“MCQs Electricity Class 10 Chapter 11” enlists some of the most important MCQs from Chapter 11.

The MCQs below are a mix of both numerical and conceptual questions. These questions will test not just memory but also your analytical skills and the application of the concepts in real-life situations, and that is what CBSE demands now.

To get the best out of the MCQs, I suggest doing the following:

• Read the complete chapter from your NCERT textbook.
• Read our short notes for the same chapter.

Electric Current And Circuit

Electric Potential And Potential Difference

Ohm’s Law

Factors On Which The Resistance Of A Conductor Depends

Resistance Of A System Of Resistors

Resistors in Series

Resistors in Parallel

Advantages And Disadvantages
Of Series And Parallel Circuits

Heating Effect Of Electric Current

Practical Applications of Heating Effect of
Electric Current

Electric Power

I have listed some of the questions of the MCQs below,

MCQs Electricity Class 10 Chapter 11

The questions below are just for review and I suggest you to practice from the MCQs quiz itself to master the chapter.

Q1. A charge of 6 coulombs flows through a conductor in 2 seconds. What is the electric current?
A) 12 A
B) 3 A
C) 0.33 A
D) 6 A

Answer: B) 3 A
Using I = Q/t, I = 6/2 = 3 A. Current is the rate of flow of charge.

Q2. Why is the conventional direction of electric current taken as opposite to the direction of electron flow?
A) Electrons move faster than positive charges
B) Positive charges are heavier than electrons
C) Electrons were discovered later; the convention was already established before their discovery
D) Electric current only exists in metallic wires

Answer: C) Electrons were discovered later; the convention was already established before their discovery
When electricity was first studied, electrons were unknown. Current was assumed to flow from positive to negative terminal. This convention was retained even after electrons were discovered moving the other way.

Q3. Which of the following correctly defines one ampere?
A) One joule of energy per second
B) One watt of power per volt
C) One electron passing per microsecond
D) One coulomb of charge flowing per second

Answer: D) One coulomb of charge flowing per second
By definition, 1 A = 1 C/s. It is the flow of one coulomb of charge through a cross-section in one second.

Q4. In a metallic conductor, what actually constitutes the electric current?
A) Flow of protons from the nucleus
B) Flow of free electrons
C) Flow of positive ions
D) Simultaneous flow of both positive and negative charges

Answer: B) Flow of free electrons
In metallic conductors, positive ions remain fixed in the lattice. Only free (conduction) electrons move, and they constitute the current in metal wires.

Q5. A student connects an ammeter in parallel with a bulb. What will most likely happen?
A) The bulb glows brighter than normal
B) The reading will be correct but doubled
C) The ammeter may get damaged as most current bypasses the bulb through the low-resistance ammeter
D) Nothing changes; parallel or series doesn’t matter for an ammeter

Answer: C) The ammeter may get damaged as most current bypasses the bulb through the low-resistance ammeter
An ammeter has very low resistance. In parallel, it creates a near short-circuit — almost all current flows through it, potentially damaging it and the circuit. It must always be connected in series.

Q6. How many electrons approximately constitute 1 coulomb of charge?
A) 1.6 × 10⁻¹⁹ electrons
B) 6 × 10⁻¹⁹ electrons
C) 1.6 × 10¹⁸ electrons
D) 6 × 10¹⁸ electrons

Answer: D) 6 × 10¹⁸ electrons
Since each electron carries 1.6 × 10⁻¹⁹ C, the number of electrons in 1 C = 1 / (1.6 × 10⁻¹⁹) ≈ 6.25 × 10¹⁸ electrons.

Q7. If 3 × 10¹⁹ electrons pass through a cross-section in 5 seconds, what is the magnitude of the current?
A) 4.8 A
B) 0.48 A
C) 0.96 A
D) 9.6 A

Answer: C) 0.96 A
Total charge Q = 3 × 10¹⁹ × 1.6 × 10⁻¹⁹ = 4.8 C. Then I = Q/t = 4.8/5 = 0.96 A.

Q8. A current of 500 μA flows through a wire. What is this expressed in amperes?
A) 5 × 10⁻³ A
B) 5 × 10⁻⁶ A
C) 5 × 10⁻² A
D) 5 × 10⁻⁴ A

Answer: D) 5 × 10⁻⁴ A
1 μA = 10⁻⁶ A. So 500 μA = 500 × 10⁻⁶ = 5 × 10⁻⁴ A.

Q9. Which of the following correctly ranks the units from smallest to largest?
A) 1 mA < 1 μA < 1 A
B) 1 A < 1 mA < 1 μA
C) 1 μA < 1 mA < 1 A
D) 1 μA < 1 A < 1 mA

Answer: C) 1 μA < 1 mA < 1 A
1 μA = 10⁻⁶ A (smallest), 1 mA = 10⁻³ A (middle), 1 A (largest). The order is 1 μA < 1 mA < 1 A.

Q10. A current of 2 A flows for 30 seconds, then 4 A flows for 15 seconds. What is the total charge that has flowed?
A) 180 C
B) 90 C
C) 60 C
D) 120 C

Answer: D) 120 C
Q₁ = 2 × 30 = 60 C. Q₂ = 4 × 15 = 60 C. Total Q = 60 + 60 = 120 C.

Q11. A milliammeter reads 250 mA. How many coulombs of charge pass through the circuit in 4 minutes?
A) 0.25 C
B) 60 C
C) 6 C
D) 1000 C

Answer: B) 60 C
I = 0.25 A, t = 4 × 60 = 240 s. Q = I × t = 0.25 × 240 = 60 C.

Q12. Which of the following is NOT a correct statement about electric current?
A) Direction of conventional current is opposite to the direction of electron flow
B) Ammeter is always connected in series in a circuit
C) Current flows from the negative terminal to the positive terminal through the external circuit
D) The SI unit of current is the ampere

Answer: C) Current flows from the negative terminal to the positive terminal through the external circuit
Conventional current flows from the positive terminal to the negative terminal through the external circuit. Electrons flow in the opposite direction (negative to positive externally).

