CBSE Class 10 Science · 2026–27

Electricity Competency Based Questions

Apply Electricity concepts through 43 original questions covering MCQs, assertion–reason, short answers, long answers and case studies. Attempt each question before opening its answer and worked solution.

Electric current and chargePotential difference and metersResistance and resistivityCombination of resistorsHeating effect and electric powerHousehold circuits and safety

Practise with question filters

Question 1 · MCQ · Apply · 1 mark

Electron flow from current

Electric current and charge

At a point in a wire, current is

0.80

A. Approximately how many electrons pass per second?

(A)

5 \times 1 0^{17}

(B)

5 \times 1 0^{18}

(C)

8 \times 1 0^{18}

(D)

1.28 \times 1 0^{20}

Show answer and solution

Answer: (B)

5 \times 1 0^{18}

Worked solution: Current

I = Q / t

. The number of electrons per second is

n = I / e = 0.80/ (1.6 \times 1 0^{- 19}) = 5 \times 1 0^{18}

. Hence option (B) is correct.

Question 2 · MCQ · Apply · 1 mark

Charge and electrons

Electric current and charge

A charge of

3.2

C passes through a wire in

8

s. The number of electrons is:

(A)

2 \times 1 0^{18}

(B)

2 \times 1 0^{19}

(C)

5 \times 1 0^{19}

(D)

1.6 \times 1 0^{20}

Show answer and solution

Answer: (B)

2 \times 1 0^{19}

Worked solution: Use

n = Q / e

. Thus

n = 3.2/ (1.6 \times 1 0^{- 19}) = 2 \times 1 0^{19}

electrons. Hence option (B) is correct.

Question 3 · MCQ · Understand · 1 mark

Electrical energy

Potential difference and meters

Which expression gives energy transferred to charge

Q

through potential difference

V

?

(A)

V / Q

(B)

V Q

(C)

Q / V

(D)

Q / t

Show answer and solution

Answer: (B)

V Q

Worked solution: Potential difference is

V = W / Q

. Therefore

W = V Q

. Hence option (B) is correct.

Question 4 · MCQ · Apply · 1 mark

Meter connections

Potential difference and meters

A student measures current through a lamp and voltage across it. Which connection is correct?

(A) Both meters in series
(B) Ammeter parallel, voltmeter series
(C) Ammeter series, voltmeter parallel
(D) Both parallel

Circuit diagram showing an ammeter in series and a voltmeter across a resistor. — Ammeter in series and voltmeter across the resistor.

Show answer and solution

Answer: (C) Ammeter series, voltmeter parallel

Worked solution: An ammeter measures current through the component, so it is in series. A voltmeter compares the two terminals of the lamp, so it is in parallel. Hence option (C) is correct.

Question 5 · MCQ · Apply · 1 mark

Voltmeter position

Potential difference and meters

6

V cell drives a

3Ω

resistor. Where should the voltmeter be connected?

(A) Across the resistor
(B) In series ahead of the cell
(C) In place of the resistor
(D) Only across the ammeter

Show answer and solution

Answer: (A) Across the resistor

Worked solution: A voltmeter must be connected across the component whose potential difference is required. Hence option (A) is correct.

Question 6 · MCQ · Analyse · 1 mark

I-V graph slope

Resistance and resistivity

On an

I

versus

V

plot, line A has slope

0.5

A/V and line B has slope

0.25

A/V. Which resistance is larger?

(A) A is twice B
(B) B is twice A
(C) They are equal
(D) Cannot be determined

I versus V graph with two straight lines of different slopes. — The steeper I-V line has lower resistance.

Show answer and solution

Answer: (B) B is twice A

Worked solution: For an

I

-versus-

V

graph, slope

I / V = 1/ R

. Thus

R_{A} = 2Ω

and

R_{B} = 4Ω

. B has twice the resistance of A.

Question 7 · MCQ · Apply · 1 mark

Wire diameter and resistance

Resistance and resistivity

A wire of given material and length is replaced by one with twice the diameter. Its new resistance is:

(A)

R /4

(B)

R /2

(C)

2 R

(D)

4 R

Show answer and solution

Answer: (A)

R /4

Worked solution: Area is proportional to diameter squared. Doubling diameter makes area

4 A

, while

R = ρ L / A

. Therefore the new resistance is

R /4

Question 8 · MCQ · Apply · 1 mark

Equal resistance wires

Resistance and resistivity

Wire B is three times as long as wire A but has the same resistance and material. What is

A_{B} : A_{A}

?

