Omega} = \frac{2}{2\,\Omega} = \frac{1}{1\,\Omega} \)
\( R_p = 1\,\Omega \)
Current in parallel circuit \( I_p = \frac{V}{R_p} = \frac{12\,V}{1\,\Omega} = 12\,A \)
Power consumed in parallel \( P_p = V \times I_p = 12\,V \times 12\,A = 144\,W \)

**Ratio of power consumed (Series to Parallel):**
\( \frac{P_s}{P_p} = \frac{36\,W}{144\,W} = \frac{1}{4} \)
The ratio of power consumed in series to parallel is 1:4.
In simple words: (a) We found the bulb uses 1.875 watts of power, has about 3.33 ohms of resistance, and uses 7.5 watt-hours of energy in 4 hours. (b) When two 2-ohm resistors are in series with a 12V battery, they use 36 watts. When they are in parallel with the same battery, they use 144 watts. So, the power used in series compared to parallel is 1 to 4.

🎯 Exam Tip: Be careful with power calculations for series and parallel circuits. Power is generally higher for parallel combinations if the components are connected to the same voltage source, as resistance is lower, leading to higher current.

Question 3. For the circuit shown in the diagram given below: Calculate: (a) the value of current through each resistor. (b) the total current in the circuit. (c) the total effective resistance of the circuit.
Answer: We are given an electric circuit with a 6 V battery and three resistors (5 Ω, 10 Ω, and 30 Ω) connected in parallel.
(a) **Current through each resistor:**
In a parallel circuit, the voltage (\( V \)) across each resistor is the same as the battery voltage, which is 6 V. Using Ohm's Law (\( I = \frac{V}{R} \)):
Current through 5 Ω resistor (\( I_1 \)):
\( I_1 = \frac{V}{R_1} = \frac{6\,V}{5\,\Omega} = 1.2\,A \)
Current through 10 Ω resistor (\( I_2 \)):
\( I_2 = \frac{V}{R_2} = \frac{6\,V}{10\,\Omega} = 0.6\,A \)
Current through 30 Ω resistor (\( I_3 \)):
\( I_3 = \frac{V}{R_3} = \frac{6\,V}{30\,\Omega} = 0.2\,A \)

(b) **Total current in the circuit:**
In a parallel circuit, the total current (\( I_{total} \)) is the sum of the currents flowing through each branch:
\( I_{total} = I_1 + I_2 + I_3 = 1.2\,A + 0.6\,A + 0.2\,A = 2.0\,A \)

(c) **Total effective resistance of the circuit:**
For resistors in parallel, the reciprocal of the total effective resistance (\( R_p \)) is the sum of the reciprocals of individual resistances:
\( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} = \frac{1}{5\,\Omega} + \frac{1}{10\,\Omega} + \frac{1}{30\,\Omega} \)
To add these fractions, the common denominator is 30:
\( \frac{1}{R_p} = \frac{6}{30\,\Omega} + \frac{3}{30\,\Omega} + \frac{1}{30\,\Omega} = \frac{6+3+1}{30\,\Omega} = \frac{10}{30\,\Omega} = \frac{1}{3\,\Omega} \)
\( R_p = 3\,\Omega \)
Alternatively, we can use Ohm's Law with the total voltage and total current:
\( R_p = \frac{V_{total}}{I_{total}} = \frac{6\,V}{2.0\,A} = 3\,\Omega \)
In simple words: (a) For each resistor, we divide the battery voltage (6V) by its resistance to find the current: 1.2 amps for 5 ohms, 0.6 amps for 10 ohms, and 0.2 amps for 30 ohms. (b) The total current in the circuit is the sum of these currents, which is 2.0 amps. (c) The total resistance of the circuit is 3 ohms, calculated by adding the reciprocals of each resistance and then flipping the sum.

🎯 Exam Tip: Always double-check your common denominator calculations when adding reciprocals for parallel resistance, as this is a common source of error.