Q13. In an electric circuit, conventional current flows from positive to negative externally. This means electrons in the external circuit flow from:
A) Positive terminal to negative terminal
B) Negative terminal to positive terminal
C) Both terminals simultaneously toward the bulb
D) Inside the cell only, not through external circuit

Answer: B) Negative terminal to positive terminal
Since conventional current is defined as flow of positive charges (positive → negative externally), electrons (being negative) move in the opposite direction — from the negative terminal to the positive terminal through the external circuit.

Q14. Why does a torch bulb glow only when the switch is on?
A) The cell’s chemical energy suddenly increases when the switch is on
B) The resistance of the bulb decreases when the switch is on
C) The switch, when closed, provides a continuous and closed conducting path for current to flow
D) Electrons are produced inside the bulb when the switch is on

Answer: C) The switch, when closed, provides a continuous and closed conducting path for current to flow
A closed switch completes the circuit. Without a complete closed path, no current flows and the bulb does not glow. Opening the switch breaks the path and stops the current.

Q15. What does the negative sign in the charge of an electron (−1.6 × 10⁻¹⁹ C) physically signify in the context of electric current?
A) The electron has less energy than a proton
B) Electrons move slower than positive charges
C) The magnitude of electron charge is very small
D) Conventional current direction is opposite to the direction in which electrons actually move

Answer: D) Conventional current direction is opposite to the direction in which electrons actually move
The negative sign indicates electrons are negative charge carriers. Since conventional current is defined as flow of positive charge, its direction is always opposite to the actual direction of electron movement.

Q16. In which scenario is the electric circuit considered complete and functional?
A) A cell connected to a bulb with one wire only, switch open
B) A cell and bulb placed close to each other without connecting wires
C) A cell, bulb, ammeter, and switch all connected in a closed loop with switch closed
D) A bulb connected to a wire with no cell

Answer: C) A cell, bulb, ammeter, and switch all connected in a closed loop with switch closed
A complete circuit requires a closed continuous conducting path including a source of EMF, the load, and control elements all connected end-to-end with the switch closed.

Q17. Why is an ammeter always connected in series and never in parallel in a circuit?
A) To measure the potential difference accurately
B) To increase the overall resistance of the circuit
C) Because it has very high internal resistance
D) So that the same current flowing in the circuit also flows through the ammeter, giving an accurate reading

Answer: D) So that the same current flowing in the circuit also flows through the ammeter, giving an accurate reading
In series, the full circuit current passes through the ammeter. In parallel, the current would split, the low-resistance ammeter would short-circuit the load, and the reading would be meaningless.

Q18. A student claims: “The more charge flowing per second, the less the electric current.” Evaluate this claim.
A) Correct, because current is inversely proportional to charge
B) Incorrect, because I = Q/t shows current is directly proportional to charge flowing per unit time
C) Correct, because resistance increases with more charge flow
D) Incorrect, because charge and current are completely unrelated quantities

Answer: B) Incorrect, because I = Q/t shows current is directly proportional to charge flowing per unit time
By definition I = Q/t. More charge flowing per second means greater current. The student’s claim is the exact opposite of the correct relationship.

Q19. An electric current of 0.5 A flows through a circuit. How long will it take for 90 coulombs of charge to pass through a cross-section of the conductor?
A) 45 s
B) 180 s
C) 45 minutes
D) 0.005 s

Answer: B) 180 s
From I = Q/t, we get t = Q/I = 90/0.5 = 180 seconds.

Q20. Which of the following best explains why electric current in a circuit stops when the switch is turned off, even though the cell is still present?
A) The cell’s voltage drops to zero when the switch is off
B) The switch absorbs all the charge when it is open
C) Electrons lose their charge when the switch is open
D) The open switch breaks the continuous closed path, so no charge can flow even though the cell maintains its potential difference

Answer: D) The open switch breaks the continuous closed path, so no charge can flow even though the cell maintains its potential difference
The cell continues to maintain a potential difference even when the switch is open. However, current requires a completely closed conducting path. The open switch removes this path, so no current flows regardless of the cell’s condition.

Q1. What is the correct formula for electric potential difference?
A) V = Q/W
B) V = W × Q
C) V = W/Q
D) V = Q × t

Answer: C) V = W/Q
Potential difference is defined as the work done to move a unit charge from one point to another. So V = W/Q, where W is work done in joules and Q is charge in coulombs.

Q2. The SI unit of potential difference is named after:
A) Andre-Marie Ampere
B) Michael Faraday
C) Alessandro Volta
D) Charles Coulomb

Answer: C) Alessandro Volta
The unit “volt” is named after Alessandro Volta (1745–1827), an Italian physicist who made significant contributions to the study of electricity.

Q3. One volt is defined as:
A) One joule of work done to move one electron between two points
B) One coulomb of charge flowing per joule of work
C) One joule of work done to move one coulomb of charge between two points
D) One watt of power dissipated per ampere of current

Answer: C) One joule of work done to move one coulomb of charge between two points
By definition, 1 V = 1 J / 1 C. When 1 joule of work is done to move 1 coulomb of charge between two points, the potential difference between those points is 1 volt.

Q4. How is a voltmeter connected in a circuit to measure potential difference?
A) In series with the component across which potential difference is to be measured
B) In parallel across the points between which potential difference is to be measured
C) In series with the ammeter only
D) At any point in the circuit; its position does not matter

Answer: B) In parallel across the points between which potential difference is to be measured
A voltmeter is always connected in parallel across the two points, so that it measures the potential difference (voltage) across that component without significantly altering the circuit current.

Q5. What is the analogy used in the text to explain why charges flow in a conductor?
A) Flow of air from high pressure to low pressure regions
B) Flow of heat from hot to cold bodies
C) Flow of water due to a pressure difference between two ends of a tube
D) Flow of sand in an hourglass due to gravity

Answer: C) Flow of water due to a pressure difference between two ends of a tube
The text uses the analogy of water in a tube — water does not flow in a perfectly horizontal tube unless there is a pressure difference. Similarly, charges flow only when there is a potential difference along the conductor.