(A)

1 : 3

(B)

1 : 1

(C)

3 : 1

(D)

9 : 1

Show answer and solution

Answer: (C)

3 : 1

Worked solution: Since

R = ρ L / A

, keeping

R

same while length becomes

3 L

requires area

3 A

. Hence

A_{B} : A_{A} = 3 : 1

Question 9 · MCQ · Analyse · 1 mark

Least resistance

Resistance and resistivity

For equal-material wires, which has the least resistance?

(A) Length

L

, diameter

d

(B) Length

L

, diameter

2 d

2 L

, diameter

2 d

(D) Length

L /2

, diameter

d

Show answer and solution

Answer: (B) Length

L

, diameter

2 d

Worked solution: Resistance varies as

L / d^{2}

. The relative values are

1

1/4

1/2

, and

1/2

. Option (B) is least.

Question 10 · MCQ · Apply · 1 mark

Cross-sectional area

Resistance and resistivity

Wire B has length

4 L

and resistance

2 R

compared with wire A of length

L

, resistance

R

and area

A

. Its cross-section is:

(A)

A /2

(B)

A

(C)

2 A

(D)

4 A

Show answer and solution

Answer: (C)

2 A

Worked solution:

2 R = ρ (4 L) / A_{B}

while

R = ρ L / A

. Comparing gives

A_{B} = 2 A

. Hence option (C) is correct.

Question 11 · MCQ · Understand · 1 mark

Temperature effect

Resistance and resistivity

At approximately constant geometry, a metallic wire is heated. Its resistance generally:

(A) Falls
(B) Rises
(C) Becomes zero
(D) Must remain identical

Show answer and solution

Answer: (B) Rises

Worked solution: Metals generally have a positive temperature coefficient of resistance. Hence option (B) is correct.

Question 12 · MCQ · Understand · 1 mark

Resistivity as a material property

Resistance and resistivity

At fixed material and temperature, a wire length doubles. What happens to resistivity

ρ

?

(A) Halves
(B) Doubles
(C) Quadruples
(D) Remains unchanged

Show answer and solution

Answer: (D) Remains unchanged

Worked solution: Resistivity is a material property at a fixed temperature. Resistance changes with length, but

ρ

does not. Hence option (D) is correct.

Question 13 · MCQ · Apply · 1 mark

Series current

Combination of resistors

4Ω

lamp and a

6Ω

resistor are in series. Current through the resistor is

0.30

A. Current through the lamp is:

(A)

0.18

A
(B)

0.30

A
(C)

0.50

A
(D)

0.75

Show answer and solution

Answer: (B)

0.30

Worked solution: There is one unbranched series path, so the same current flows through each component. Hence option (B) is correct.

Question 14 · MCQ · Apply · 1 mark

Maximum equivalent resistance

Combination of resistors

Five separate

2Ω

resistors must all be used. What is the largest achievable equivalent resistance?

(A)

0.4Ω

(B)

2Ω

(C)

5Ω

(D)

10Ω

Show answer and solution

Answer: (D)

10Ω

Worked solution: The largest equivalent resistance is obtained in series:

5 \times 2 = 10Ω

Question 15 · MCQ · Apply · 1 mark

Parallel branches

Combination of resistors

Three branches of

3Ω

6Ω

, and

6Ω

connect the same terminals X and Y. Their equivalent resistance is:

(A)

1.5Ω

(B)

2Ω

(C)

3Ω

(D)

15Ω

Three parallel resistor branches connected between terminals X and Y. — Each branch has the same terminal potential difference.

Show answer and solution

Answer: (A)

1.5Ω

Worked solution:

1/ R = 1/3 + 1/6 + 1/6 = 2/3

. Therefore

R = 1.5Ω

. Hence option (A) is correct.

Question 16 · MCQ · Apply · 1 mark

Four resistors in series

Combination of resistors

What is the maximum equivalent resistance using four

3Ω

resistors, all included?

