Selina Concise Solutions for ICSE Class 10 Physics Chapter 8 Current Electricity

 

Exercise (8 A)

 

Question 1. Define the term current and state its S.I unit.
Answer: Current is defined as the rate of flow of charge.
I = Q/t
Its S.I. unit is Ampere.
In simple words: Current means how much charge flows through a wire in one second. Think of it like water flowing through a pipe - more water flowing means more current.

📝 Teacher's Note: Use a water pipe example to explain current. Show students that current is like water flow - the faster the water flows, the more current we have. Many students forget the formula I = Q/t, so practice it often.

🎯 Exam Tip: Always write the formula I = Q/t and mention "rate of flow of charge." The unit is Ampere (symbol A). These are the key words examiners look for.

 

Question 2. Define the term electric potential. State its S.I. unit.
Answer: Electric potential at a point is defined as the amount of work done in bringing a unit positive charge from infinity to that point. Its unit is the volt.
In simple words: Electric potential is like the height of a hill. Just as you need energy to climb a hill, you need energy to bring a charge to a point with high potential.

📝 Teacher's Note: Compare electric potential to gravitational potential energy. A ball on a high table has more potential energy than one on the floor. Similarly, a charge at high potential has more electrical energy.

🎯 Exam Tip: Write "work done in bringing unit positive charge from infinity to that point" exactly. The unit is volt (V). Don't confuse potential with potential difference.

 

Question 3. How is the electric potential difference between the two points defined? State its S.I. unit.
Answer: The potential difference between two points is equal to the work done in moving a unit positive charge from one point to the other.
Its S.I. unit is Volt.
In simple words: Potential difference is like the height difference between two floors of a building. The bigger the difference, the more work you need to move from one floor to another.

📝 Teacher's Note: Use stairs to explain potential difference. Moving from ground floor to first floor needs work. Similarly, moving charge between different potentials needs work. This creates current flow.

🎯 Exam Tip: Key phrase is "work done in moving unit positive charge from one point to other." Unit is Volt. Remember potential difference causes current to flow.

 

Question 4. Explain the statement 'the potential difference between two points is 1 volt'.
Answer: One volt is the potential difference between two points in an electric circuit when 1 joule of work is done to move charge of 1 coulomb from one point to other.
In simple words: 1 volt means you need 1 joule of energy to move 1 coulomb of charge between two points. It's like saying you need 1 joule to lift 1 coulomb up a "voltage hill."

📝 Teacher's Note: Emphasize the numbers: 1 volt = 1 joule per 1 coulomb. Students often forget these exact values. Use the formula V = W/Q to show this relationship clearly.

🎯 Exam Tip: Write "1 joule of work" and "1 coulomb of charge" clearly. This definition question often comes in exams and these exact numbers give you full marks.

 

Question 5. Explain the analogy between the flow of charge (or current) in a conductor under a potential difference with the free fall of a body under gravity.
Answer: If a body is free to fall, on releasing it from a height, it falls downwards towards the earth's surface. For this one end has to be at higher level and other at lower level, so that gravity could effect on this difference and body could freely fall. Same way to make flow of the charge through a conductor, the gravity of course has no role to play; there should be difference of electric potential. This difference gives the flow of charge in a conductor.
In simple words: Just like a ball rolls down a hill due to gravity, electric charges flow from high potential to low potential. Both need a "difference" to make things move.

📝 Teacher's Note: Drop a ball in class to show gravity effect. Then explain that potential difference acts like "electrical gravity" - it pulls charges from high potential to low potential, creating current.

🎯 Exam Tip: Compare both situations clearly: ball needs height difference and gravity, charges need potential difference and electric field. Write "potential difference causes current flow" as the main conclusion.

 

Question 6. Define the term resistance. State its S.I. unit.
Answer: It is the property of a conductor to resist the flow of charges through it. Its S.I. unit is Ohm.
In simple words: Resistance is like friction for electric current. Just as rough roads slow down cars, resistance slows down the flow of electric charges.

📝 Teacher's Note: Use a crowded hallway example. Students move slowly in a crowded hallway due to resistance. Similarly, electrons move slowly in a wire with high resistance. The symbol for Ohm is Ω.

🎯 Exam Tip: Write "property of conductor to resist flow of charges" and "unit is Ohm (Ω)." Don't write "opposition" - use "resist" as it's the standard term.

 

Question 7. Name the particles which are responsible for the flow of current in a metal. Explain the flow of current in a metal on the basis of movement of the particles named by you.
Answer: In a metal, the charges responsible for the flow of current are the free electrons. The direction of flow of current is conventionally taken opposite to the direction of motion of electrons.
In simple words: Free electrons are like tiny balls that can move freely inside metals. When they move in one direction, we say current flows in the opposite direction. This is just a convention scientists follow.

📝 Teacher's Note: Explain that electrons are negative, so they move towards positive terminal. But current direction was decided before electrons were discovered, so we still follow the old convention of current flowing from positive to negative.

🎯 Exam Tip: Write "free electrons" as the particles and "current direction is opposite to electron motion." This conventional direction point often gets marks in exams.

 

Question 8. How does the resistance of a wire depend on its radius? Explain your answer.
Answer: Resistance of a wire is inversely proportional to the area of cross-section of the wire.
R ∝ \( \frac{1}{A} \)
R ∝ \( \frac{1}{\pi r^2} \)
This means if a wire of same length, but of double radius is taken, its resistance is found to be one-fourth.
In simple words: Thicker wires have less resistance. Think of a wide road - more cars can pass easily. Similarly, in thick wires, more electrons can flow easily, so resistance is less.

📝 Teacher's Note: Show students thick and thin wires. Explain that thick wire is like a wide highway - less traffic jams. Thin wire is like a narrow road - more traffic jams (resistance).

🎯 Exam Tip: Write the formula R ∝ 1/A and R ∝ 1/πr². Also mention "inversely proportional" clearly. The example of double radius giving one-fourth resistance shows you understand the concept.

 

Question 9. How does the resistance of a wire depend on its length? Give a reason for your answer with reason.
Answer: Resistance of a wire is directly proportional to the length of the wire.
R ∝ l
The resistance of a conductor depends on the number of collisions which the electrons suffer with the fixed positive ions while moving from one end to the other end of the conductor. Obviously the number of collisions will be more in a longer conductor as compared to a shorter conductor. Therefore, a longer conductor offers more resistance.
In simple words: Longer wires have more resistance. It's like a long obstacle course - the longer the course, the more hurdles you face. Electrons face more obstacles in longer wires.

📝 Teacher's Note: Compare to a long corridor versus short corridor. In a long corridor, you might bump into more people. Similarly, electrons collide more in longer wires, increasing resistance.

🎯 Exam Tip: Write "R ∝ l" and explain collisions clearly. The word "collisions" is important - it shows the physical reason for increased resistance in longer wires.

 

Question 10. How does the resistance of a metallic wire depend on its temperature? Explain with reason.
Answer: With the increase in temperature of conductor, both the random motion of electrons and the amplitude of vibration of fixed positive ions increase. As a result, the number of collisions increases. Hence, the resistance of a conductor increases with the increase in its temperature.
The resistance of filament of a bulb is more when it is glowing (i.e., when it is at a high temperature) as compared to when it is not glowing (i.e., when it is cold).
In simple words: Hot wires have more resistance than cold wires. When you heat a metal, atoms vibrate more and create more obstacles for electrons, just like people walking faster create more chaos in a crowded room.

📝 Teacher's Note: Touch a cold bulb and then a hot glowing bulb (carefully). The hot bulb has higher resistance. Students can relate this to the bulb getting dimmer when voltage drops.

🎯 Exam Tip: Write "resistance increases with temperature" and explain increased collisions. The bulb example is a good practical example to include for extra marks.

 

Question 11. Two wires one of copper and other of iron, are of the same length and same radius. Which will have more resistance? Give reason.
Answer: Iron wire will have more resistance than copper wire of the same length and same radius because resistivity of iron is more than that of copper.
In simple words: Iron wire has more resistance because iron is not as good at conducting electricity as copper. It's like comparing a smooth road (copper) with a rough road (iron) - the rough road slows down traffic more.

📝 Teacher's Note: Explain that resistivity is a material property. Different materials have different resistivities. That's why copper is used in electrical wires and iron is not commonly used.

🎯 Exam Tip: Write "resistivity of iron is more than copper" clearly. This shows you understand that resistance depends on the material (resistivity) even when size is same.

 

Question 12. Name three factors on which resistance of a given wire depends and state how is it affected by the factors stated by you.
Answer:
(i) Resistance of a wire is directly proportional to the length of the wire means with the increase in length resistance also increases.
R ∝ l
(ii) Resistance of a wire is inversely proportional to the area of cross-section of the wire. If area of cross-section of the wire is more, then resistance will be less and vice versa.
R ∝ \( \frac{1}{A} \)
(iii) Resistance increases with the increase in temperature since with increase in temperature the number of collisions increases.
(iv) Resistance depends on the nature of conductor because different substances have different concentration of free electrons. Substances such as silver, copper etc. offer less resistance and are called good conductors; but substances such as rubber, glass etc. offer very high resistance and are called insulators.
In simple words: Four things affect resistance: longer wire means more resistance, thicker wire means less resistance, hotter wire means more resistance, and different materials have different resistance.

📝 Teacher's Note: Use the formula R = ρl/A to show all factors together. ρ is resistivity (material), l is length, A is area, and temperature affects ρ. This covers all factors in one equation.

🎯 Exam Tip: Write all formulas clearly: R ∝ l, R ∝ 1/A, and mention temperature and material effects. Give examples of good conductors (copper) and insulators (rubber) for full marks.

 

Question 13. State Ohm's law and draw a neat labelled circuit diagram containing a battery, a key, a voltmeter, an ammeter, a rheostat and an unknown resistance to verify it.
Answer: It states that electric current flowing through a metallic wire is directly proportional to the potential difference V across its ends provided its temperature remains the same. This is called Ohm's law.
V = IR

[Diagram: Circuit diagram showing a battery connected to a key, rheostat, ammeter in series with an unknown resistance R, and a voltmeter connected in parallel across the resistance R]

In simple words: Ohm's law says that more voltage makes more current flow, but only if temperature stays the same. It's like pressing harder on a water tap - more pressure gives more water flow.

📝 Teacher's Note: Emphasize that temperature must be constant. Draw the circuit step by step, showing series connection for ammeter and parallel connection for voltmeter. Students often connect voltmeter in series by mistake.

🎯 Exam Tip: State the law clearly with the condition "temperature remains constant." Write V = IR formula. In the circuit, ammeter is in series and voltmeter is in parallel - this distinction gets marks.

 

Question 14. What is the necessary condition for a conductor to obey Ohm's law?
Answer: Ohm's law is obeyed only when the physical conditions and the temperature of the conductor remain constant.
In simple words: For Ohm's law to work, the wire must stay at the same temperature and nothing physical about it should change. If the wire gets hot, Ohm's law won't work properly.