Q6. What happens when no potential difference exists across the ends of a conducting wire?
A) Current flows but in random directions
B) Electrons accelerate continuously
C) Charges do not flow, just as water does not flow in a perfectly horizontal tube
D) The wire gets heated due to static charges

Answer: C) Charges do not flow, just as water does not flow in a perfectly horizontal tube
Without a potential difference (electric pressure difference), there is no net force on the electrons, and no current flows — analogous to water remaining stationary in a level tube.

Q7. A charge of 4 coulombs is moved between two points and 20 joules of work is done. What is the potential difference between the two points?
A) 80 V
B) 0.2 V
C) 5 V
D) 16 V

Answer: C) 5 V
V = W/Q = 20/4 = 5 V. This directly applies the definition of potential difference.

Q8. What is the source of the potential difference produced by a cell?
A) The flow of electrons through the external circuit
B) The resistance of the connecting wires
C) The gravitational pull on the charge carriers inside the cell
D) The chemical action occurring within the cell

Answer: D) The chemical action occurring within the cell
Chemical reactions inside the cell generate the potential difference across its terminals. This potential difference exists even when no current is drawn from the cell.

Q9. If 60 joules of work is done to move a charge through a potential difference of 12 V, what is the magnitude of the charge moved?
A) 720 C
B) 48 C
C) 0.2 C
D) 5 C

Answer: D) 5 C
From V = W/Q, Q = W/V = 60/12 = 5 C.

Q10. What energy does a cell expend to maintain continuous current in a circuit?
A) Electrical energy stored in the wires
B) Kinetic energy of the electrons
C) Chemical energy stored in the cell
D) Nuclear energy from the cell’s electrodes

Answer: C) Chemical energy stored in the cell
To maintain a potential difference and hence a continuous current, the cell converts its stored chemical energy into electrical energy. This is why a cell eventually “runs out.”

Q11. A potential difference of 6 V is applied across a conductor. How much work is done in moving a charge of 3 coulombs from one end to the other?
A) 2 J
B) 0.5 J
C) 9 J
D) 18 J

Answer: D) 18 J
W = V × Q = 6 × 3 = 18 J. Work done equals potential difference multiplied by the charge moved.

Q12. Which of the following correctly expresses 1 volt in base SI units?
A) 1 V = 1 C J⁻¹
B) 1 V = 1 J A⁻¹
C) 1 V = 1 J C⁻¹
D) 1 V = 1 W C⁻¹

Answer: C) 1 V = 1 J C⁻¹
By definition, potential difference = Work / Charge = Joule / Coulomb. So 1 V = 1 J C⁻¹. This is the standard SI expression.

Q13. A cell maintains a potential difference across its terminals even when no current is drawn. What does this tell us about the nature of potential difference?
A) Potential difference requires current to exist
B) Potential difference is produced by the flow of electrons through the external circuit
C) Potential difference is an intrinsic property generated by chemical action, independent of current flow
D) Potential difference only exists when the circuit is complete

Answer: C) Potential difference is an intrinsic property generated by chemical action, independent of current flow
The text explicitly states that chemical action within a cell generates potential difference across the terminals even when no current is drawn. It is the cause of current, not its result.

Q14. Why do electrons in a copper wire not flow by themselves without a cell or battery?
A) Copper is an insulator that does not allow electron movement
B) Electrons in copper are tightly bound to the nucleus and cannot move
C) There is no potential difference to provide the electric pressure needed to drive the electrons
D) Gravity prevents electrons from moving in a horizontal wire

Answer: C) There is no potential difference to provide the electric pressure needed to drive the electrons
Just as water does not flow in a level tube without pressure difference, electrons do not have a net directed flow without a potential difference. Copper has free electrons, but they need an electric pressure (potential difference) to flow in a definite direction.

Q15. The potential difference between two points in a circuit is 9 V. How much work is done in moving a charge of 500 mC between the two points?
A) 18 J
B) 0.045 J
C) 4.5 J
D) 45 J

Answer: C) 4.5 J
500 mC = 0.5 C. W = V × Q = 9 × 0.5 = 4.5 J.

Q16. Which of the following best distinguishes a voltmeter connection from an ammeter connection in a circuit?
A) Voltmeter is in series; ammeter is in parallel
B) Both voltmeter and ammeter are connected in series
C) Voltmeter is in parallel; ammeter is in series
D) Both voltmeter and ammeter are connected in parallel

Answer: C) Voltmeter is in parallel; ammeter is in series
A voltmeter measures potential difference and is connected in parallel. An ammeter measures current and is connected in series. Interchanging these connections would give wrong readings and could damage the instruments.

Q17. Work done in moving a charge of 2 C between two points in a circuit is 10 J. Another circuit requires 30 J of work to move 3 C between two similar points. Which circuit has a higher potential difference and by how much?
A) Both are equal at 5 V
B) Second circuit is higher by 5 V
C) First circuit is higher by 5 V
D) First circuit is higher by 3 V

Answer: B) Second circuit is higher by 5 V
V₁ = 10/2 = 5 V. V₂ = 30/3 = 10 V. Wait — V₂ = 10 V, V₁ = 5 V, so the second circuit is higher by 5 V. Answer: B) Second circuit is higher by 5 V.

Q18. When a cell is connected to a conducting circuit element, what does the potential difference do?
A) It increases the mass of charge carriers
B) It reduces the resistance of the conductor
C) It sets the charges in motion in the conductor, producing electric current
D) It converts kinetic energy of electrons into chemical energy

Answer: C) It sets the charges in motion in the conductor, producing electric current
The potential difference acts as the driving force. When the cell is connected, the potential difference exerts an electric force on the free charges in the conductor, setting them in motion and producing a current.

Q19. A student measures potential difference using a voltmeter connected in series. What error will occur?
A) The voltmeter will measure current instead of voltage
B) The reading will be double the actual potential difference
C) The voltmeter, having high resistance in series, will drastically reduce the current and give incorrect readings for the circuit
D) No error occurs; both series and parallel work for a voltmeter

Answer: C) The voltmeter, having high resistance in series, will drastically reduce the current and give incorrect readings for the circuit
A voltmeter has very high resistance. In series, it would significantly reduce or nearly stop the current in the circuit, making it non-functional as a measuring device and altering the circuit’s behaviour.