(A)

3/4Ω

(B)

3Ω

(C)

9Ω

(D)

12Ω

Show answer and solution

Answer: (D)

12Ω

Worked solution: Connect all four resistors in series:

3 + 3 + 3 + 3 = 12Ω

Question 17 · MCQ · Apply · 1 mark

Series-parallel ratio

Combination of resistors

Four

3Ω

resistors give

R_{s}

in series and

R_{p}

in parallel. What is

R_{s} : R_{p}

?

(A)

4 : 1

(B)

8 : 1

(C)

16 : 1

(D)

1 : 16

Show answer and solution

Answer: (C)

16 : 1

Worked solution:

R_{s} = 12Ω

and

R_{p} = 3/4Ω

. Thus

R_{s} : R_{p} = 12 : (3/4) = 16 : 1

Question 18 · MCQ · Apply · 1 mark

Cells in series

Potential difference and meters

Three identical

1.5

V cells are connected in series aiding. Their net emf is:

(A)

1.5

V
(B)

3.0

V
(C)

4.5

V
(D)

0

Show answer and solution

Answer: (C)

4.5

Worked solution: In series aiding, emfs add:

1.5 + 1.5 + 1.5 = 4.5

V. Hence option (C) is correct.

Question 19 · MCQ · Apply · 1 mark

Heating at fixed voltage

Heating effect and electric power

An ideal fixed-voltage supply feeds a resistor. If its resistance becomes

3 R

while time stays

t

, the new heat compared with initial heat

H

is:

(A)

H /3

(B)

H

(C)

3 H

(D)

9 H

Show answer and solution

Answer: (A)

H /3

Worked solution: For fixed voltage and time,

H = V^{2} t / R

. Replacing

R

3 R

gives

H /3

Question 20 · MCQ · Apply · 1 mark

Heat in a series resistor

Heating effect and electric power

3Ω

and

9Ω

resistor are in series across

12

V for

4

s. Heat in the

9Ω

resistor is:

(A)

9

J
(B)

18

J
(C)

36

J
(D)

48

Show answer and solution

Answer: (C)

36

Worked solution: Total resistance is

12Ω

, so

I = 1

A. Heat in

9Ω

I^{2} R t = 1^{2} \times 9 \times 4 = 36

Question 21 · MCQ · Analyse · 1 mark

Greatest energy draw

Combination of resistors

Across the same

12

V supply for equal time, which draws the most electrical energy?

(A) One

4Ω

resistor
(B) Two

4Ω

resistors in series
(C) Two

4Ω

resistors in parallel
(D) All are equal

Show answer and solution

Answer: (C) Two

4Ω

resistors in parallel

Worked solution: Equivalent resistances are

4Ω

8Ω

, and

2Ω

. Since

W = V^{2} t / R

, the smallest resistance draws the most energy.

Question 22 · MCQ · Apply · 1 mark

Voltage from power

Heating effect and electric power

5Ω

resistor converts

45

J to heat in

1

s. Voltage across it is:

(A)

5

V
(B)

10

V
(C)

15

V
(D)

25

Show answer and solution

Answer: (C)

15

Worked solution: Power is

45

W. Using

P = V^{2} / R

45 = V^{2} /5

, so

V^{2} = 225

and

V = 15

Question 23 · MCQ · Apply · 1 mark

Fuse rating

Household circuits and safety

900

W device operates at

230

V. Among idealised ratings

2

4

6

A and

10

A, choose the smallest rating above normal current.

(A)

2

A
(B)

4

A
(C)

6

A
(D)

10

Show answer and solution

Answer: (B)

4

Worked solution: Normal current is

I = P / V = 900/230 \approx 3.91

A. The smallest listed rating above this is

4

Question 24 · MCQ · Understand · 1 mark

High-power appliance current

Household circuits and safety

At the same household voltage, which appliance draws greater current?

(A)

150

W fan
(B)

1200

W heater
(C) Both equal
(D) Impossible to tell

Show answer and solution

Answer: (B)

1200

W heater

Worked solution: At fixed voltage,

I = P / V

. The higher rated power draws greater current. Hence option (B) is correct.