📝 Teacher's Note: Explain that when current increases, wire gets hot, temperature changes, resistance changes, so Ohm's law fails. This is why we say "provided temperature remains constant" in the law.

🎯 Exam Tip: Write "temperature and physical conditions remain constant" clearly. This condition is often tested separately in exams because many students forget it.

 

Question 15. (a) draw a V-I graph for a conductor obeying Ohm's law. (b) what does the slope of V–I graph for a conductor represent?
Answer:

[Diagram: A straight line graph passing through origin with Potential difference (V) on Y-axis and Current (I) on X-axis, showing a linear relationship]

Slope of V-I graph represents the Resistance.
In simple words: The graph is a straight line through zero. The steepness (slope) of this line tells us the resistance - steeper line means higher resistance.

📝 Teacher's Note: Show that slope = rise/run = V/I = R. A steep graph means high resistance, a gentle slope means low resistance. The line must pass through origin for Ohm's law.

🎯 Exam Tip: Draw a straight line through origin. Label axes clearly as "Current (I)" and "Potential difference (V)". Write "slope = resistance" clearly for full marks.

 

Question 16. Draw a I – V graph for a linear resistor. What does its slope represent?
Answer:

[Diagram: A straight line graph passing through origin with Current (I) on Y-axis and Voltage (V) on X-axis, showing a linear relationship]

The slope of I-V graph (= \( \frac{\Delta I}{\Delta V} \)) is equal to the reciprocal of the resistance of the conductor, i.e.
Slope = \( \frac{\Delta I}{\Delta V} = \frac{1}{\text{Resistance of conductor}} \) = Conductance
In simple words: This graph also gives a straight line, but now current is on Y-axis. The slope tells us conductance, which is the opposite of resistance - how easily current flows.

📝 Teacher's Note: Compare with previous graph - axes are swapped. Slope is now 1/R = conductance. High conductance means low resistance (good conductor), low conductance means high resistance.

🎯 Exam Tip: Draw straight line through origin with correct axis labels. Write "slope = 1/R = conductance" clearly. Don't confuse with V-I graph where slope is R.

 

Question 17. What is an ohmic resistor? Give one example of an ohmic resistor. Draw a graph to show its current voltage relationship. How is the resistance of the resistor determined from this graph.
Answer: An ohmic resistor is a resistor that obeys Ohm's law, meaning current is directly proportional to voltage at constant temperature. Example: Metal wire resistors, carbon resistors. The I-V graph is a straight line passing through origin. Resistance is determined from the slope of V-I graph (R = V/I) or as reciprocal of slope of I-V graph (R = 1/slope).

[Diagram: A straight line graph passing through origin showing linear relationship between current and voltage]

In simple words: Ohmic resistors follow Ohm's law perfectly. Their graph is always a straight line. You can find resistance by dividing voltage by current at any point on the line.

📝 Teacher's Note: Examples include metal wires, carbon resistors. Non-ohmic examples are LED, diode. The graph must be straight line for ohmic behavior. Any curved graph means non-ohmic.

🎯 Exam Tip: Define as "resistor obeying Ohm's law" and give one clear example. Draw straight line graph and explain how to find R from slope. Both V-I and I-V graphs can be drawn.

 

Question 17. What is an ohmic resistor? Give examples and explain how resistance is determined from its graph.
Answer: An ohmic resistor is a resistor that obeys Ohm's law. For example: all metallic conductors (such as silver, aluminium, copper, iron etc.) From the graph, resistance is determined in the form of slope.
In simple words: Ohmic resistors follow Ohm's law perfectly. This means voltage and current have a straight line relationship. The slope of this straight line gives us the resistance value.

[Diagram: This diagram shows a V-I graph with a straight line going up from left to right, showing that voltage increases directly with current for an ohmic resistor.]

📝 Teacher's Note: Show students a graph on the board. Draw a straight line and explain that slope means how steep the line is. In ohmic resistors, this slope never changes.

🎯 Exam Tip: Always write "obeys Ohm's law" and "V-I graph is linear". These are key words examiners look for.

 

Question 18. What are non-ohmic resistors? Give one example and draw a graph to show its current-voltage relationship.
Answer: The conductors which do not obey Ohm's Law are called non-ohmic resistors. Example: diode valve.
In simple words: Non-ohmic resistors do not follow Ohm's law. Their resistance changes as voltage or current changes. The graph is not a straight line but a curve.

[Diagram: This diagram shows a V-I graph with a curved line that starts slowly and then rises sharply, showing non-linear behavior of a diode.]

📝 Teacher's Note: Compare with a light bulb filament. When you increase voltage, the bulb gets hot and its resistance changes. That's why it's non-ohmic.

🎯 Exam Tip: Write "do not obey Ohm's law" and "V-I graph is curved/non-linear". Give diode as the example.

 

Question 19. Give two difference between an ohmic and non-ohmic resistor.
Answer:

1. Ohmic resistor obeys ohm's law i.e., V/I is constant for all values of V or I; whereas Non-ohmic resistor does not obey ohm's law i.e., V/I is not same for all values of V or I.
2. In Ohmic resistor, V-I graph is linear in nature whereas in non-ohmic resistor, V-I graph is non-linear in nature.

In simple words: Ohmic resistors give a straight line graph and follow Ohm's law. Non-ohmic resistors give a curved graph and do not follow Ohm's law perfectly.

📝 Teacher's Note: Draw both graphs side by side on the board. Students can easily see the difference between straight line and curved line.

🎯 Exam Tip: Always mention "V/I is constant" for ohmic and "linear graph" vs "non-linear graph". These phrases get full marks.

 

Question 20. Fig 8.13 below shows the I-V characteristic curves for four resistors. Identify the ohmic and non-ohmic resistors. Give a reason for your answer.
Answer: Ohmic: (d), Non-Ohmic: (a), (b) and (c). Only for (d) the I-V graph is a straight line or linear while for (a), (b) and (c), the graph is a curve.
In simple words: Look at the shape of each graph. Only graph (d) is a straight line, so it is ohmic. The other three are curved, so they are non-ohmic.

[Diagram: This diagram shows four I-V graphs labeled (a), (b), (c), and (d). Graphs (a), (b), and (c) show curved lines while graph (d) shows a straight line.]

📝 Teacher's Note: Point to each graph and ask students "Is this line straight or curved?" This makes identification very easy for them.

🎯 Exam Tip: Write clearly: "straight line = ohmic" and "curved line = non-ohmic". Give the letter names correctly.

 

Question 21. Draw a V-I graph for a conductor at two different temperature. What conclusion do you draw from your graph for the variation of resistance of conductor with temperature?
Answer: In the above graph, T₁ > T₂. The straight line A is steeper than the line B, which leads us to conclude that the resistance of conductor is more at high temperature T₁ than at low temperature T₂. Thus, we can say that resistance of a conductor increases with the increase in temperature.
In simple words: When temperature goes up, resistance also goes up. The steeper line shows higher resistance. This happens because heat makes electrons move less freely through the conductor.

[Diagram: This diagram shows two straight lines on a V-I graph. Line A is steeper than line B, showing that resistance increases with temperature.]

📝 Teacher's Note: Heat a wire with a battery and bulb. As the wire gets hot, the bulb dims because resistance increases. This is a great demonstration.

🎯 Exam Tip: Write "steeper slope = higher resistance" and conclude "resistance increases with temperature". This gets full marks.

 

Question 22. Define the term resistivity and state its S.I unit.
Answer: The resistivity of a material is the resistance of a wire of that material of unit length and unit area of cross-section. Its S.I. unit is ohm metre.
In simple words: Resistivity tells us how much a material fights against electric current. It is measured for a standard piece - 1 meter long with 1 square meter area.

📝 Teacher's Note: Explain that resistivity is a property of the material itself. A copper wire and iron wire of same size will have different resistivity.

🎯 Exam Tip: Write the definition with "unit length and unit area" clearly. Units must be "ohm metre" not just "ohm".

 

Question 23. Write an expression connecting the resistance and resistivity. State the meaning of symbols used.
Answer: Expression: \( R = \rho \frac{l}{A} \)
ρ - resistivity
R - resistance
l - length of conductor
A - area of cross-section
In simple words: This formula shows that resistance depends on the material (ρ), length (l), and thickness (A). Longer wire has more resistance. Thicker wire has less resistance.

📝 Teacher's Note: Use a water pipe example. Longer pipe has more resistance to water flow. Thicker pipe has less resistance. Same idea works for electricity.

🎯 Exam Tip: Write the formula clearly and define all four symbols. This is a standard 4-mark question format.

 

Question 24. State the order of resistivity of (i) a metal, (ii) a semiconductor and (iii) an insulator.
Answer: Metal < Semiconductor < Insulator
In simple words: Metals have the lowest resistivity (electricity flows easily). Insulators have the highest resistivity (electricity cannot flow). Semiconductors are in between.

📝 Teacher's Note: Think of it like traffic. Metal is a highway (easy flow). Insulator is a wall (no flow). Semiconductor is a narrow road (some flow).

🎯 Exam Tip: Write the inequality clearly with < signs. Remember: lowest resistivity means best conductor.

 

Question 25. Name a substance of which the resistance remains almost unchanged by the increase in temperature.
Answer: Manganin
In simple words: Manganin is a special alloy. Even when it gets hot or cold, its resistance stays almost the same. This makes it very useful for making standard resistors.

📝 Teacher's Note: Explain that this property makes manganin perfect for making resistors that work the same in all weather conditions.

🎯 Exam Tip: Just write "Manganin". This is a one-word answer question. Some books also accept "Constantan".

 

Question 26. Name the material used for making the connection wires. Give a reason for your answer. Why should a connection wire be thick?
Answer: 'Copper or Aluminium' is used as a material for making connection wires because the resistivity of these materials is very small, and thus, wires made of these materials possess negligible resistance. The connection wires are made thick so that their resistance can be considered as negligible. \( R = \rho \frac{l}{A} \) Therefore, greater the area of cross-section, lesser shall be the resistance.
In simple words: Copper and aluminium let electricity flow very easily (low resistivity). Thick wires have bigger area, so resistance becomes very small. We want connection wires to have almost zero resistance.

📝 Teacher's Note: Show students thin and thick wires. Explain that we want the resistance in our devices (bulbs, heaters) not in the connecting wires.

🎯 Exam Tip: Write both materials, give the reason (low resistivity), and explain thick wire using the formula. Connect area and resistance clearly.

 

Question 27. Name a material which is used for making the standard resistor. Give a reason for your answer.
Answer: Manganin is used for making the standard resistor because its resistivity is quite large and the effect of change in temperature on their resistance is negligible.
In simple words: Standard resistors need to have the exact same resistance always. Manganin is perfect because temperature changes do not affect its resistance value.