Q20. Two charges Q₁ = 2 C and Q₂ = 5 C are moved between the same two points in a circuit. The work done for Q₁ is 14 J. How much work is done for Q₂, assuming the potential difference remains the same?
A) 35 J
B) 5.6 J
C) 70 J
D) 28 J

Answer: A) 35 J
V = W/Q = 14/2 = 7 V. For Q₂: W = V × Q = 7 × 5 = 35 J.

Q1. Ohm’s Law states that the potential difference across a metallic conductor is directly proportional to the current through it, provided:
A) The length of the conductor remains the same
B) The cross-sectional area remains constant
C) Its temperature remains the same
D) The voltage source remains unchanged

Answer: C) Its temperature remains the same
Ohm’s Law holds only when the temperature of the conductor is constant. If temperature changes, resistance changes, and the linear V–I relationship no longer holds strictly.

Q2. Who discovered the relationship between potential difference and current in a metallic wire, and in which year?
A) Alessandro Volta, 1800
B) Andre-Marie Ampere, 1820
C) Georg Simon Ohm, 1827
D) Michael Faraday, 1831

Answer: C) Georg Simon Ohm, 1827
Georg Simon Ohm (1787–1854), a German physicist, discovered this relationship in 1827. The unit of resistance — ohm — is named in his honour.

Q3. The V–I graph for a conductor obeying Ohm’s Law is:
A) A curve passing through the origin
B) A straight line parallel to the current axis
C) A straight line that passes through the origin
D) A parabola with the vertex at the origin

Answer: C) A straight line that passes through the origin
Since V ∝ I, the V–I graph is a straight line through the origin. The slope of this line gives the resistance R = V/I, which remains constant.

Q4. The SI unit of electrical resistance is ohm. One ohm is defined as:
A) The resistance when 1 coulomb of charge flows per second per volt
B) The resistance when a potential difference of 1 V causes a current of 1 A to flow
C) The resistance when 1 joule of work is done per ampere
D) The resistance when 1 watt of power is dissipated per volt

Answer: B) The resistance when a potential difference of 1 V causes a current of 1 A to flow
By definition, 1 Ω = 1 V / 1 A. When a 1 V potential difference produces exactly 1 A of current through a conductor, its resistance is 1 ohm.

Q5. A conductor has a resistance of 10 Ω. If the potential difference across it is 50 V, what is the current through it?
A) 500 A
B) 0.2 A
C) 5 A
D) 2 A

Answer: C) 5 A
Using I = V/R = 50/10 = 5 A. This is a direct application of Ohm’s Law rearranged to find current.

Q6. If the resistance of a conductor is doubled while the potential difference across it remains the same, what happens to the current?
A) It doubles
B) It remains the same
C) It becomes four times
D) It gets halved

Answer: D) It gets halved
From I = V/R, current is inversely proportional to resistance. If R doubles and V stays constant, I = V/2R, which is half the original current.

Q7. A current of 2 A flows through a conductor when a potential difference of 8 V is applied. What is the resistance of the conductor?
A) 16 Ω
B) 0.25 Ω
C) 10 Ω
D) 4 Ω

Answer: D) 4 Ω
R = V/I = 8/2 = 4 Ω. Resistance is the ratio of potential difference to current through the conductor.

Q8. What physical reason causes a conductor to offer resistance to the flow of electrons?
A) Electrons repel each other and slow down
B) The conductor absorbs kinetic energy from electrons directly
C) Electrons are restrained by the attraction of atoms within the conductor, which retards their motion
D) The potential difference forces electrons to slow down near the negative terminal

Answer: C) Electrons are restrained by the attraction of atoms within the conductor, which retards their motion
Electrons are not completely free — they interact with the positive ions of the conductor’s atomic lattice. These interactions retard the electron flow, giving rise to electrical resistance.

Q9. What is a rheostat used for in an electric circuit?
A) To measure the potential difference across a component
B) To increase the voltage supplied by the cell
C) To change the resistance in the circuit and thereby regulate the current without changing the voltage source
D) To convert alternating current to direct current

Answer: C) To change the resistance in the circuit and thereby regulate the current without changing the voltage source
A rheostat is a variable resistance device. By changing its resistance, the current in the circuit can be increased or decreased without altering the voltage source.

Q10. A potential difference of 6 V is applied across a resistor and the current is 0.5 A. If the potential difference is increased to 9 V (temperature unchanged), what is the new current?
A) 0.5 A
B) 1.5 A
C) 0.75 A
D) 1 A

Answer: C) 0.75 A
R = V/I = 6/0.5 = 12 Ω (constant, same temperature). New I = 9/12 = 0.75 A.

Q11. Which of the following correctly expresses 1 ohm in SI base units?
A) 1 Ω = 1 C A⁻¹
B) 1 Ω = 1 J C⁻²
C) 1 Ω = 1 V A⁻¹
D) 1 Ω = 1 W A⁻²

Answer: C) 1 Ω = 1 V A⁻¹
From R = V/I, the unit of resistance = volt/ampere = V A⁻¹. This is the definition of the ohm.

Q12. Which of the following lists correctly orders components by increasing resistance for the same size?
A) Insulator < Poor conductor < Resistor < Good conductor
B) Good conductor < Resistor < Poor conductor < Insulator
C) Resistor < Good conductor < Poor conductor < Insulator
D) Poor conductor < Good conductor < Insulator < Resistor

Answer: B) Good conductor < Resistor < Poor conductor < Insulator
Good conductors offer low resistance; resistors offer appreciable resistance; poor conductors offer higher resistance; insulators offer the highest resistance — all for the same physical size.

Q13. The V/I ratio for a given metallic wire at constant temperature gives:
A) The power dissipated in the wire
B) The charge flowing through the wire
C) A variable value that depends on the applied voltage
D) A constant value equal to the resistance of the wire

Answer: D) A constant value equal to the resistance of the wire
According to Ohm’s Law, V/I = R = constant for a given metallic wire at constant temperature. This constant ratio is the resistance of the conductor.