Question 25 · MCQ · Apply · 1 mark

Power change with current

Heating effect and electric power

Through a fixed resistor, current becomes

3 I

. What is the percentage increase in power?

(A)

200%

(B)

300%

(C)

800%

(D)

900%

Show answer and solution

Answer: (C)

800%

Worked solution:

P_{2} = (3 I)^{2} R = 9 P_{1}

. The increase is

8 P_{1}

, which is

800%

Question 26 · A-R · Apply · 1 mark

Length and resistance

Resistance and resistivity

Assertion (A): At constant cross-section, material and temperature, tripling a wire's length triples its resistance.
Reason (R):

R = ρ L / A

.

(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Show answer and solution

Answer: (A) Both A and R are true, and R correctly explains A.

Worked solution: With all other factors fixed,

R

is directly proportional to

L

. The stated formula directly explains the assertion.

Question 27 · A-R · Analyse · 1 mark

Copper lead heat loss

Heating effect and electric power

Assertion (A): A copper lead typically wastes less energy than an identical high-resistivity alloy lead carrying the same current.
Reason (R): For fixed current and time,

H = I^{2} R t

.

(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Show answer and solution

Answer: (A) Both A and R are true, and R correctly explains A.

Worked solution: Copper has lower resistance for the same dimensions. At fixed current and time, lower

R

produces lower heat loss by

H = I^{2} R t

Question 28 · A-R · Analyse · 1 mark

Temperature and short circuit

Household circuits and safety

Assertion (A): Every increase in a wire's temperature necessarily causes a short circuit.
Reason (R): A metal wire's resistance generally increases with temperature.

(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Show answer and solution

Answer: (D) A is false, but R is true.

Worked solution: A short circuit is an unintended low-resistance path. Heating a metal wire does not itself create that path, although its resistance generally increases.

Question 29 · A-R · Understand · 1 mark

Nichrome heating elements

Heating effect and electric power

Assertion (A): Nichrome can be suitable for a heating element.
Reason (R): It has relatively high resistivity, a high melting point and tolerates high operating temperature.

(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Show answer and solution

Answer: (A) Both A and R are true, and R correctly explains A.

Worked solution: Nichrome's high resistivity helps generate heat and its high melting point helps it withstand the temperature. The reason explains the assertion.

Question 30 · A-R · Understand · 1 mark

Heating element survival

Heating effect and electric power

Assertion (A): A heating element should withstand the temperature reached during operation.
Reason (R): A material with a suitable high melting point can remain intact when heated strongly.

(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Show answer and solution

Answer: (A) Both A and R are true, and R correctly explains A.

Worked solution: An element must not melt during use. A suitable high melting point directly supports this requirement.

Question 31 · A-R · Apply · 1 mark

Bulb rating comparison

Heating effect and electric power

Assertion (A): A

60

W and a

100

W incandescent bulb connected in parallel across their common rated voltage: the

100

W bulb consumes more power.
Reason (R): Their rated powers at that voltage are

100

W and

60

W respectively.

(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Show answer and solution

Answer: (A) Both A and R are true, and R correctly explains A.

Worked solution: In parallel each bulb receives its rated voltage. Their stated ratings therefore show that the

100

W bulb consumes more power.

Question 32 · A-R · Apply · 1 mark

Measured V and I

Potential difference and meters

Assertion (A): Measuring

V

across and

I

through an ohmic resistor permits calculation of its resistance and power.
Reason (R): For this resistor,

R = V / I

and

P = V I

.

(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Show answer and solution

Answer: (A) Both A and R are true, and R correctly explains A.

Worked solution: The formulas use the simultaneously measured potential difference and current. The reason directly explains the assertion.

Question 33 · Case Study · Apply · 5 marks

Workstation circuit

Combination of resistors

12

V workstation supply feeds a

2Ω

resistor in series with a parallel pair of

6Ω

and

3Ω

. Take the battery as ideal.

Answer the following:
(i) Find the parallel-pair resistance. [1 mark]
(ii) Find total resistance and source current. [1 mark]
(iii) Find the potential difference across the parallel pair. [1 mark]
(iv) Find currents through the

6Ω

and

3Ω

branches. [1 mark]
(v) If the

6Ω

branch is disconnected, find the new total current. [1 mark]

Mixed circuit with a 2 ohm series resistor followed by 6 ohm and 3 ohm parallel branches on a 12 volt supply. — A 2 ohm series resistor is followed by 6 ohm and 3 ohm parallel branches.