📝 Teacher's Note: Standard resistors are like measuring scales - they must give the same reading always. Temperature should not change their value.

🎯 Exam Tip: Write "Manganin" and give both reasons: "high resistivity" and "temperature independent". Both points are needed for full marks.

 

Question 28. Name the material used for making a fuse wire. Give a reason.
Answer: Generally fuse wire is made of an alloy of lead and tin because its resistivity is high and melting point is low.
In simple words: Fuse wire must melt quickly when too much current flows. Lead-tin alloy melts easily (low melting point) and has high resistance to create heat when current is high.

📝 Teacher's Note: Explain that fuse is like a safety guard. When current is too high, it sacrifices itself by melting to protect other devices.

🎯 Exam Tip: Write "alloy of lead and tin" and give both properties: "high resistivity" and "low melting point". Both are essential.

 

Question 29. Name the material used for (i) filament of an electric bulb, (ii) heating element of a room heater.
Answer: (i) A wire made of tungsten is used for filament of electric bulb because it has a high melting point and high resistivity. (ii) A nichrome wire is used as a heating element for a room heater because the resistivity of nichrome is high and increase in its value with increase in temperature is high.
In simple words: Tungsten can get very hot without melting, perfect for light bulbs. Nichrome gets hotter as current increases, making it great for heaters that need to produce heat.

📝 Teacher's Note: Show students that bulb filaments glow white hot but don't melt. Heater elements turn red hot. Both need materials that can handle high temperatures.

🎯 Exam Tip: For bulb: write "tungsten" with "high melting point". For heater: write "nichrome" with "high resistivity". Give reasons for both parts.

 

Question 30. What is a superconductor? Give one example of it.
Answer: A superconductor is a substance of zero resistance at a very low temperature. Example: Mercury at 4.2 K.
In simple words: Superconductors have no resistance at all when made very cold. Electricity can flow through them without any loss. Mercury becomes a superconductor at very cold temperature.

📝 Teacher's Note: Explain that 4.2 K is extremely cold - much colder than ice. At this temperature, mercury allows perfect flow of electricity.

🎯 Exam Tip: Write "zero resistance" and "very low temperature" clearly. Give mercury with the temperature value 4.2 K.

 

Question 31. A substance has zero resistance below 1 k. what is such a substance called?
Answer: Superconductor
In simple words: Any substance that has zero resistance at very low temperatures is called a superconductor. Below 1 K means extremely cold temperature.

📝 Teacher's Note: Connect this to the previous question. When any material shows zero resistance at low temperature, we call it a superconductor.

🎯 Exam Tip: Simply write "Superconductor". The definition is already given in the question.

 

Multiple Choice Type

 

Question 1. Which of the following is an ohmic resistance?
(a) diode valve
(b) junction diode
(c) filament of a bulb
(d) nichrome
Answer: (d) nichrome
In simple words: Nichrome follows Ohm's law perfectly. Its V-I graph is a straight line. Diodes and bulb filaments do not follow Ohm's law, so they are non-ohmic.

📝 Teacher's Note: Remind students that metals and alloys like nichrome are usually ohmic. Electronic devices like diodes are usually non-ohmic.

🎯 Exam Tip: Remember that nichrome is an alloy used in heating elements. It obeys Ohm's law, making it ohmic.

 

Question 2. For which of the following substance, resistance decreases with increase in temperature?
(a) copper
(b) mercury
(c) carbon
(d) platinum
Answer: (c) carbon
In simple words: Carbon is a semiconductor. When semiconductors get hot, their resistance goes down. This is opposite to metals like copper where resistance goes up with temperature.

📝 Teacher's Note: Explain that metals and semiconductors behave opposite to each other. Metals: more heat = more resistance. Semiconductors: more heat = less resistance.

🎯 Exam Tip: Remember that carbon and silicon are semiconductors. For semiconductors, resistance decreases with temperature increase.

 

Numericals

 

Question 1. In a conductor 6.25 × 10¹⁶ electrons flow from its end A to B in 2 s. Find the current flowing through the conductor (e = 1.6 × 10⁻¹⁹ C)
Answer:
Given:
Number of electrons flowing through the conductor, N = 6.25 × 10¹⁶ electrons
Time taken, t = 2 s
Charge of electron, e = 1.6 × 10⁻¹⁹ C

Step 1: Use the formula for current.
Current \( I = \frac{ne}{t} \)

Step 2: Substitute the values.
\[ I = \frac{(6.25 \times 10^{16}) \times (1.6 \times 10^{-19})}{2} \]
\[ I = \frac{10 \times 10^{-3}}{2} = 5 \times 10^{-3} \text{ A} \]

Step 3: Convert to milliamperes.
I = 5 mA

Thus, 5 mA current flows from B to A.
In simple words: We count how many electrons move and multiply by their charge. Then divide by time to get current. Current flows opposite to electron flow.

📝 Teacher's Note: Show students that electrons are negative, so they move from A to B, but current (positive charge flow) goes from B to A. Use water flow analogy - electrons are like water drops, current is like the stream direction.

🎯 Exam Tip: Always write the formula I = ne/t first. Remember current direction is opposite to electron flow. Show all steps clearly to get full marks.

 

Question 2. A current of 1.6 mA flows through a conductor. If charge on an electron is -1.6 × 10⁻¹⁹ coulomb, find the number of electrons that will pass each second through the cross section of that conductor.
Answer:
Given:
Current, I = 1.6 mA = 1.6 × 10⁻³ A
Charge on electron, Q = -1.6 × 10⁻¹⁹ coulomb
Time, t = 1 sec

Step 1: Find total charge using I = Q/t.
Q = I × t
Q = 1.6 × 10⁻³ × 1 = 1.6 × 10⁻³ C

Step 2: Find number of electrons.
Number of electrons = \( \frac{\text{Total charge}}{\text{Charge per electron}} \)
Number of electrons = \( \frac{1.6 \times 10^{-3}}{1.6 \times 10^{-19}} = 10^{16} \)

In simple words: We find how much total charge flows in one second. Then we divide by the charge of one electron to count how many electrons that is.

📝 Teacher's Note: Explain that current tells us charge per second. To find electrons per second, divide total charge by charge of one electron. Use counting analogy - like counting coins when you know total money.

🎯 Exam Tip: First convert mA to A. Then use Q = It to find total charge. Finally divide by electron charge. Don't forget the minus sign doesn't affect the count.

 

Question 3. Find the potential difference required to pass a current of 0.2 A in a wire of resistance 20Ω
Answer:
Given:
Current (I) = 0.2 A
Resistance (R) = 20 ohm
Potential Difference (V) = ?

Step 1: Use Ohm's Law.
V = IR

Step 2: Substitute values.
V = 0.2 × 20 = 4 V

In simple words: Ohm's law says voltage equals current times resistance. It's like pressure needed to push water through a pipe - more resistance needs more pressure.

📝 Teacher's Note: Use water pipe example - voltage is like water pressure, current is like water flow, resistance is like pipe narrowness. More pressure needed for narrow pipes.

🎯 Exam Tip: Always write V = IR formula first. Check units - current in A, resistance in Ω gives voltage in V. Show calculation clearly.

 

Question 4. An electric bulb draws 1.2 A current at 6.0 V. Find the resistance of filament of bulb while glowing.
Answer:
Given:
Current (I) = 1.2 A
Potential Difference/Voltage (V) = 6.0 V
Resistance (R) = ?

Step 1: Use Ohm's Law.
V = IR
Therefore, R = V/I

Step 2: Substitute values.
R = 6/1.2 = 5 Ohm

In simple words: We use Ohm's law to find how much the bulb resists current flow. Higher resistance means the wire fights more against current.

📝 Teacher's Note: Explain that hot bulb filament has higher resistance than cold filament. Use analogy of thick honey (high resistance) vs thin water (low resistance) flowing through a straw.

🎯 Exam Tip: Write R = V/I clearly. Always check if answer makes sense - typical bulb resistance is few ohms. Units must be correct.

 

Question 5. A car bulb connected to a 12 volt battery draws 2 A current when glowing. What is the resistance of the filament of the bulb? Will the resistance be more same or less when the bulb is not glowing?
Answer:
Given:
Potential Difference/Voltage (V) = 12 V
Current (I) = 2 A
Resistance (R) = ?

Step 1: Use Ohm's Law.
V = IR
Therefore, R = V/I

Step 2: Calculate resistance.
R = 12/2 = 6 Ohm

Step 3: Answer about cold bulb.
Resistance will be less when the bulb is not glowing.

In simple words: When the bulb glows, the filament gets very hot. Hot metal has more resistance than cold metal. So cold bulb has less resistance.

📝 Teacher's Note: Show students a cold vs hot metal wire. Hot metal atoms vibrate more and block electron flow more. Like trying to walk through a busy crowd vs empty room.

🎯 Exam Tip: First calculate resistance using R = V/I. Then explain that hot metal has higher resistance than cold metal. This is a common exam question.

 

Question 6. Calculate the current flowing through a wire of resistance 5 Ω connected to a battery of potential difference 3 V.
Answer:
Given:
Potential Difference/Voltage (V) = 3 V
Resistance (R) = 5 ohm
Current (I) = ?

Step 1: Use Ohm's Law.
V = IR
Therefore, I = V/R

Step 2: Substitute values.
I = 3/5 = 0.6 A

In simple words: Higher voltage pushes more current. Higher resistance allows less current. It's like water pressure and pipe thickness.

📝 Teacher's Note: Use garden hose example - more water pressure gives more flow, but narrower hose reduces flow. Voltage is pressure, resistance is narrowness.

🎯 Exam Tip: Write I = V/R formula first. Check that higher voltage gives higher current, higher resistance gives lower current. Show division clearly.

 

Question 7. In an experiment of verification of Ohm's law following observations are obtained.

Potential difference V (in volt)	0.5	1.0	1.5	2.0	2.5
Current I (in ampere)	0.2	0.4	0.6	0.8	1.0

Draw a characteristic V-I graph and use this graph to find:
(a) potential difference V when the current I is 0.5 A,
(b) current I when the potential difference V is 0.75 V,
(c) resistance in circuit.

Answer:
[Diagram: This shows a V-I graph with voltage on Y-axis (0 to 2.5 V) and current on X-axis (0 to 1.0 A). The graph is a straight line passing through origin, showing Ohm's law relationship.]

(a) From the graph: V = 1.25 V when I = 0.5 A
(b) From the graph: I = 0.3 A when V = 0.75 V
(c) The graph is linear so resistance can be found from any value of the given table.
When V = 2.5 Volt and I = 1.0 amp
Using Ohm's law: R = V/I
R = 2.5/1.0 = 2.5 ohm

In simple words: The straight line graph shows that voltage and current are directly related. The slope of the line gives us the resistance value.