Q14. Two conductors A and B are tested. For conductor A, V/I = 5 Ω. For conductor B, V/I = 20 Ω. Which conductor allows easier flow of current and why?
A) B, because higher resistance means more current flows
B) Both allow the same current since voltage is the same
C) A, because lower resistance means less opposition to current flow
D) A, because higher resistance means more current flows

Answer: C) A, because lower resistance means less opposition to current flow
From I = V/R, lower resistance means higher current for the same voltage. Conductor A with R = 5 Ω offers less opposition than conductor B with R = 20 Ω.

Q15. A student plots a V–I graph for two resistors X and Y. The graph for X has a steeper slope than Y. What can be concluded?
A) X has lower resistance than Y
B) X has the same resistance as Y
C) X has higher resistance than Y
D) X conducts more current than Y for the same voltage

Answer: C) X has higher resistance than Y
In a V–I graph, slope = V/I = R. A steeper slope means a higher value of V/I, which means higher resistance. So resistor X has greater resistance than Y.

Q16. A resistor carries a current of 3 A when connected to a 12 V supply. What potential difference would be needed to drive a current of 5 A through the same resistor at the same temperature?
A) 20 V
B) 7.2 V
C) 60 V
D) 15 V

Answer: A) 20 V
R = V/I = 12/3 = 4 Ω. For 5 A: V = IR = 5 × 4 = 20 V.

Q17. What is the fundamental difference between a good conductor and an insulator in terms of resistance?
A) Good conductors have zero resistance; insulators have infinite resistance
B) Good conductors offer low resistance to current flow; insulators of the same size offer very high resistance
C) Good conductors have high resistance at low temperatures only
D) Insulators allow current but slow it down; conductors stop current completely

Answer: B) Good conductors offer low resistance to current flow; insulators of the same size offer very high resistance
The text clearly states that a component offering low resistance is a good conductor, while an insulator of the same size offers even higher resistance than a poor conductor. The difference is one of degree, not absolute zero or infinity in practical terms.

Q18. The resistance of a conductor is a property that:
A) Depends only on the potential difference applied
B) Changes with the amount of current flowing through it
C) Resists the flow of charges through the conductor
D) Helps charges to flow more easily through the conductor

Answer: C) Resists the flow of charges through the conductor
Resistance is defined as the property of a conductor to resist the flow of charges through it. It is an intrinsic property of the material and its physical dimensions (at a given temperature).

Q19. A variable resistance (rheostat) in a circuit is increased from 4 Ω to 12 Ω. If the supply voltage is 24 V throughout, by how much does the current change?
A) Current increases by 4 A
B) Current decreases by 4 A
C) Current decreases by 2 A
D) Current increases by 2 A

Answer: B) Current decreases by 4 A
Initial I = 24/4 = 6 A. Final I = 24/12 = 2 A. Change = 6 − 2 = 4 A decrease.

Q20. According to Ohm’s Law, if the potential difference across a conductor is tripled while its temperature and resistance remain unchanged, what happens to the current?
A) It becomes one-third
B) It doubles
C) It triples
D) It remains unchanged

Answer: C) It triples
From I = V/R, with R constant, current is directly proportional to voltage. If V is tripled, I is also tripled.

Q1. Which of the following factors does the resistance of a uniform metallic conductor depend on?
A) Colour and density of the material only
B) Length, area of cross-section, and nature of material
C) Length and temperature only
D) Mass and volume of the conductor

Answer: B) Length, area of cross-section, and nature of material
The resistance of a conductor depends on its length, cross-sectional area, and the nature (resistivity) of the material it is made of. These are the three primary factors stated in the text.

Q2. What happens to the resistance of a conductor when its length is doubled?
A) It becomes four times
B) It remains the same
C) It gets halved
D) It doubles

Answer: D) It doubles
Since R ∝ l, doubling the length doubles the resistance. This is why the ammeter reading (current) decreases to one-half when the wire length is doubled — more resistance means less current.

Q3. What is the SI unit of electrical resistivity?
A) Ω A⁻¹
B) Ω m
C) Ω m⁻¹
D) Ω cm

Answer: B) Ω m
The SI unit of resistivity (ρ) is ohm-metre (Ω m). It is a characteristic property of the material of the conductor, independent of its shape or size.

Q4. The formula for resistance of a uniform conductor in terms of resistivity is:
A) R = ρA/l
B) R = ρ/lA
C) R = ρl/A
D) R = l/ρA

Answer: C) R = ρl/A
The resistance R = ρl/A, where ρ is the resistivity, l is the length, and A is the area of cross-section. This combines both the direct proportionality with length and inverse proportionality with area.

Q5. Why are alloys commonly used in electrical heating devices rather than pure metals?
A) Alloys have very low resistivity and heat up slowly
B) Alloys have lower melting points than pure metals
C) Alloys have higher resistivity than pure metals and do not oxidise readily at high temperatures
D) Alloys are cheaper and more flexible than pure metals

Answer: C) Alloys have higher resistivity than pure metals and do not oxidise readily at high temperatures
Alloys generally have higher resistivity than their constituent metals, generating more heat. They also resist oxidation at high temperatures, making them ideal for electric irons, toasters, and other heating appliances.

Q6. Which material is used almost exclusively for filaments of electric bulbs, and why?
A) Copper, because it has very low resistance
B) Aluminium, because it is lightweight
C) Nichrome, because it is an alloy with high resistivity
D) Tungsten, because of its very high melting point and suitability for high-temperature operation

Answer: D) Tungsten, because of its very high melting point and suitability for high-temperature operation
Tungsten is used for bulb filaments because it can withstand the extremely high temperatures needed to produce light without melting. Its high melting point makes it uniquely suitable.