Question 4. (a) What is meant by saying that potential difference between two points is 1 volt? Name a device that helps to measure the potential difference across a conductor.
(b) Why does the connecting cord of an electric heater not glow hot while the heating element does?
(c) Electrical resistivities of some substances at 20°C are given below:

Which one is the best conductor of electricity out of them?
Answer the following questions in relation to them:
(i) Among silver and copper, which one is better conductor?
(ii) Which material would you advice to be used in electrical heating devices? Why?
Answer:
(a) When we say the potential difference between two points is 1 volt, it means that 1 Joule of work is done to move a unit positive charge (1 Coulomb) from one point to the other. A voltmeter is the device used to measure potential difference. This instrument helps in finding the voltage drop across different parts of a circuit.
(b) The connecting cord of an electric heater does not glow hot because it is made of copper, which has very low resistance. The heating element inside the heater, however, is made of a material like nichrome (an alloy) which has high resistance. When current flows through the high resistance heating element, it produces a lot of heat and glows. This high resistance is crucial for the heating effect.
(c) Silver and copper are the best conductors of electricity because they have very low resistivity, allowing current to flow easily.
(i) Silver is a better conductor than copper because its resistivity is slightly lower, even though both are excellent conductors. A lower resistivity means less opposition to current flow.
(ii) For electrical heating devices, nichrome (an alloy) should be used. This is because nichrome has a very high resistivity compared to metals like copper or silver. This high resistance means it heats up significantly when current passes through it, which is the desired effect for heating devices.
In simple words: 1 volt means 1 Joule of energy is used for 1 Coulomb of charge. A voltmeter measures this. Heater cords stay cool because they are good conductors (low resistance), but the heating part gets hot because it is a poor conductor (high resistance). Silver and copper conduct electricity well. Nichrome is best for heaters because it resists electricity a lot and gets very hot.

🎯 Exam Tip: Remember Ohm's law and how resistance affects heat production (\( H = I^2Rt \)). Materials with low resistivity are good conductors, while those with high resistivity are used for heating elements.

Question 5. (a) Name an instrument that measures electric current in a circuit. Define the unit of electric current.
(b) What do the following symbols mean in circuit diagram?
(i) \( \text{wwww} \)
(ii) \( \text{—(•)—} \)
(c) An electric circuit consisting of a 0.5 m long nichrome wire, an ammeter, a voltmeter, four cells of 1.5 V each and a plug key was set up.
(i) Draw a diagram of the electric circuit according to given details.
(ii) Following graph was plotted between V and I values:

What would be the values of V/I ratios when the potential difference is 0.5 V, 1.5 V and 2.0 V respectively? What conclusion do you draw from these values?
Answer:
(a) An ammeter is the instrument used to measure electric current in a circuit. The S.I. unit of electric current is ampere (A). One ampere is defined as the current flowing through a resistor of 1 \( \Omega \) when a potential difference of 1 volt is applied across its ends. This means that 1A is the flow of 1 Coulomb of charge per second.
(b)
(i) \( \text{wwww} \) This symbol represents a resistor, which is a component that opposes the flow of electric current.
(ii) \( \text{—(•)—} \) This symbol represents a plug key (or switch), used to open or close an electric circuit.
(c)
(i) Here is the circuit diagram as described:

(ii) To find the V/I ratio for each point on the graph:
When \( V = 0.5 \text{ V} \), from the graph, \( I = 0.2 \text{ A} \). So, \( \frac{V}{I} = \frac{0.5}{0.2} = 2.5 \, \Omega \)
When \( V = 1.5 \text{ V} \), from the graph, \( I = 0.6 \text{ A} \). So, \( \frac{V}{I} = \frac{1.5}{0.6} = 2.5 \, \Omega \)
When \( V = 2.0 \text{ V} \), from the graph, \( I = 0.8 \text{ A} \). So, \( \frac{V}{I} = \frac{2.0}{0.8} = 2.5 \, \Omega \)
From these values, we can conclude that the ratio \( \frac{V}{I} \) remains constant at \( 2.5 \, \Omega \) for different potential differences. This constant ratio represents the resistance of the nichrome wire, confirming Ohm's Law for a metallic conductor at a constant temperature. This means the material's resistance does not change with voltage or current.
In simple words: An ammeter measures current, and its unit is ampere. A resistor symbol looks like a zigzag line, and a plug key is a switch. For the circuit, the battery, ammeter, and key are in a loop, and the voltmeter is hooked across the wire. The graph shows that when voltage changes, the current changes in a way that their division always gives the same number (2.5 ohms). This means the wire's resistance stays the same, as per Ohm's Law.