Show answer and solution

Answer: (i)

2Ω

; (ii)

4Ω

3

A; (iii)

6

V; (iv)

1

A and

2

A; (v)

2.4

Worked solution: (i)

1/ R_{p} = 1/6 + 1/3 = 1/2

, so

R_{p} = 2Ω

. (ii)

R_{t o t a l} = 2 + 2 = 4Ω

, so

I = 12/4 = 3

A. (iii) The series

2Ω

drops

I R = 6

V, so the parallel pair has

6

V. (iv)

I_{6} = 6/6 = 1

A and

I_{3} = 6/3 = 2

A. (v) With only

3Ω

branch, total resistance is

2 + 3 = 5Ω

, so current is

12/5 = 2.4

Question 34 · Case Study · Analyse · 5 marks

Testing a solar lamp

Resistance and resistivity

An ohmic lamp test at nearly constant temperature gives

(V, I)

readings

(2 V, 0.1 A)

(4 V, 0.2 A)

, and

(6 V, 0.3 A)

. Treat the device as ohmic in this narrow range.

Answer the following:
(i) Find its resistance. [1 mark]
(ii) For an I-on-vertical versus V-on-horizontal graph, find the slope. [1 mark]
(iii) If

V

becomes

8

V at constant resistance, find current. [1 mark]
(iv) At

6

V, find power. [1 mark]
(v) If both

V

and

R

double, what happens to current? [1 mark]

I versus V graph showing straight ohmic lines through the origin. — An ohmic I-V graph has constant slope.

Show answer and solution

Answer: (i)

20Ω

; (ii)

0.05

A/V; (iii)

0.4

A; (iv)

1.8

W; (v) Current stays unchanged.

Worked solution: (i)

R = V / I = 2/0.1 = 20Ω

. (ii) Slope

Δ I /Δ V = 0.1/2 = 0.05

A/V, which is

1/ R

. (iii)

I = 8/20 = 0.4

A. (iv)

P = V I = 6 \times 0.3 = 1.8

W. (v)

I^{'} = 2 V / (2 R) = V / R

, so it remains the same.

Question 35 · VSA · Apply · 2 marks

Charge transferred

Electric current and charge

0.5

A current flows for

30

s. Find the charge transferred.

Show answer and solution

Answer:

15

Worked solution: Use

Q = I t = 0.5 \times 30 = 15

Question 36 · VSA · Understand · 2 marks

Unit of resistivity

Resistance and resistivity

Give the SI unit of resistivity.

Show answer and solution

Answer:

Ω

Worked solution: From

R = ρ L / A

, we get

ρ = R A / L

. Its SI unit is ohm metre, written

Ω

Question 37 · VSA · Understand · 2 marks

Current relation in series

Combination of resistors

In a series circuit, two lamps carry currents

I_{1}

and

I_{2}

. State their relation.

Show answer and solution

Answer:

I_{1} = I_{2}

Worked solution: A series circuit has one unbranched path, so the same current passes through every component.

Question 38 · SA · Apply · 3 marks

Measuring resistance

Potential difference and meters

Describe how to measure the resistance of a

5Ω

resistor using a DC supply, ammeter and voltmeter. With readings

V = 10

V and

I = 2

A, verify the resistance.

Circuit with a series ammeter and parallel voltmeter for resistance measurement. — Standard ammeter-voltmeter arrangement.

Show answer and solution

Answer: Connect the ammeter in series and the voltmeter across the resistor. The resistance is

5Ω

Worked solution: Place the ammeter in series with the resistor and the voltmeter across it. Connect positive meter terminals toward the positive side of the cell, close the switch and read both meters.

R = V / I = 10/2 = 5Ω

Question 39 · SA · Apply · 3 marks

Stretched wire

Resistance and resistivity

A wire of resistance

8Ω

is stretched to twice its length while volume stays constant. Find new resistance, assuming unchanged resistivity and temperature.