Q7. Copper and aluminium are used for electrical transmission lines because:
A) They have high resistivity and reduce current loss
B) They are insulators with very low resistance
C) They are good conductors with very low resistivity, minimising energy loss during transmission
D) They have high resistivity in the range of 10¹² to 10¹⁷ Ω m

Answer: C) They are good conductors with very low resistivity, minimising energy loss during transmission
Copper and aluminium have very low resistivity (in the range of 10⁻⁸ to 10⁻⁶ Ω m), meaning they offer very little resistance and minimise energy loss as heat during electrical transmission over long distances.

Q8. A wire has resistance R. If its length is halved and its cross-sectional area is also halved, what is the new resistance?
A) R/2
B) 2R
C) R/4
D) R

Answer: D) R
R = ρl/A. New R = ρ(l/2)/(A/2) = ρl/A = R. Both changes cancel each other out exactly, leaving the resistance unchanged.

Q9. The resistivity of metals and alloys ranges from 10⁻⁸ Ω m to 10⁻⁶ Ω m, while insulators have resistivity of 10¹² to 10¹⁷ Ω m. What does this difference tell us?
A) Insulators carry current at very high voltages
B) Metals have more electrons than insulators per atom
C) The difference in resistivity spans roughly 18 to 25 orders of magnitude, explaining why insulators essentially block current flow
D) Resistivity is not a reliable measure of conducting ability

Answer: C) The difference in resistivity spans roughly 18 to 25 orders of magnitude, explaining why insulators essentially block current flow
The ratio of resistivities between insulators and metals is approximately 10¹⁷/10⁻⁸ = 10²⁵ — an enormous difference. This is why even a thin layer of insulator can completely prevent current flow that would easily pass through a metal wire.

Q10. A conductor of resistivity ρ, length l, and cross-sectional area A carries a current I under potential difference V. If the same material is drawn out to double its length (volume remains constant), what is the new resistance in terms of the original resistance R?
A) R/2
B) 4R
C) 2R
D) R/4

Answer: B) 4R
When length doubles (l → 2l) at constant volume, area halves (A → A/2). New R = ρ(2l)/(A/2) = 4ρl/A = 4R. Both the increase in length and decrease in area contribute to a fourfold increase in resistance.

Q11. Two wires X and Y are made of the same material. Wire X has twice the length and twice the diameter of wire Y. What is the ratio of resistance of X to resistance of Y?
A) 1 : 1
B) 2 : 1
C) 1 : 2
D) 4 : 1

Answer: C) 1 : 2
Area ∝ diameter². So A_X = 4A_Y. R_X = ρ(2l)/(4A) = ρl/2A. R_Y = ρl/A. Ratio R_X/R_Y = (ρl/2A)/(ρl/A) = 1/2, so R_X : R_Y = 1 : 2.

Q12. A student claims that resistivity is just another name for resistance. Identify the fundamental flaw in this claim.
A) Resistivity is measured in ohms, while resistance is measured in volts
B) Resistivity is a material property independent of shape and size, whereas resistance depends on the dimensions of the specific conductor
C) Resistivity applies only to insulators, while resistance applies to conductors
D) There is no flaw; resistivity and resistance are the same quantity

Answer: B) Resistivity is a material property independent of shape and size, whereas resistance depends on the dimensions of the specific conductor
Resistivity (ρ) is an intrinsic property of the material — it is the same for all samples of copper regardless of their shape. Resistance (R) depends on both the material’s resistivity and its physical dimensions (length and cross-sectional area). Two copper wires of different sizes have the same resistivity but different resistances.

Q13. Three wires P, Q, and R are made of the same material and have the same volume. Their lengths are in the ratio 1 : 2 : 3. What is the ratio of their resistances?
A) 1 : 4 : 9
B) 1 : 2 : 3
C) 3 : 2 : 1
D) 9 : 4 : 1

Answer: A) 1 : 4 : 9
At constant volume V = Al, so A = V/l. R = ρl/A = ρl/(V/l) = ρl²/V. Since V and ρ are constant, R ∝ l². For lengths in ratio 1 : 2 : 3, resistances are in ratio 1² : 2² : 3² = 1 : 4 : 9.

Q14. A wire of resistance 12 Ω is stretched uniformly to three times its original length. What is the new resistance? (Assume volume remains constant.)
A) 36 Ω
B) 4 Ω
C) 108 Ω
D) 144 Ω

Answer: C) 108 Ω
Volume = Al is constant. When l becomes 3l, A becomes A/3. New R = ρ(3l)/(A/3) = 9ρl/A = 9R = 9 × 12 = 108 Ω. Stretching to n times original length increases resistance by n² times.

Q15. The resistance of a conductor is found to be 5 Ω at a certain temperature. At a higher temperature it is found to be 6 Ω for the same conductor. What does this tell us about the relationship between resistance and temperature?
A) Resistance decreases with increase in temperature for all materials
B) Temperature has no effect on resistance; the change must be due to a change in length
C) Both resistance and resistivity vary with temperature, and for metals, resistance generally increases with temperature
D) Only resistivity changes with temperature; resistance remains constant

Answer: C) Both resistance and resistivity vary with temperature, and for metals, resistance generally increases with temperature
The text explicitly states that both resistance and resistivity vary with temperature. For metals, as temperature rises, atomic vibrations increase, causing more collisions with electrons and hence greater resistance.

Q16. A nichrome wire and a copper wire have the same length and cross-sectional area. Given that nichrome has much higher resistivity than copper, what happens if both are connected to the same potential difference?
A) Both carry the same current since their dimensions are equal
B) Nichrome carries more current due to its higher resistivity
C) Copper carries more current because its lower resistivity means lower resistance
D) Nichrome carries more current because higher resistivity means lower resistance

Answer: C) Copper carries more current because its lower resistivity means lower resistance
For the same dimensions, R = ρl/A means higher resistivity → higher resistance. By I = V/R, higher resistance → lower current. So nichrome, with higher resistivity, carries less current than copper for the same applied voltage.

Q17. A wire of resistance R is cut into 5 equal pieces. These pieces are then connected in parallel. What is the resistance of the parallel combination?
A) R/5
B) R/25
C) 5R
D) 25R

Answer: B) R/25
Each piece has resistance R/5. For 5 equal resistors in parallel, the combined resistance = (R/5)/5 = R/25. The resistance drops by a factor of n² (here 25) when a wire is cut into n pieces and reconnected in parallel.