📝 Teacher's Note: Show students how to read values from graph by drawing horizontal and vertical lines. Explain that straight line means ohmic behavior - constant resistance.

🎯 Exam Tip: Draw neat graph with proper labels and units. For resistance, use any point on line and apply R = V/I. Straight line through origin confirms Ohm's law.

 

Question 8. Two wires of the same material and same length have radii r₁ and r₂ respectively compare: (i) their resistances, (ii) their resistivities.
Answer:
(i) For wire of radius r₁:
\( R_1 = \rho \frac{l}{A_1} \)
\( R_1 = \rho \frac{l}{\pi r_1^2} \)

For wire of radius r₂:
\( R_2 = \rho \frac{l}{A_2} \)
\( R_2 = \rho \frac{l}{\pi r_2^2} \)

\( R_1 : R_2 = \frac{\rho l}{\pi r_1^2} : \frac{\rho l}{\pi r_2^2} \)
\( R_1 : R_2 = r_2^2 : r_1^2 \)

(ii) Since the material of the two wires is same, so their resistivities will also be same i.e.,
ρ₁ : ρ₂ = 1:1

In simple words: Thicker wire (bigger radius) has less resistance. But resistivity depends only on material, not on size. Same material means same resistivity.

📝 Teacher's Note: Use water pipe analogy - thicker pipe allows more water flow (less resistance). But the pipe material properties don't change with thickness.

🎯 Exam Tip: Write resistance formula with area. Remember area of circle is πr². Resistivity is material property - same material means same resistivity.

 

Question 9. A given wire of resistance 1 Ω is stretched to double its length. What will be its new resistance?
Answer:
Given:
Let 'l' be the length and 'a' be the area of cross-section of the resistor with resistance, R = 1Ω

Step 1: When wire is stretched to double its length.
New length l' = 2l
New area a' = a/2 (volume remains constant)

Step 2: Calculate new resistance.
\( R' = \rho \frac{l'}{a'} = \rho \frac{2l}{a/2} \)
\( R' = 4\rho \frac{l}{a} = 4R \)
\( R' = 4 \times 1 = 4Ω \)

In simple words: When we stretch a wire, it becomes longer but thinner. Both changes increase resistance. The new resistance becomes 4 times the original.

📝 Teacher's Note: Show students with rubber band or clay - when stretched, it becomes longer and thinner. Both effects make it harder for current to flow.

🎯 Exam Tip: Remember volume stays constant when stretching. If length doubles, area becomes half. Write R = ρl/a formula and substitute new values.

 

Question 10. A wire 3 ohm resistance and 10 cm length is stretched to 30 cm length. Assuming that it has a uniform cross section, what will be its new resistance?
Answer:
Given:
Resistance (R) = 3 ohm
Length l = 10 cm
New Length (l') = 30 cm = 3 × l

\( R = \rho \frac{l}{A} \)

Step 1: Find new dimensions.
With stretching, length increases and area decreases in same proportion.
New area A' = A/3 (since length is tripled)

Step 2: Calculate new resistance.
\( R' = \rho \frac{3l}{A/3} = 9\rho \frac{l}{A} = 9R \)
\( R' = 9 \times 3 = 27Ω \)

In simple words: The wire becomes 3 times longer and 3 times thinner. This makes resistance 9 times bigger (3 × 3 = 9).

📝 Teacher's Note: Explain that both length increase and area decrease contribute to higher resistance. The total effect multiplies - like double trouble.

🎯 Exam Tip: If length increases by factor n, area decreases by factor n. So resistance increases by factor n². Here n = 3, so R' = 9R.

 

Question 11. A wire of 9 ohm resistance having 30 cm length is tripled on itself. What is its new resistance?
Answer:
Given:
Resistance (R) = 9 ohm
Length l = 30 cm
New Length (l') = 30/3 = 10 cm (tripled means folded in 3 parts)

\( R = \rho \frac{l}{A} \)

Step 1: Understand tripling.
When tripled on itself, the wire becomes 3 times thicker and 3 times shorter.
New length l' = l/3
New area A' = 3A

Step 2: Calculate new resistance.
\( R' = \rho \frac{l/3}{3A} = \frac{1}{9} \rho \frac{l}{A} = \frac{R}{9} \)
\( R' = \frac{9}{9} = 1Ω \)

In simple words: Folding the wire makes it shorter and thicker. Both changes reduce resistance. The new resistance becomes 9 times smaller.

📝 Teacher's Note: Show with a rope or wire - when folded in 3 parts, it becomes 3 times thicker and 3 times shorter. Both help current flow better.

🎯 Exam Tip: "Tripled on itself" means folded into 3 layers. Length becomes l/3, area becomes 3A. So resistance becomes R/9.

 

New Resistance:
With change in length, there will be change in area of cross-section also in the same order.
\( R' = \rho \frac{l/3}{3A} \)
\( R' = \frac{1}{9} \rho \frac{l}{A} \)
\( R' = \frac{1}{9}R \)
\( R' = 1 \text{ ohm} \)

 

Question 12. What length of copper wire of resistivity 1.7 × 10⁻⁸ Ω m and radius 1 mm is required so that its resistance is 1Ω?
Answer:
Given:
Resistance (R) = 1 ohm
Resistivity (ρ) = 1.7 × 10⁻⁸ ohm metre
Radius (r) = 1 mm = 10⁻³ m
Length (l) = ?

Step 1: Use the resistance formula.
\( R = \rho \frac{l}{A} \)

Step 2: Find length.
\( l = \frac{RA}{\rho} \)

Step 3: Find area of cross-section.
\( A = \pi r^2 \)
\( l = \frac{R \pi r^2}{\rho} \)

Step 4: Substitute values.
\( l = \frac{1 × \pi × (10^{-3})^2}{1.7 × 10^{-8}} \)
\( l = \frac{1 × \pi × 10^{-6}}{1.7 × 10^{-8}} \)
\( l = 184.7 \text{ m} \)

Length of copper wire = 184.7 m
In simple words: We used the resistance formula to find how long the wire should be. We need almost 185 meters of thin copper wire to get 1 ohm resistance.

📝 Teacher's Note: Show students that a very long thin wire has more resistance than a short thick wire. It is like water flowing through a long narrow pipe versus a short wide pipe.

🎯 Exam Tip: Always write "Given:" first, then the formula. Convert mm to m before calculation. Write the final answer with correct units (meters).

 

Exercise (8B)

 

Question 1. Explain the meaning of the terms e.m.f., terminal voltage and internal resistance of a cell.
Answer:
e.m.f.: When no current is drawn from a cell, the potential difference between the terminals of the cell is called its electro-motive force (or e.m.f.).

Terminal voltage: When current is drawn from a cell, the potential difference between the electrodes of the cell is called its terminal voltage.

Internal Resistance: The resistance offered by the electrolyte inside the cell to the flow of electric current through it is called the internal resistance of the cell.
In simple words: e.m.f. is the full voltage when cell is not used. Terminal voltage is less voltage when cell is being used. Internal resistance is like friction inside the cell that reduces voltage.

📝 Teacher's Note: Compare with a water tank. e.m.f. is like full pressure when tap is closed. Terminal voltage is less pressure when tap is open and water flows out.

🎯 Exam Tip: Write all three definitions clearly. Remember: e.m.f. is when "no current flows" and terminal voltage is when "current flows". This difference gets marks.

 

Question 2. State two differences between the e.m.f and terminal voltage of a cell.
Answer:

e.m.f. of cell	Terminal voltage of cell
1. It is measured by the amount of work done in moving a unit positive charge in the complete circuit inside and outside the cell.	1. It is measured by the amount of work done in moving a unit positive charge in the circuit outside the cell.
2. It is the characteristic of the cell i.e., it does not depend on the amount of current drawn from the cell	2. It depends on the amount of current drawn from the cell. More the current is drawn from the cell, less is the terminal voltage.
3. It is equal to the terminal voltage when cell is not in use, while greater than the terminal voltage when cell is in use.	3. It is equal to the emf of cell when cell is not in use, while less than the emf when cell is in use.

In simple words: e.m.f. stays the same always. Terminal voltage changes - it becomes less when we use more current from the cell.

📝 Teacher's Note: Use a battery example. New battery shows full voltage (e.m.f.). When you connect a bulb, voltage drops (terminal voltage) because current flows.

🎯 Exam Tip: Make a neat table. Write that e.m.f. is constant but terminal voltage depends on current. This comparison gets full marks.

 

Question 3. Name two factors on which the internal resistance of a cell depends and state how does it depend on the factors stated by you.
Answer:
Internal resistance of a cell depends upon the following factors:
(i) The surface area of the electrodes: Larger the surface area of the electrodes, less is the internal resistance.
(ii) The distance between the electrodes: More the distance between the electrodes, greater is the internal resistance.
In simple words: Big electrode plates give less resistance. Far apart plates give more resistance. It is like a wide short road versus a narrow long road.

📝 Teacher's Note: Show two different sized batteries. Bigger battery has bigger plates and less internal resistance. That is why car batteries are big.

🎯 Exam Tip: Write both factors clearly. State the relationship - "larger area = less resistance" and "more distance = more resistance". Use these exact words.

 

Question 4. A cell of e.m.f ε and internal resistance r is used to send current to an external resistance R. write expressions for (a) the total resistance of circuit, (b) the current drawn from the cell. (c) the p.d across the cell. And (d) voltage drop inside the cell.
Answer:
(a) Total resistance = R + r

(b) Current drawn from the circuit:
As we know that,
ε = V + v
= IR + Ir
= I(R + r)
\( I = \frac{\varepsilon}{R + r} \)

(c) p.d. across the cell: \( \frac{\varepsilon}{R + r} × R \)

(d) voltage drop inside the cell: \( \frac{\varepsilon}{R + r} × r \)
In simple words: Total resistance adds up. Current depends on total voltage divided by total resistance. External voltage is current times external resistance.

📝 Teacher's Note: Draw a simple circuit diagram. Show that internal resistance r is in series with external resistance R. So they add up.

🎯 Exam Tip: Write all four formulas clearly. Remember the current formula I = ε/(R+r). This is the most important formula in this chapter.

 

Question 5. A cell is used to send current to an external circuit. (a) How does the voltage across its terminals compare with its e.m.f? (b) under what condition is the e.m.f of a cell equal to its terminal voltage?
Answer:
(a) Terminal voltage is less than the emf: Terminal Voltage < e.m.f.
(b) e.m.f. is equal to the terminal voltage when no current is drawn.
In simple words: When cell is used, terminal voltage is always less than e.m.f. They are equal only when cell is not being used at all.

📝 Teacher's Note: Test this with a real battery and voltmeter. Measure voltage when nothing is connected, then when a bulb is connected. Students will see the voltage drop.

🎯 Exam Tip: Write the inequality sign clearly: Terminal voltage < e.m.f. For part (b), write "when no current flows" or "open circuit condition".