Show answer and solution

Answer:

32Ω

Worked solution: Constant volume gives

A L = A^{'} L^{'}

, so when length doubles, area becomes

A /2

. Thus

R^{'} = ρ (2 L) / (A /2) = 4 R = 32Ω

Question 40 · SA · Apply · 3 marks

Power and heat

Heating effect and electric power

10Ω

resistor receives

20

V for

30

s. Find current, power and heat produced.

Show answer and solution

Answer: Current

2

A; power

40

W; heat

1200

Worked solution:

I = V / R = 20/10 = 2

P = V I = 20 \times 2 = 40

W. Heat

H = P t = 40 \times 30 = 1200

Question 41 · LA · Apply · 5 marks

Series-parallel supply

Combination of resistors

24

V ideal source is connected to a

4Ω

resistor in series with a parallel pair of

12Ω

and

6Ω

. Calculate (i) total resistance, (ii) source current, (iii) branch currents, (iv) total power and (v) verify that branch currents add correctly.

Show answer and solution

Answer:

R_{t o t a l} = 8Ω

; source current

3

A; branch currents

1

A and

2

A; total power

72

Worked solution:

1/ R_{p} = 1/12 + 1/6 = 3/12

, so

R_{p} = 4Ω

and

R_{t o t a l} = 8Ω

. Source current is

24/8 = 3

A. The series

4Ω

drops

12

V, so branch currents are

12/12 = 1

A and

12/6 = 2

A. Total power is

V I = 24 \times 3 = 72

W. Finally,

1 + 2 = 3

A, matching source current.

Question 42 · LA · Analyse · 5 marks

Changing an element

Heating effect and electric power

A device contains a

6Ω

element on a fixed

12

V supply for

5

minutes. It is replaced with a

12Ω

element while voltage and duration remain fixed. Calculate original and new current, power and electrical energy; state the energy ratio.

Show answer and solution

Answer: Original:

2

24

7200

J. New:

1

12

3600

J. Energy ratio

1 : 2

Worked solution: Time is

300

s. Initially

I_{1} = 12/6 = 2

P_{1} = 1 2^{2} /6 = 24

W and

E_{1} = 24 \times 300 = 7200

J. With

12Ω

I_{2} = 1

P_{2} = 1 2^{2} /12 = 12

W and

E_{2} = 12 \times 300 = 3600

J. Thus

E_{2} / E_{1} = 1/2

Question 43 · LA · Analyse · 5 marks

Comparing two conductors

Resistance and resistivity

Two ohmic conductors give points on an I-versus-V graph: conductor A

(3 V, 0.30 A)

and conductor B

(3 V, 0.15 A)

. (i) Calculate both resistances and slopes. (ii) At common

6

V, find currents and their ratio. (iii) Explain which graph is steeper.

Two I versus V straight lines with different slopes. — Lower resistance gives a steeper I-V graph.

Show answer and solution

Answer:

R_{A} = 10Ω

R_{B} = 20Ω

; slopes

0.10

A/V and

0.05

A/V; currents

0.60

A and

0.30

A, ratio

2 : 1

; A is steeper.

Worked solution:

R_{A} = 3/0.30 = 10Ω

and

R_{B} = 3/0.15 = 20Ω

. Their I-V slopes are

0.10

A/V and

0.05

A/V. At

6

V, currents are

6/10 = 0.60

A and

6/20 = 0.30

A, giving ratio

2 : 1

. A is steeper because I-V slope equals

1/ R

Electricity Competency Based Questions

Electric current and chargePotential difference and metersResistance and resistivityCombination of resistorsHeating effect and electric powerHousehold circuits and safety

Practise with question filters

Question 1 · MCQ · Apply · 1 mark

Electron flow from current

Electric current and charge

At a point in a wire, current is

0.80

A. Approximately how many electrons pass per second?

(A)

5 \times 1 0^{17}

(B)

5 \times 1 0^{18}

(C)

8 \times 1 0^{18}

(D)

1.28 \times 1 0^{20}

Show answer and solution

Answer: (B)

5 \times 1 0^{18}

Worked solution: Current

I = Q / t

. The number of electrons per second is

n = I / e = 0.80/ (1.6 \times 1 0^{- 19}) = 5 \times 1 0^{18}

. Hence option (B) is correct.