Q18. Two materials A and B have resistivities of 2 × 10⁻⁶ Ω m and 8 × 10⁻⁶ Ω m respectively. Wires of equal length are made from each material. To make both wires have identical resistance, the cross-sectional area of wire B must be how many times that of wire A?
A) 4 times
B) 1/4 times
C) 2 times
D) 16 times

Answer: A) 4 times
For equal R: ρ_A × l/A_A = ρ_B × l/A_B → A_B/A_A = ρ_B/ρ_A = 8×10⁻⁶ / 2×10⁻⁶ = 4. Wire B must have 4 times the cross-sectional area of wire A to compensate for its higher resistivity.

Q19. A student has two wires of the same material — wire 1 has length l and radius r, wire 2 has length 2l and radius 2r. The student claims both wires have the same resistance. Is the student correct?
A) No, wire 2 has twice the resistance of wire 1
B) No, wire 2 has half the resistance of wire 1
C) Yes, because doubling length and doubling radius cancel out
D) No, wire 2 has four times the resistance of wire 1

Answer: B) No, wire 2 has half the resistance of wire 1
R = ρl/A = ρl/πr². R₁ = ρl/πr². R₂ = ρ(2l)/π(2r)² = 2ρl/4πr² = ρl/2πr² = R₁/2. Since area grows as r², doubling radius quadruples the area, which more than compensates for the doubled length, resulting in half the original resistance.

Q20. A conducting wire has resistivity ρ, length l, and circular cross-section of radius r. It carries current I under voltage V. Without changing the material or the voltage V, an engineer wants to reduce the current to I/4. Which single change achieves this?
A) Double the length of the wire only
B) Halve the radius of the wire only
C) Double the length and halve the radius simultaneously
D) Either doubling the length only, or halving the radius only — both individually achieve the same result

Answer: B) Halve the radius of the wire only
Original R = ρl/πr². To get I/4, we need R to become 4R (since I = V/R). If radius is halved: new A = π(r/2)² = πr²/4. New R = ρl/(πr²/4) = 4ρl/πr² = 4R. ✓ If length is doubled: new R = ρ(2l)/πr² = 2R only — current becomes I/2, not I/4. So only halving the radius achieves the required fourfold increase in resistance.

Q1. Three resistors of 4 Ω, 6 Ω, and 10 Ω are connected in series to a battery of 20 V. Find (a) the equivalent resistance, (b) the current through the circuit, and (c) the potential difference across the 6 Ω resistor.

Answer:
(a) Rs = R1 + R2 + R3 = 4 + 6 + 10 = 20 Ω
(b) I = V/Rs = 20/20 = 1 A
(c) V₂ = I × R2 = 1 × 6 = 6 V

Explanation: In series, the same current (1 A) flows through every resistor. The potential difference across each resistor is found individually using Ohm’s Law (V = IR). Notice that the sum of individual potential differences = 6 + 4 + 10 = 20 V, which equals the supply voltage — confirming V = V1 + V2 + V3.

Q2. Two resistors are connected in series to a 12 V battery. The potential difference across the first resistor is 8 V and the current through the circuit is 2 A. Find the resistance of each resistor.

Answer:
V1 = 8 V, so V2 = 12 − 8 = 4 V
R1 = V1/I = 8/2 = 4 Ω
R2 = V2/I = 4/2 = 2 Ω

Explanation: Since the resistors are in series, the same current (2 A) flows through both. The total voltage is split between them. Using V = IR individually for each resistor gives the respective resistances. Check: Rs = 4 + 2 = 6 Ω, and V = IRs = 2 × 6 = 12 V ✓

Q3. Four identical resistors each of resistance 8 Ω are connected in series to a battery. An ammeter reads 0.5 A. Find (a) the terminal voltage of the battery, (b) the potential difference across each resistor, and (c) what ammeter reading would be obtained if one of the four resistors is removed from the series circuit (voltage unchanged).

Answer:
(a) Rs = 4 × 8 = 32 Ω. V = IRs = 0.5 × 32 = 16 V
(b) V per resistor = I × R = 0.5 × 8 = 4 V
(c) New Rs = 3 × 8 = 24 Ω. New I = 16/24 = 0.67 A

Explanation: Removing one resistor decreases the total series resistance, which increases the current for the same voltage. This illustrates that in a series circuit, adding more resistors reduces current and removing them increases it.

Q4. Three resistors R1, R2, and R3 are in series. It is given that R2 = 2R1 and R3 = 3R1. The total power dissipated in the circuit is 36 W when a current of 2 A flows. Find the values of R1, R2, and R3.

Answer:
Total power P = I²Rs → 36 = (2)² × Rs → Rs = 36/4 = 9 Ω
Rs = R1 + 2R1 + 3R1 = 6R1 = 9 Ω → R1 = 1.5 Ω
R2 = 3 Ω, R3 = 4.5 Ω

Explanation: Using P = I²R to find total resistance, then expressing all resistances in terms of R1 using the given ratios.

Q5. A series circuit consists of two resistors. When a 24 V battery is connected, the current is 3 A. When an additional unknown resistor Rx is added in series, the current drops to 2 A. Find the value of Rx.

Answer:
Original Rs = V/I = 24/3 = 8 Ω
New Rs = V/I = 24/2 = 12 Ω
Rx = New Rs − Original Rs = 12 − 8 = 4 Ω

Explanation: Adding a resistor in series increases total resistance. Since voltage remains the same, the drop in current reflects the increase in resistance.

Q6. In a series circuit, it is often said that “the current has no choice.” If one bulb in a series string of decorative lights fuses and creates an open circuit, which of the following correctly explains why all other bulbs also go out?
A) The fused bulb absorbs all the remaining voltage, leaving zero voltage for other bulbs
B) The battery’s chemical energy is completely used up by the fused bulb, leaving no energy for others
C) The open circuit at the fused bulb breaks the only conducting path available, so no current can flow anywhere in the circuit regardless of how intact the other bulbs are
D) The resistance of the fused bulb becomes zero, causing a short circuit that prevents current from reaching other bulbs

Answer: C) The open circuit at the fused bulb breaks the only conducting path available, so no current can flow anywhere in the circuit regardless of how intact the other bulbs are
In a series circuit, there is only one conducting path. Every electron must pass through every element. When a filament breaks, it creates a gap — equivalent to infinite resistance. No current flows anywhere.