 

Question 6. Explain why the p.d across the terminals of a cell is more in an open circuit and reduced in a closed circuit.
Answer:
When the electric cell is in a closed circuit the current flows through the circuit. There is a fall of potential across the internal resistance of the cell. So, the p.d. across the terminals in a closed circuit is less than the p.d. across the terminals in an open circuit by an amount equal to the potential drop across the internal resistance of the cell.
In simple words: When current flows, some voltage is lost inside the cell due to internal resistance. So terminal voltage becomes less than e.m.f.

📝 Teacher's Note: Compare with water flowing through a pipe with friction. Some pressure is lost due to friction inside the pipe. Same way, some voltage is lost inside the cell.

🎯 Exam Tip: Mention "potential drop across internal resistance" in your answer. This key phrase shows you understand the concept and gets marks.

 

Question 7. Write the expressions for the equivalent resistance R of three resistors R₁, R₂ and R₃ joined in (a) parallel (b) series
Answer:
(a) Total Resistance in series:
R = R₁ + R₂ + R₃

(b) Total Resistance in parallel:
\( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \)
In simple words: In series, resistances add up. In parallel, we add up the reciprocals (1/R values) then take reciprocal of the sum.

📝 Teacher's Note: Series is like people standing in a line - they add up. Parallel is like multiple roads - easier path, so less total resistance.

🎯 Exam Tip: Learn both formulas by heart. Series: simple addition. Parallel: reciprocal formula. Do not mix them up in exam.

 

Question 8. How would you connect two resistors in series? Draw a diagram. Calculate the total equivalent resistance.
Answer:

[Diagram: Circuit diagram showing two resistors R₁ and R₂ connected one after the other in a single loop with a battery, with current I flowing through both resistors]

If current I is drawn from the battery, the current through each resistor will also be I.
On applying Ohm's law to the two resistors separately, we further have
V₁ = I R₁
V₂ = I R₂
V = V₁ + V₂
IR = I R₁ + I R₂
R = R₁ + R₂

Total Resistance in series R = R₁ + R₂ + R₃
In simple words: Connect resistors one after another in a line. Same current flows through all. Total resistance is sum of all resistances.

📝 Teacher's Note: Use Christmas lights as example. Old Christmas lights were in series - if one bulb failed, all bulbs went off because same current flows through all.

🎯 Exam Tip: Draw neat diagram. Show current I same everywhere. Write V = V₁ + V₂ and derive R = R₁ + R₂. Show all steps.

 

Question 9. Show by a diagram how two resistors R₁ and R₂ are joined in parallel. Obtain an expression for the total resistance of combination.
Answer:

[Diagram: Circuit diagram showing two resistors R₁ and R₂ connected in parallel branches between points A and B, with currents I₁ and I₂ flowing through each branch respectively, and total current I from the battery]

On applying Ohm's law to the two resistors separately, we further have
I₁ = V / R₁
I₂ = V / R₂
I = I₁ + I₂
\( \frac{V}{R} = \frac{V}{R_1} + \frac{V}{R_2} \)
\( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} \)
In simple words: Connect resistors side by side between same two points. Same voltage across all. Total current splits between branches.

📝 Teacher's Note: Use water pipes analogy. Two parallel pipes carry more water than one pipe. Similarly, parallel resistors carry more current, so less total resistance.

🎯 Exam Tip: Draw neat diagram showing parallel connection. Write I = I₁ + I₂ and derive the reciprocal formula step by step.

 

Question 10. State how are the two resistors joined with a battery in each of the following cases when:
(a) same current flows in each resistor
(b) potential difference is same across each resistor
(c) equivalent resistance is less than either of the two resistances
(d) equivalent resistance is more than either of the two resistances.
Answer:
(a) series
(b) parallel
(c) parallel
(d) series
In simple words: Series means same current but different voltages. Parallel means same voltage but different currents. Series increases total resistance. Parallel decreases total resistance.

📝 Teacher's Note: Make students remember: Series = Same current, More resistance. Parallel = Same voltage, Less resistance. This helps them answer quickly.

🎯 Exam Tip: Learn the key differences: Series has same current everywhere. Parallel has same voltage across each branch. Series adds resistance, parallel reduces it.

 

Question 11. The V-I graph for a series combination and for a parallel combination of two resistors is shown in Fig – 8.38. Which of the two, A or B, represents the parallel combination? Give a reason for your answer.
Answer: For the same change in I, change in V is less for the straight line A than for the straight line B (i.e., the straight line A is less steeper than B), so the straight line A represents small resistance, while the straight line B represents more resistance. In parallel combination, the resistance decreases while in series combination, the resistance increases. So A represents the parallel combination.
In simple words: Line A is less steep, which means less resistance. Parallel circuits have less total resistance than series circuits. So A is the parallel combination.

📝 Teacher's Note: Draw two lines on the board - one steep, one gentle. Show students that the gentle slope means less resistance. Like having more paths for water to flow makes it easier.

🎯 Exam Tip: Write "A represents parallel because it has lower resistance (less steep line)." Always explain why parallel has lower resistance.

 

Multiple Choice Type

 

Question 1. In series combination of resistances:
(a) p.d is same across each resistance
(b) total resistance is reduced
(c) current is same in each resistance
(d) all above are true
Answer: (c) current is same in each resistance
In simple words: In series, there is only one path for current to flow. So the same current passes through all resistors, like water flowing through a single pipe.

📝 Teacher's Note: Use a chain of students holding hands. The same "current" (squeeze) passes through each student. This shows series connection clearly.

🎯 Exam Tip: Remember "Series = Same current." Write that current has only one path to flow through all resistors.

 

Question 2. In parallel combination of resistances:
(a) p.d is same across each resistance
(b) total resistance is increased
(c) current is same in each resistance
(d) all above are true
Answer: (a) p.d is same across each resistance
In simple words: In parallel, all resistors are connected to the same battery terminals. So they all get the same voltage, like all bulbs in your house getting 230V.

📝 Teacher's Note: Show parallel connection like ladder rungs - each rung connects the same two sides. All resistors connect to same two points, so same voltage.

🎯 Exam Tip: Write "Parallel = Same voltage (p.d)" and explain that all resistors connect to same battery terminals.

 

Question 3. Which of the following combinations have the same equivalent resistance between X and Y?
Answer: (a) and (d)
Solution:
In fig (a), the resistors are connected in parallel between X and Y.
Let R' be their equivalent resistance.
Then, \( \frac{1}{R'} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} \)
Or, R' = 1 Ω ………………. (i)

In fig (d) a series combination of two 1Ω resistors is in parallel with another series combination of two 1Ω resistors.
Series resistance of two 1 Ohm resistors, R = (1 + 1)Ω = 2 Ω
Thus, we can say that across X and Y, two 2Ω resistors are connected in parallel.
Let R' be the net resistance across X and Y.
Then, \( \frac{1}{R'} = \frac{1}{2} + \frac{1}{2} = \frac{2}{2} \)
Or, R' = 1Ω ……………………(ii)

From (i) and (ii), it is clear that (a) and (d) have the same equivalent resistance between X and Y.
In simple words: Both circuits end up giving 1Ω total resistance. Even though they look different, the math gives the same answer.

[Diagram: Circuit diagrams showing different combinations of 1Ω and 2Ω resistors in series and parallel arrangements between points X and Y]

📝 Teacher's Note: Show students that different looking circuits can have same total resistance. Use building blocks to show different arrangements giving same result.

🎯 Exam Tip: Calculate step by step. First find series combinations, then parallel. Show all working clearly to get full marks.

 

Numericals

 

Question 1. The diagram below in Fig. 8.40 shows a cell of e.m.f. ε = 2 volt and internal resistance r = 1 ohm to an external resistance R = 4 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when (i) the key K is open, (ii) the key K is closed.
Answer:
(i) Ammeter reading = 0 because of no current
Voltage V = ε − Ir
V = 2 − 0 × 1 = 2 volt

(ii) Ammeter reading:
I = ε/(R + r)
I = 2/(4+1) = 2/5 = 0.4 amp
Voltage reading:
Voltage V = ε - Ir
V = 2 - 0.4 × 1 = 2 - 0.4 = 1.6 V
In simple words: When switch is open, no current flows so ammeter reads zero. When closed, current flows and voltage drops due to internal resistance.

[Diagram: Circuit showing a cell with internal resistance connected to external resistance through a key, with ammeter and voltmeter]

📝 Teacher's Note: Explain that real batteries have internal resistance like a small resistor inside. This causes voltage to drop when current flows.

🎯 Exam Tip: Always use formula V = ε - Ir for terminal voltage. Show both cases clearly - open circuit (I=0) and closed circuit.

 

Question 2. A battery of e.m.f 3.0 V supplies current through a circuit in which the resistance can be changed. A high resistance voltmeter is connected across the battery. When the current is 1.5 A, the voltmeter reads 2.7 V. Find the internal resistance of the battery.
Answer:
ε = 3 volt
I = 1.5 A
V = 2.7 V
V = ε − Ir
r = (ε-V)/I
= (3 – 2.7)/1.5 = 0.2 ohm
In simple words: The battery voltage drops from 3V to 2.7V when current flows. This 0.3V drop is due to internal resistance.

📝 Teacher's Note: Show that real batteries are not perfect. They have some resistance inside that causes voltage to drop when we draw current.

🎯 Exam Tip: Write the formula V = ε - Ir first, then rearrange to find r. Always show units in your final answer.

 

Question 3. A cell of e.m.f. 1.8V and internal resistance 2Ω is connected in series with an ammeter of resistance 0.7Ω and a resistor of 4.5Ω as shown in Fig. 8.41
(a) what would be the reading of the ammeter?
(b) what is the potential difference across the terminals of the cell?
Answer:
(a) ε = 1.8 V
Total Resistance = 2 + 4.5 + 0.7 = 7.2 Ω
I = ε/R (total resistance)
I = 1.8/7.2 = 0.25 A

(b) Current (calculated in (a) part) I = 0.25 A
Now, total resistance excluding internal resistance = 4.5 + 0.7 = 5.2 ohm
V = IR = 0.25 × 5.2 = 1.3 V
In simple words: Current flows through all resistances in series. Terminal voltage is the voltage across external resistances only.

[Diagram: Circuit showing cell with internal resistance connected in series with ammeter and external resistor]

📝 Teacher's Note: Clarify that terminal voltage means voltage across external circuit only, not including voltage drop across internal resistance.

🎯 Exam Tip: For terminal voltage, multiply current by external resistance only. Don't include internal resistance in this calculation.