🎯 Exam Tip: Always draw circuit diagrams clearly, labeling all components and their connections (series or parallel). For graphs, ensure you accurately read values and interpret the relationship between the plotted quantities, especially for Ohm's law experiments.

Question 7. A piece of wire of resistance 2 \( \Omega \) is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new situation.
Answer:
Let the original length of the wire be \( l_1 \) and its original cross-sectional area be \( A_1 \). The initial resistance is \( R_1 = 2 \, \Omega \).
The resistance of a wire is given by the formula: \( R = \rho \frac{l}{A} \), where \( \rho \) is resistivity.
So, for the original wire: \( R_1 = \rho \frac{l_1}{A_1} \)
When the wire is drawn out, its length is increased to twice its original length, so the new length \( l_2 = 2l_1 \).
When a wire is drawn, its volume remains constant. Volume \( V = A \times l \).
So, \( A_1 l_1 = A_2 l_2 \)
Since \( l_2 = 2l_1 \), we have \( A_1 l_1 = A_2 (2l_1) \).
This means \( A_1 = 2A_2 \), or the new cross-sectional area \( A_2 = \frac{A_1}{2} \).
Now, calculate the new resistance \( R_2 \) using the new length and area:
\( R_2 = \rho \frac{l_2}{A_2} \)
Substitute \( l_2 = 2l_1 \) and \( A_2 = \frac{A_1}{2} \):
\( R_2 = \rho \frac{2l_1}{A_1/2} \)
\( R_2 = \rho \frac{4l_1}{A_1} \)
We know that \( R_1 = \rho \frac{l_1}{A_1} \). So, \( R_2 = 4R_1 \).
Given \( R_1 = 2 \, \Omega \).
\( R_2 = 4 \times 2 \, \Omega \)
\( R_2 = 8 \, \Omega \)
Therefore, the resistance of the wire in the new situation is 8 \( \Omega \). Stretching a wire makes it longer and thinner, both of which increase its resistance.
In simple words: If you stretch a wire to be twice as long, it also becomes half as thick. Because resistance depends on both length and thickness, the new resistance will be four times the original resistance. So, 4 multiplied by the initial 2 ohms gives a new resistance of 8 ohms.

🎯 Exam Tip: Remember that when a wire is stretched, its volume remains constant. This means if length changes by a factor 'n', area changes by '1/n', and resistance changes by 'n²'.

Question 8. (a) What is a magnetic field? State the properties of magnetic field lines.
(b) State the rule for direction of the magnetic field produced around a current carrying conductor. Draw a sketch of the pattern of field lines due to a current flowing through a straight conductor.
Answer:
(a) A magnetic field is the region or space around a magnet or a current-carrying conductor where magnetic force can be felt by other magnetic substances. It is an invisible field that exerts a force on magnetic materials and moving electric charges. The S.I. unit of magnetic field strength is Tesla (T) or Weber per square meter (\( \text{Wb/m}^2 \)).
The properties of magnetic field lines are:
1. Magnetic field lines emerge from the North pole and merge at the South pole outside the magnet.
2. Inside the magnet, the direction of field lines is from the South pole to the North pole, forming continuous closed loops.
3. Magnetic field lines never intersect each other. If they did, it would mean two directions of the magnetic field at one point, which is impossible.
4. The closeness of field lines indicates the strength of the magnetic field. Where lines are closer, the field is stronger.
5. The tangent to a field line at any point gives the direction of the magnetic field at that point.
(b) The rule for determining the direction of the magnetic field produced around a current-carrying straight conductor is the **Right-Hand Thumb Rule**.
**Right-Hand Thumb Rule:** Imagine holding the current-carrying straight conductor in your right hand so that your thumb points in the direction of the electric current. Then, the direction in which your fingers curl around the conductor gives the direction of the magnetic field lines.
**Sketch of magnetic field lines due to a current flowing through a straight conductor:**