Q7. A student argues: “If I keep adding more resistors in series, the equivalent resistance keeps increasing — so at some point the current will become exactly zero.” Which of the following most precisely evaluates this argument?
A) The argument is completely correct
B) The argument is correct in direction but flawed in conclusion
C) The argument is wrong because adding resistors decreases resistance
D) The argument is partially correct

Answer: B) The argument is correct in direction but flawed in conclusion
Since I = V/R, current approaches zero as resistance increases but never becomes exactly zero unless resistance is infinite (open circuit).

Q8. Three resistors R1 < R2 < R3 are connected in series to a battery. Which resistor has the largest voltage drop?
A) R1
B) R2
C) R3
D) All equal

Answer: C) R3
In series, current is same. Since V = IR, voltage drop is proportional to resistance. Largest resistance gets largest voltage.

Q9. A student replaces one resistor in a series circuit with a wire of zero resistance. What happens to the voltage distribution?
A) Voltage law fails
B) Voltage disappears
C) The wire has zero voltage drop, others adjust
D) Battery voltage changes

Answer: C) The wire has zero voltage drop, others adjust
Voltage law still holds. The wire contributes 0 V, remaining resistors share total voltage.

Q10. A battery is connected to R1 and R2 in series. A resistor R3 is connected in parallel with R2. What happens to the voltage across R2?
A) Increases
B) Decreases
C) Remains the same
D) Becomes zero

Answer: B) Decreases
Parallel combination reduces resistance → increases current → more drop across R1 → less voltage left for R2.

Q1. Three resistors of 4 Ω, 6 Ω, and 10 Ω are connected in series to a battery of 20 V. Find (a) the equivalent resistance, (b) the current through the circuit, and (c) the potential difference across the 6 Ω resistor.

Answer:
(a) Rs = R1 + R2 + R3 = 4 + 6 + 10 = 20 Ω
(b) I = V/Rs = 20/20 = 1 A
(c) V₂ = I × R2 = 1 × 6 = 6 V

Explanation: In series, the same current (1 A) flows through every resistor. The potential difference across each resistor is found individually using Ohm’s Law (V = IR). Notice that the sum of individual potential differences = 6 + 4 + 10 = 20 V, which equals the supply voltage — confirming V = V1 + V2 + V3.

Q2. Two resistors are connected in series to a 12 V battery. The potential difference across the first resistor is 8 V and the current through the circuit is 2 A. Find the resistance of each resistor.

Answer:
V1 = 8 V, so V2 = 12 − 8 = 4 V
R1 = V1/I = 8/2 = 4 Ω
R2 = V2/I = 4/2 = 2 Ω

Explanation: Since the resistors are in series, the same current (2 A) flows through both. The total voltage is split between them. Using V = IR individually for each resistor gives the respective resistances. Check: Rs = 4 + 2 = 6 Ω, and V = IRs = 2 × 6 = 12 V ✓

Q3. Four identical resistors each of resistance 8 Ω are connected in series to a battery. An ammeter reads 0.5 A. Find (a) the terminal voltage of the battery, (b) the potential difference across each resistor, and (c) what ammeter reading would be obtained if one of the four resistors is removed from the series circuit (voltage unchanged).

Answer:
(a) Rs = 4 × 8 = 32 Ω. V = IRs = 0.5 × 32 = 16 V
(b) V per resistor = I × R = 0.5 × 8 = 4 V
(c) New Rs = 3 × 8 = 24 Ω. New I = 16/24 = 0.67 A

Explanation: Removing one resistor decreases the total series resistance, which increases the current for the same voltage.

Q4. Three resistors R1, R2, and R3 are in series. It is given that R2 = 2R1 and R3 = 3R1. The total power dissipated in the circuit is 36 W when a current of 2 A flows. Find the values of R1, R2, and R3.

Answer:
Total power P = I²Rs → 36 = (2)² × Rs → Rs = 36/4 = 9 Ω
Rs = R1 + 2R1 + 3R1 = 6R1 = 9 Ω → R1 = 1.5 Ω
R2 = 3 Ω, R3 = 4.5 Ω

Explanation: Using P = I²R to find total resistance, then expressing all resistances in terms of R1.

Q5. A series circuit consists of two resistors. When a 24 V battery is connected, the current is 3 A. When an additional unknown resistor Rx is added in series, the current drops to 2 A. Find the value of Rx.

Answer:
Original Rs = V/I = 24/3 = 8 Ω
New Rs = V/I = 24/2 = 12 Ω
Rx = New Rs − Original Rs = 12 − 8 = 4 Ω

Explanation: Adding resistance in series reduces current.

Q6. In a series circuit, it is often said that “the current has no choice.” If one bulb fuses, why do all bulbs go out?

A) Voltage absorbed
B) Energy used
C) Open circuit breaks path
D) Short circuit

Answer: C) Open circuit breaks path

Explanation: Series has one path → break stops all current.

Q7. Adding many resistors in series makes current zero?

A) Yes
B) Approaches zero but never zero
C) Wrong
D) Depends

Answer: B) Approaches zero but never zero

Explanation: Only infinite resistance gives zero current.

Q8. In series R1 < R2 < R3, which has largest voltage?

A) R1
B) R2
C) R3
D) Equal

Answer: C) R3

Explanation: V ∝ R in series.

Q9. Replacing resistor with wire (0 Ω) affects voltage rule?

A) Breaks rule
B) Still holds
C) Depends
D) Voltage reduces

Answer: B) Still holds

Explanation: Wire has 0 V drop, rest adjust.

Q10. Adding parallel resistor across R2 changes its voltage because:

A) Resistance increases
B) Voltage drops
C) Current increases in R1
D) No change

Answer: C) Current increases in R1