 

Question 4. A battery of e.m.f. 15 V and internal resistance 3 ohm is connected to two resistors of resistances 3 ohm and 6 ohm in series Find:
(a) the current through the battery
(b) the p.d. between the terminals of the battery.
Answer:
(a) ε = 15 V
R = 6 + 3 = 9 ohm
r = 3 ohm
I = ε/(R + r)
I = 15/(9 + 3) = 15/12 = 1.25 A

(b) Current (calculated in (a) part) I = 1.25 A
External Resistance R = 6 + 3 = 9 ohm
V = IR = 1.25 × 9 = 11.25 V
In simple words: Total current flows through battery and external resistors. Terminal voltage is voltage across external resistors only.

📝 Teacher's Note: Show that in series, same current flows through battery and all external resistors. Terminal voltage excludes internal voltage drop.

🎯 Exam Tip: Add all external resistances first, then find current. Terminal voltage = current × external resistance only.

 

Question 5. A cell of e.m.f. ε and internal resistance r sends current 1.0 A when it is connected to an external resistance 1.9Ω. But it sends current 0.5 A when it is connected to an external resistance 3.9 Ω. Calculate the values of ε and r.
Answer:
In first case
I = 1 A, R = 1.9 ohm
ε = I(R + r) = 1(1.9+r)
ε = 1.9 + r------------(1)

In second case
I = 0.5 A, R = 3.9 ohm
ε = I(R + r) = 0.5 (3.9 + r)
ε = 1.95 + 0.5r ----------------(2)

From eq. (1) and (2),
1.9 + r = 1.95 + 0.5r
r = 0.05/0.5 = 0.1 ohm

Substituting value of r
ε = 1.9 + r = 1.9 + 0.1 = 2 V
In simple words: We make two equations using the two given conditions, then solve them together to find both unknown values.

📝 Teacher's Note: This is simultaneous equations. Show students to write both cases first, then eliminate one variable to find the other.

🎯 Exam Tip: Label your equations (1) and (2). Show substitution step clearly. Write both final answers with correct units.

 

Question 6. Two resistors having resistance 4Ω and 6Ω are connected in parallel. Find their equivalent resistance.
Answer:
Let R' be their equivalent resistance of the 4Ω and 6Ω resistors connected in parallel.
Then, \( \frac{1}{R'} = \frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12} \)
Or, \( R' = \frac{12}{5} = 2.4Ω \)
In simple words: In parallel, we add reciprocals of resistances, then take reciprocal of the sum. The answer is always less than the smallest resistance.

📝 Teacher's Note: Show that parallel resistance is always smaller than any individual resistance. Like having more lanes on a road reduces traffic resistance.

🎯 Exam Tip: Write the parallel formula \( \frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_2} \) first. Take LCM correctly to avoid calculation errors.

 

Question 7. Four resistors each of resistance 2Ω are connected in parallel. What is the effective resistance?
Answer:
R₁ = 2 ohm
R₂ = 2 ohm
R₃ = 2 ohm
R₄ = 2 ohm

For parallel connection:
\( \frac{1}{R'} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = \frac{4}{2} = 2 \)
Or, R' = 1/2 = 0.5Ω
In simple words: When identical resistors are in parallel, divide the resistance by the number of resistors. 2Ω ÷ 4 = 0.5Ω.

📝 Teacher's Note: For identical resistors in parallel, use the shortcut: R' = R/n where n is number of resistors. Much faster than the full formula.

🎯 Exam Tip: When all resistors are equal in parallel, use R' = R/n. This saves time and reduces errors in calculations.

 

Question 8. You have three resistors of values 2Ω, 3Ω and 5Ω. How will you join them so that the total resistance is less than 1Ω? Draw diagram and find the total resistance.
Answer: The three resistors should be connected in parallel to get a total resistance less than 1Ω.

Let R' be the total resistance.
Then, \( \frac{1}{R'} = \frac{1}{2} + \frac{1}{3} + \frac{1}{5} = \frac{15 + 10 + 6}{30} = \frac{31}{30} \)

Or, \( R' = \frac{30}{31} = 0.97Ω \)

[Diagram: This diagram shows three resistors of 2Ω, 3Ω and 5Ω connected in parallel between two points, with a battery connected to the circuit.]

In simple words: When we connect resistors in parallel, the total resistance becomes very small. It is like making the path wider for current to flow. So we get less than 1Ω.

📝 Teacher's Note: Draw the parallel connection clearly. Show students that parallel means all resistors connect to the same two points. Use the formula 1/R = 1/R₁ + 1/R₂ + 1/R₃.

🎯 Exam Tip: Always write "parallel connection" clearly. Show the calculation step by step. Write the final answer with proper units (Ω).

 

Question 9. Three resistors each of 2Ω are connected together so that their total resistance is 3Ω. Draw a diagram to show this arrangement and check it by calculation.
Answer: A parallel combination of two resistors, in series with one resistor.

Let R₁ = R₂ = R₃ = 2 ohm

Step 1: Find resistance of parallel combination of R₁ and R₂
\( \frac{1}{R'} = \frac{1}{R₁} + \frac{1}{R₂} = \frac{1}{2} + \frac{1}{2} = 1 \)

R' = 1 ohm

Step 2: Add R₃ in series
R = R' + R₃ = 1 + 2 = 3 ohm

[Diagram: This diagram shows two 2Ω resistors connected in parallel, and this parallel combination is connected in series with a third 2Ω resistor.]

In simple words: We put two resistors side by side (parallel), then connect the third one in a line (series). This gives us exactly 3Ω total.

📝 Teacher's Note: Show students the mixed connection clearly. First calculate parallel part, then add the series part. This is a common exam question type.

🎯 Exam Tip: Draw the diagram first. Show both calculations clearly - parallel first, then series addition. Write "mixed connection" in your answer.

 

Question 10. Calculate the equivalent resistance of the following combination of resistors r₁, r₂, r₃ and r₄ if r₁ = r₂ = r₃ = r₄ = 2.0Ω, between the points A and B in Fig. 8.42
Answer:
r₁ = r₂ = r₃ = r₄ = 2.0 ohm

Step 1: Find resistance of series combination r₁ and r₂
r' = r₁ + r₂ = 2 + 2 = 4 ohm

Step 2: Find resistance of parallel combination r₃ and r₄
\( \frac{1}{r''} = \frac{1}{r₃} + \frac{1}{r₄} = \frac{1}{2} + \frac{1}{2} = 1 \)

r'' = 1 ohm

Step 3: Find total resistance between A and B
r = r' + r'' = 4 + 1 = 5 ohm

[Diagram: This diagram shows resistors r₁ and r₂ in series in the upper branch, and resistors r₃ and r₄ in parallel in the lower branch, with both branches connected between points A and B.]

In simple words: The circuit has two parts - top part has resistors in series, bottom part has resistors in parallel. Then we add both parts together.

📝 Teacher's Note: Teach students to identify series and parallel sections first. Draw the equivalent circuit after each step. This makes complex circuits easier to solve.

🎯 Exam Tip: Break the circuit into smaller parts. Solve each part separately, then combine. Write "equivalent resistance = 5Ω" as your final answer.

 

Question 11. A combination consists of three resistors in series. Four similar sets are connected in parallel. If the resistance of each resistor is 2 ohm, find the resistance of the combination.
Answer:
Resistance of each set:
r₁ = 2 + 2 + 2 = 6 ohm
r₂ = 2 + 2 + 2 = 6 ohm
r₃ = 2 + 2 + 2 = 6 ohm
r₄ = 2 + 2 + 2 = 6 ohm

Now these resistances are arranged in parallel:
\( \frac{1}{r} = \frac{1}{r₁} + \frac{1}{r₂} + \frac{1}{r₃} + \frac{1}{r₄} \)

\( \frac{1}{r} = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{4}{6} \)

\( r = \frac{6}{4} = 1.5 \text{ ohm} \)

In simple words: First we make 4 identical sets, each with 3 resistors in a line. Then we connect all 4 sets side by side. This gives us 1.5Ω total.

📝 Teacher's Note: Show students that we solve this in two steps. First find resistance of one set (series), then find total of all sets (parallel). Use the formula carefully.

🎯 Exam Tip: Write "each set has 6Ω" first. Then write "4 sets in parallel give 1.5Ω". Show all calculation steps clearly for full marks.

 

Question 12. In the circuit shown below in Fig 8.43, calculate the value of x if the equivalent resistance between A and B is 4Ω.
Answer:
r₁ = 4 ohm
r₂ = 8 ohm
r₃ = x ohm
r₄ = 5 ohm
r = 4 ohm

Step 1: Find resistance of upper series branch
r' = r₁ + r₂ = 4 + 8 = 12 ohm

Step 2: Find resistance of lower series branch
r'' = r₃ + r₄ = (x + 5) ohm

Step 3: Use parallel formula for total resistance
\( \frac{1}{r} = \frac{1}{r'} + \frac{1}{r''} \)

\( \frac{1}{4} = \frac{1}{12} + \frac{1}{5 + x} \)

\( \frac{1}{4} - \frac{1}{12} = \frac{1}{5 + x} \)

\( \frac{3 - 1}{12} = \frac{1}{5 + x} \)

\( \frac{2}{12} = \frac{1}{5 + x} \)

\( \frac{1}{6} = \frac{1}{5 + x} \)

Therefore, 5 + x = 6
x = 1 ohm

[Diagram: This diagram shows two parallel branches between points A and B. Upper branch has 4Ω and 8Ω in series. Lower branch has xΩ and 5Ω in series.]

In simple words: We have two paths from A to B. We know the total resistance is 4Ω. We use the parallel formula to find the missing value x.

📝 Teacher's Note: Teach students to identify the parallel branches first. Then use algebra to solve for the unknown. Practice this type of question often.

🎯 Exam Tip: Write "Given: total resistance = 4Ω" at the start. Show all algebra steps clearly. Write "x = 1Ω" as your final answer.

 

Question 13. Calculate the effective resistance between the points A and B in the circuit shown in Fig 8.44.
Answer:
In the figure above,
Resistance between XAY = (1 + 1 + 1) = 3Ω
Resistance between XY = 2Ω
Resistance between XBY = 6Ω

Let R' be the net resistance between points X and Y

Then, \( \frac{1}{R'} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1 \)

Or, R' = 1Ω

Thus, we can say that between points A and B, three 1Ω resistors are connected in series.

Let R_AB be the net resistance between points A and B.
Then, R_AB = (1 + 1 + 1)Ω = 3Ω

[Diagram: This diagram shows a complex network with resistors arranged between points A and B, with intermediate points X and Y.]

In simple words: This circuit has a middle section (between X and Y) with parallel resistors. We first solve that part, then add the remaining resistors in series.

📝 Teacher's Note: Show students how to identify the parallel section first. Solve the middle part, then treat the whole circuit as three resistors in series. Draw the simplified circuit.

🎯 Exam Tip: Write "resistance between X and Y = 1Ω" clearly. Then write "total resistance A to B = 3Ω". Show the parallel calculation step by step.