In simple words: A magnetic field is the area around a magnet where its push or pull can be felt. Magnetic lines always form closed loops, go from North to South outside the magnet, never cross, and show a stronger field where they are close. The Right-Hand Thumb Rule helps find the direction of the magnetic field around a wire: point your thumb with the current, and your fingers show the field's direction. The field lines are circles around the wire.

🎯 Exam Tip: Clearly distinguish between magnetic field properties for bar magnets and current-carrying conductors. For diagrams, ensure arrows correctly indicate field direction and that lines are concentric circles around the wire.

Question 9. (a) An electric bulb is rated as 50 W, 220 V, calculate the energy consumed by the bulbs in 20 minutes. Express your answer in commercial units of electric energy.
(b) What is the difference in construction of a DC generator than that of an AC generator.
(c) Explain the working of commutator in DC generator.
Answer:
(a) Given:
Power of the bulb (\( P \)) = 50 W
Voltage (\( V \)) = 220 V
Time (\( t \)) = 20 minutes
First, convert time from minutes to hours for commercial units (kilowatt-hour):
\( t = 20 \text{ minutes} = \frac{20}{60} \text{ hours} = \frac{1}{3} \text{ hour} \)
Energy consumed (\( E \)) = Power (\( P \)) \( \times \) Time (\( t \))
\( E = 50 \text{ W} \times \frac{1}{3} \text{ h} \)
To express in commercial units (kWh), convert power from Watts to kilowatts:
\( P = 50 \text{ W} = \frac{50}{1000} \text{ kW} = 0.05 \text{ kW} \)
Now, calculate energy:
\( E = 0.05 \text{ kW} \times \frac{1}{3} \text{ h} \)
\( E = \frac{0.05}{3} \text{ kWh} \approx 0.0166 \text{ kWh} \)
So, the energy consumed is approximately 0.017 kWh.
(b) The main construction difference between a DC generator and an AC generator lies in their ring systems. An AC generator uses **slip rings**, which are two full metallic rings that maintain continuous contact with the brushes. In contrast, a DC generator uses a **split-ring commutator**, which is a single metallic ring divided into two halves (C1 and C2), separated by an insulating material. This split ring allows the current direction in the external circuit to remain consistent.
(c) **Working of a Commutator in a DC Generator:**
The split-ring commutator acts as a device that reverses the direction of current flowing through the external circuit after every half rotation of the coil. When the armature coil rotates, the current induced in it changes direction every half cycle. The commutator ensures that the brush connected to one half of the ring always stays positive, and the brush connected to the other half always stays negative, relative to the external circuit. This is achieved because as the coil rotates, the halves of the split ring (C1 and C2) switch contact with the external brushes (B1 and B2) at precisely the point where the current in the coil reverses. This way, the current delivered to the external circuit always flows in one single direction, making it direct current (DC). This switching of contacts maintains a unidirectional flow of current in the external circuit.

In simple words: (a) A 50 W bulb used for 20 minutes (which is 1/3 of an hour) consumes about 0.017 kilowatt-hours of energy. (b) AC generators have full slip rings, but DC generators have split rings (commutators). (c) The commutator in a DC generator is like a special switch that flips the connections every half turn of the coil. This makes sure that electricity always flows out in one direction, giving a steady direct current instead of one that keeps changing direction.

🎯 Exam Tip: When calculating energy, always ensure units are consistent. For commercial energy, convert power to kilowatts and time to hours. For generators, focus on the role of the commutator in DC output and slip rings in AC output.