 

Question 14. A wire of uniform thickness with a resistance of 27Ω is cut into three equal pieces and they are joined in parallel. Find the equivalent resistance of the parallel combination.
Answer:
Wire cut into three pieces means new resistance = 27/3 = 9Ω

Now three resistances connected in parallel:
\( \frac{1}{r} = \frac{1}{r₁} + \frac{1}{r₂} + \frac{1}{r₃} = \frac{1}{9} + \frac{1}{9} + \frac{1}{9} = \frac{3}{9} = \frac{1}{3} \)

r = 9/3 = 3 ohm

In simple words: When we cut a wire into 3 equal parts, each part has 1/3 of the original resistance. When we connect them in parallel, we get even less resistance.

📝 Teacher's Note: Explain that cutting a wire reduces its length, so resistance also reduces. Then parallel connection reduces it further. Use the formula step by step.

🎯 Exam Tip: Write "each piece has 9Ω" first. Then use parallel formula. Write "final answer = 3Ω" clearly. Remember resistance of each piece = total resistance ÷ number of pieces.

 

Question 15. A circuit consists of a 1 ohm resistor in series with a parallel arrangement of 6 ohm and 3 ohm resistors. Calculate the total resistance of the circuit. Draw a diagram.
Answer:
Step 1: Find resistance of parallel combination of 6Ω and 3Ω
\( \frac{1}{r} = \frac{1}{6} + \frac{1}{3} = \frac{1 + 2}{6} = \frac{3}{6} = \frac{1}{2} \)

R = 2 ohm

Step 2: Add 1Ω resistor in series
R = 2 + 1 = 3 ohm

[Diagram: This diagram shows a 1Ω resistor in series with a parallel combination of 3Ω and 6Ω resistors.]

In simple words: First we find the resistance of the parallel part (3Ω and 6Ω together). Then we add the 1Ω resistor that is in series with this parallel part.

📝 Teacher's Note: Teach students to always solve parallel parts first, then add series parts. Draw the circuit clearly showing which resistors are in parallel and which are in series.

🎯 Exam Tip: Write "parallel resistance = 2Ω" first. Then write "total resistance = 3Ω". Draw the diagram clearly. Show both calculation steps.

 

Question 16. Calculate the effective resistance between the points A and B in the network shown below in Fig 8.45
Answer:
R₁ = 1 + 2 = 3 ohm
R₂ = 1.5 ohm

[Diagram: This diagram shows a network with resistors arranged between points A and B, including 1Ω, 2Ω, and 1.5Ω resistors in a specific configuration.]

In simple words: This circuit has resistors arranged in a special network pattern. We need to identify which are in series and which are in parallel to solve it step by step.

📝 Teacher's Note: The solution appears incomplete in the source. Teach students to first identify the circuit topology, then apply series and parallel formulas systematically.

🎯 Exam Tip: Always start by redrawing the circuit in a clearer form. Identify series and parallel sections first. Show each calculation step clearly with proper units.

 

Question 17. Calculate the equivalent resistance between A and B in the adjacent diagram in Fig 8.46.
Answer:
Given:
From the circuit diagram:
- Top path: 3Ω + 2Ω = 5Ω (R₁)
- Middle path: 30Ω (R₂)
- Bottom path: 6Ω + 4Ω = 10Ω (R₃)

Step 1: Calculate total resistance for each path.
R₁ = 3 + 2 = 5 ohm
R₂ = 30 ohm
R₃ = 6 + 4 = 10 ohm

Step 2: Since R₁, R₂ and R₃ are connected in parallel:
\( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \)

\( \frac{1}{R} = \frac{1}{5} + \frac{1}{30} + \frac{1}{10} = \frac{6 + 1 + 3}{30} = \frac{10}{30} \)

Step 3: Calculate final resistance.
R = 30/10 = 3 ohm

Equivalent resistance between A and B = 3 ohm
In simple words: We first add resistors in each separate path. Then we use the parallel formula to combine all three paths. Three parallel paths make it easier for current to flow.

📝 Teacher's Note: Draw the circuit clearly and mark each path separately. Students often forget to add series resistors in each path first. Use different colors for each path.

🎯 Exam Tip: Always write "Given" and list all values first. Show the parallel formula clearly. Final answer must have units (ohm).

 

Question 18. In the network shown in adjacent Fig. 8.47, calculate the equivalent resistance between the points.
(a) A and B
(b) A and C
Answer:
(a) Between A and B:
R₁ = 2 + 2 + 2 = 6 ohm (top path)
R₂ = 2 ohm (bottom path)

R₁ and R₂ are connected in parallel:
\( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6} + \frac{1}{2} = \frac{1 + 3}{6} = \frac{4}{6} \)

R = 6/4 = 1.5 ohm

(b) Between A and C:
R₁ = 2 + 2 = 4 ohm (one path)
R₂ = 2 + 2 = 4 ohm (other path)

R₁ and R₂ are connected in parallel:
\( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \)

R = 2 ohm

In simple words: First we find the total resistance of each separate path by adding resistors. Then we combine parallel paths using the parallel formula.

📝 Teacher's Note: Show students how to trace each path from start to finish. Mark which resistors are in series (same path) and which paths are in parallel (different routes).

🎯 Exam Tip: Label each path clearly as R₁, R₂ etc. Show series addition first, then parallel combination. Answer both parts completely.

 

Question 19. Five resistors, each 3Ω, are connected as shown in Fig 8.48. Calculate the resistance (a) between the points P and Q. (b) between the points X and Y.
Answer:
(a) Between P and Q:
R₁ = 3 + 3 = 6 ohm (one path)
R₂ = 3 ohm (other path)

R₁ and R₂ are connected in parallel:
\( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{6} + \frac{1}{3} = \frac{1 + 2}{6} = \frac{3}{6} = \frac{1}{2} \)

R = 2 ohm

(b) Between X and Y:
From part (a), resistance between P and Q = 2 ohm
R₃ = 3 ohm (additional resistor)
R₄ = 3 ohm (additional resistor)

Total resistance = R + R₃ + R₄ = 2 + 3 + 3 = 8 ohm

In simple words: For P to Q, we have two parallel paths. For X to Y, we add the P-Q resistance plus two more resistors in series.

📝 Teacher's Note: Draw the circuit step by step. Show how P-Q is just one part of the X-Y circuit. Use building blocks approach - solve smaller parts first.

🎯 Exam Tip: Use the result from part (a) in part (b). Show that X-Y includes P-Q plus extra resistors. This saves time and shows clear thinking.

 

Question 20. Two resistors of 2.0Ω and 3.0Ω are connected (a) in series (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through the battery.
Answer:
(a) Series connection:
[Diagram: Circuit shows battery connected to R₁ and R₂ in series]

R₁ = 2 ohm
R₂ = 3 ohm
R = R₁ + R₂ = 2 + 3 = 5 ohm
V = 6 V

Current: \( I = \frac{V}{R} = \frac{6}{5} = 1.2 \text{ A} \)

(b) Parallel connection:
[Diagram: Circuit shows battery connected to R₁ and R₂ in parallel]

R₁ and R₂ are connected in parallel:
\( \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6} \)

R = 6/5 = 1.2 ohm
V = 6 V

Current: \( I = \frac{V}{R} = \frac{6}{1.2} = 5 \text{ A} \)

In simple words: In series, total resistance is more, so current is less. In parallel, total resistance is less, so current is more. Parallel gives more current from same battery.

📝 Teacher's Note: Draw both circuits clearly on the board. Show that series adds resistance but parallel reduces it. Use water pipe analogy - parallel pipes allow more flow.

🎯 Exam Tip: Always draw the circuit diagram first. Show all calculations step by step. Remember: series adds resistance, parallel reduces it.

 

Question 21. A resistor of 6Ω is connected in series with another resistor of 4Ω. A potential difference of 20 V is applied across the combination. Calculate (a) the current in the circuit and (b) the potential difference across the 6Ω resistor.
Answer:
(a) Current in circuit:
R₁ = 6 ohm
R₂ = 4 ohm
Total resistance: R = R₁ + R₂ = 6 + 4 = 10 ohm
V = 20 V

Current: \( I = \frac{V}{R} = \frac{20}{10} = 2 \text{ A} \)

(b) Potential difference across 6Ω resistor:
R = 6 ohm
I = 2 A (same current flows through both resistors in series)

Voltage: V = IR = 6 × 2 = 12 V

In simple words: In series, same current flows through both resistors. Each resistor gets some voltage based on its resistance value.

📝 Teacher's Note: Emphasize that current is same in series but voltage divides. The bigger resistor gets more voltage. Use the water pipe analogy with different sized sections.

🎯 Exam Tip: For series circuits, always find total resistance first, then current. Use V = IR to find voltage across individual resistors. Current is same everywhere in series.

 

Question 22. In fig 8.50, calculate:
(a) the total resistance of the circuit
(b) the value if R, and
(c) the current flowing in R.
Answer:
For resistor A:
R = 1 ohm, V = 2 V
Current: \( I = \frac{V}{R} = \frac{2}{1} = 2 \text{ A} \)

For resistor B:
R = 2 ohm, V = 2 V
Current: \( I = \frac{V}{R} = \frac{2}{2} = 1 \text{ A} \)

(a) Total resistance of circuit:
V = 4 V, I = 0.4 A
Total resistance: \( R' = \frac{V}{I} = \frac{4}{0.4} = 10 \text{ ohm} \)

(b) Value of R:
R₁ = 20 ohm, R' = 10 ohm
Using parallel formula: \( \frac{1}{R'} = \frac{1}{R} + \frac{1}{R_1} \)
\( \frac{1}{10} = \frac{1}{R} + \frac{1}{20} \)
\( \frac{1}{R} = \frac{1}{10} - \frac{1}{20} = \frac{2-1}{20} = \frac{1}{20} \)
R = 20 ohm

(c) Current flowing in R:
R = 20 ohm, V = 4 V
Current: \( I = \frac{V}{R} = \frac{4}{20} = 0.2 \text{ A} \)

In simple words: We use Ohm's law to find current in each part. Total current splits between parallel branches based on their resistance.

📝 Teacher's Note: Show how total current splits in parallel circuits. Current goes more through the path with less resistance. Use river splitting into streams as analogy.

🎯 Exam Tip: For parallel circuits, voltage is same across all branches. Use V = IR for each branch separately. Check that branch currents add up to total current.

 

Question 23. A wire of length 5 m has a resistance of 2.0Ω calculate:
(a) the resistance of wire of length 1 m
(b) the equivalent resistance if two such wires each of length 2 m are joined in parallel.
(c) the resistance of 1 m length of wire of same material but of half diameter.
Answer:
(a) Resistance of 1 m wire:
5 m wire has 2.0 Ω resistance
1 m wire has: \( \frac{2.0}{5} = 0.4 \text{ Ω} \)

(b) Two 2 m wires in parallel:
2 m wire has: 2 × 0.4 = 0.8 Ω each
Two 0.8 Ω resistors in parallel:
\( \frac{1}{R} = \frac{1}{0.8} + \frac{1}{0.8} = \frac{2}{0.8} = 2.5 \)
R = 1/2.5 = 0.4 Ω

(c) Wire with half diameter:
When diameter becomes half, area becomes \( \frac{1}{4} \) (Area ∝ diameter²)
Resistance ∝ \( \frac{1}{\text{Area}} \)
So resistance becomes 4 times
New resistance = 4 × 0.4 = 1.6 Ω

In simple words: Longer wire has more resistance. Thinner wire has more resistance. Parallel wires share the current, so total resistance is less.

📝 Teacher's Note: Use the water pipe analogy. Longer pipe has more resistance to flow. Thinner pipe also has more resistance. Two parallel pipes allow more flow.

🎯 Exam Tip: Remember: Resistance ∝ Length and Resistance ∝ 1/Area. When diameter halves, area becomes 1/4, so resistance becomes 4 times.

 

Question 24. A particular resistance wire has a resistance of 3.0 ohm per meter. Find:
(a) The total resistance of three lengths of this wire each 1.5 m long, joined in parallel.
(b) The potential difference of the battery which gives a current od 2.0 A in each of the 1.5 m length when connected in parallel to the battery (assume that the resistance of battery is negligible)
(c) the resistance of 5 m length of a wire of the same material, but with twice the area of cross section.
Answer:
(a) Three 1.5 m wires in parallel:
Each 1.5 m wire has resistance: 3.0 × 1.5 = 4.5 Ω
Three identical 4.5 Ω resistors in parallel:
\( \frac{1}{R} = \frac{1}{4.5} + \frac{1}{4.5} + \frac{1}{4.5} = \frac{3}{4.5} \)
R = 4.5/3 = 1.5 Ω

(b) Battery voltage:
Each wire carries 2.0 A and has resistance 4.5 Ω
Voltage across each wire: V = IR = 2.0 × 4.5 = 9.0 V
Since parallel connection, battery voltage = 9.0 V

(c) 5 m wire with twice the area:
Normal 5 m wire would have: 3.0 × 5 = 15 Ω
With twice the area, resistance becomes half: 15/2 = 7.5 Ω

In simple words: Three parallel wires share the current, so total resistance is less. In parallel, all wires get same voltage. Thicker wire (more area) has less resistance.

📝 Teacher's Note: Show that n identical resistors in parallel give total resistance = R/n. In parallel, voltage is same but current adds up.

🎯 Exam Tip: For parallel identical resistors, total resistance = individual resistance ÷ number of resistors. Remember: resistance ∝ 1/area.

 

Question 24:
Answer:
(a)
Resistance of 1m of wire = 3 ohm
Resistance of 1.5 m of wire = 3 × 1.5 = 4.5 Ω
\( \frac{1}{R} = \frac{1}{4.5} + \frac{1}{4.5} + \frac{1}{4.5} = \frac{3}{4.5} \)
R = 1.5 ohm

(b)
I = 2 A
V = IR = 2 × 4.5 = 9 V

(c)
R = 3 ohm for 1 m
For 5 m: R = 3 × 5 = 15 ohm
But Area A is double i.e. 2A and Resistance is inversely proportional to area so Resistance will be half.
R = 15/2 = 7.5 ohm
In simple words: When wires are connected in parallel, the total resistance becomes less. When the area of wire doubles, resistance becomes half because more space for current to flow.

📝 Teacher's Note: Show students how parallel wires are like having more lanes on a road - more paths means less resistance. When area doubles, it's like making the road twice as wide.

🎯 Exam Tip: For parallel resistors, use the formula 1/R = 1/R₁ + 1/R₂ + 1/R₃. Remember that resistance is inversely proportional to area - write this clearly.

 

Question 25. A cell supplies a current of 1.2 A through two 2 Ω resistors connected in parallel. When the resistors are connected in series, it supplies a current of 0.4 A. Calculate: (i) the internal resistance and (ii) e.m.f. of the cell.
Answer:
In parallel R = 1/2 + 1/2 = 1 ohm
I = 1.2 A
ε = I(R + r) = 1.2(1 + r) = 1.2 + 1.2 r

In series R = 2+2 = 4 ohm
I = 0.4 A
ε = I(R + r) = 0.4(4 + r) = 1.6 + 0.4 r

It means:
1.2 + 1.2 r = 1.6 + 0.4 r
0.8 r = 0.4
r = 0.4 / 0.8 = 1/2 = 0.5 ohm

(i) Internal resistance r = 0.5 ohm
(ii) ε = I(R+r) = 1.2(1+0.5) = 1.8 V
In simple words: The cell has internal resistance that reduces voltage. By comparing parallel and series connections, we can find both internal resistance and total voltage of the cell.

📝 Teacher's Note: Explain that every battery has internal resistance like a small resistor inside it. This is why battery voltage drops when we draw more current.

🎯 Exam Tip: Always write ε = I(R + r) formula first. Set up two equations for parallel and series cases, then solve. Show all steps clearly.

 

Question 26. A battery of e.m.f 15 V and internal resistance 3 Ω is connected to two resistors 3Ω and 6Ω connected in parallel. Find: (a) the current through the battery. (b) p.d. between the terminals of the battery, (c) the current in 3 Ω resistors, (d) the current in 6 Ω resistor.
Answer:
(a) In parallel 1/R = 1/3 + 1/6 = 1/2
So R = 2 ohm
r = 3 Ω
ε = 15 V
ε = I(R + r)
15 = I(2 + 3)
I = 15/5 = 3 A

(b) V = ?
R = 2 ohm
V = IR = 3 × 2 = 6 V

(c) V = 6 V
R = 3 ohm
I = V/R = 6/3 = 2 A

(d) R = 6 ohm
V = 6 V
I = V/R = 6/6 = 1 A
In simple words: First find total resistance of parallel resistors. Then use Ohm's law to find currents. The voltage across parallel resistors is same for both.

📝 Teacher's Note: Draw the circuit diagram. Show students that in parallel, voltage is same across each resistor but current divides. Total current equals sum of individual currents.

🎯 Exam Tip: Find equivalent resistance first, then battery current, then terminal voltage, then individual currents. Follow this order step by step for full marks.

 

Question 27. The following circuit diagram (Fig. 8.51) shows three resistors 2Ω, 4Ω and RΩ connected to a battery of e.m.f 2 V and internal resistance 3Ω. A main current of 0.25A flows through the circuit.
(a) What is the p.d. across the 4Ω resistor?
(b) Calculate the p.d. across the internal resistance of the cell.
(c) What is the p.d. across the RΩ or 2Ω resistor?
(d) calculate the value of R.

[Diagram: Circuit diagram showing 2Ω and RΩ resistors in parallel, connected in series with 4Ω resistor and battery with internal resistance 3Ω]

Answer:
(a) R = 4 Ω
I = 0.25 A
V = IR = 0.25 × 4 = 1 V

(b) Internal Resistance r = 3 ohm
I = 0.25 A
V = IR = 0.25 × 3 = 0.75 V

(c) Effective resistance of parallel combination of two 2 ohm resistances = 1 ohm
V = IR = 0.25 × 1 = 0.25 V

(d) I = 0.25 A
ε = 2V, r = 3 ohm
ε = I(R' + r)
2 = 0.25(R' + 3)
R' = 5 Ω
\( \frac{2R}{2 + R} + 4 = 5 \)
R = 2 ohm
In simple words: Use V = IR to find voltage across each resistor. The same current flows through series parts. Find total resistance first, then individual resistance.

📝 Teacher's Note: Explain that in series, current is same everywhere. In parallel, voltage is same. Use these rules to solve step by step.

🎯 Exam Tip: For mixed circuits, identify series and parallel parts first. Use V = IR for each part. Show the equation setup clearly for finding unknown resistance.

 

Question 28. Three resistors of 6.0 Ω, 2.0Ω and 4.0Ω are joined to an ammeter A and a cell of e.m.f. 6.0 V as shown in fig 8.52 Calculate:
(a) the effective resistance of the circuit and
(b) the reading of ammeter.

[Diagram: Circuit showing 6.0Ω resistor in parallel with series combination of 2.0Ω and 4.0Ω resistors, connected to 6.0V battery with ammeter]

Answer:
(a) R₁ = 6 Ω
R' = R₂ + R₃ = 2 + 4 = 6 Ω
R₁ and R' in parallel:
\( \frac{1}{R} = \frac{1}{R₁} + \frac{1}{R'} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} \)
R = 3 ohm

(b) R = 3 ohm
V = 6 V
I = ?
I = V/R = 6/3 = 2 A
In simple words: First add the series resistors (2Ω + 4Ω = 6Ω). Then find parallel combination with 6Ω resistor. Use Ohm's law to find current.

📝 Teacher's Note: Draw the circuit step by step. Show how series resistors add up, then how to find parallel combination. Make sure students see the difference between series and parallel.

🎯 Exam Tip: Always simplify series parts first, then parallel parts. Write each step clearly. Use I = V/R for final current calculation.

 

Question 29. The diagram below in Fig. 8.53 shows the arrangement of five different resistances connected to a battery of e.m.f. 1.8V Calculate:
(a) the total resistance of the circuit, and
(b) the reading of ammeter A.

[Diagram: Complex circuit with resistors R₁=10Ω, R₂=40Ω in parallel between points X-Y, then R₃=30Ω, R₄=20Ω, R₅=60Ω in parallel between points A-B, all connected to 1.8V battery with ammeter]

Answer:
(a) In the figure above,
Let resistance between X and Y be R_{xy}
Then, \( \frac{1}{R_{xy}} = \frac{1}{10} + \frac{1}{40} = \frac{4+1}{40} = \frac{5}{40} \)
Or, R_{xy} = 8Ω

Let R_{AB} be the net resistance between points A and B.
Then, \( \frac{1}{R_{AB}} = \frac{1}{30} + \frac{1}{20} + \frac{1}{60} = \frac{2+3+1}{60} = \frac{6}{60} \)
Or, R_{AB} = 10Ω
∴ Total resistance of the circuit = 8Ω + 10Ω = 18Ω

(b) Current, I = \( \frac{Voltage}{Total\ resistance} = \frac{1.8}{18} \) A
Or, I = 0.1 A
Thus, 0.1 A shall be the reading of the ammeter
In simple words: Find resistance of each parallel group separately. Then add them in series. Use V = IR to find total current through the ammeter.

📝 Teacher's Note: Break complex circuits into simpler parts. Solve parallel parts first, then treat them as single resistors in series. This makes difficult circuits easy to solve.

🎯 Exam Tip: Identify parallel groups first. Use 1/R = 1/R₁ + 1/R₂ + 1/R₃ for parallel. Then add series resistances normally. Show each step of simplification.