Question 2 · MCQ · Apply · 1 mark

Charge and electrons

Electric current and charge

A charge of

3.2

C passes through a wire in

8

s. The number of electrons is:

(A)

2 \times 1 0^{18}

(B)

2 \times 1 0^{19}

(C)

5 \times 1 0^{19}

(D)

1.6 \times 1 0^{20}

Show answer and solution

Answer: (B)

2 \times 1 0^{19}

Worked solution: Use

n = Q / e

. Thus

n = 3.2/ (1.6 \times 1 0^{- 19}) = 2 \times 1 0^{19}

electrons. Hence option (B) is correct.

Question 3 · MCQ · Understand · 1 mark

Electrical energy

Potential difference and meters

Which expression gives energy transferred to charge

Q

through potential difference

V

?

(A)

V / Q

(B)

V Q

(C)

Q / V

(D)

Q / t

Show answer and solution

Answer: (B)

V Q

Worked solution: Potential difference is

V = W / Q

. Therefore

W = V Q

. Hence option (B) is correct.

Question 4 · MCQ · Apply · 1 mark

Meter connections

Potential difference and meters

Show answer and solution

Answer: (C) Ammeter series, voltmeter parallel

Worked solution: An ammeter measures current through the component, so it is in series. A voltmeter compares the two terminals of the lamp, so it is in parallel. Hence option (C) is correct.

Question 5 · MCQ · Apply · 1 mark

Voltmeter position

Potential difference and meters

6

V cell drives a

3Ω

resistor. Where should the voltmeter be connected?

(A) Across the resistor
(B) In series ahead of the cell
(C) In place of the resistor
(D) Only across the ammeter

Show answer and solution

Answer: (A) Across the resistor

Worked solution: A voltmeter must be connected across the component whose potential difference is required. Hence option (A) is correct.

Question 6 · MCQ · Analyse · 1 mark

I-V graph slope

Resistance and resistivity

On an

I

versus

V

plot, line A has slope

0.5

A/V and line B has slope

0.25

A/V. Which resistance is larger?

(A) A is twice B
(B) B is twice A
(C) They are equal
(D) Cannot be determined

Show answer and solution

Answer: (B) B is twice A

Worked solution: For an

I

-versus-

V

graph, slope

I / V = 1/ R

. Thus

R_{A} = 2Ω

and

R_{B} = 4Ω

. B has twice the resistance of A.

Question 7 · MCQ · Apply · 1 mark

Wire diameter and resistance

Resistance and resistivity

A wire of given material and length is replaced by one with twice the diameter. Its new resistance is:

(A)

R /4

(B)

R /2

(C)

2 R

(D)

4 R

Show answer and solution

Answer: (A)

R /4

Worked solution: Area is proportional to diameter squared. Doubling diameter makes area

4 A

, while

R = ρ L / A

. Therefore the new resistance is

R /4

Question 8 · MCQ · Apply · 1 mark

Equal resistance wires

Resistance and resistivity

Wire B is three times as long as wire A but has the same resistance and material. What is

A_{B} : A_{A}

?

(A)

1 : 3

(B)

1 : 1

(C)

3 : 1

(D)

9 : 1

Show answer and solution

Answer: (C)

3 : 1

Worked solution: Since

R = ρ L / A

, keeping

R

same while length becomes

3 L

requires area

3 A

. Hence

A_{B} : A_{A} = 3 : 1

Question 9 · MCQ · Analyse · 1 mark

Least resistance

Resistance and resistivity

For equal-material wires, which has the least resistance?

(A) Length

L

, diameter

d

(B) Length

L

, diameter

2 d

2 L

, diameter

2 d

(D) Length

L /2

, diameter

d

Show answer and solution

Answer: (B) Length

L

, diameter

2 d

Worked solution: Resistance varies as

L / d^{2}

. The relative values are

1

1/4

1/2

, and

1/2

. Option (B) is least.

Question 10 · MCQ · Apply · 1 mark

Cross-sectional area

Resistance and resistivity

Wire B has length

4 L

and resistance

2 R

compared with wire A of length

L

, resistance

R

and area

A

. Its cross-section is:

(A)

A /2

(B)

A

(C)

2 A

(D)

4 A

Show answer and solution

Answer: (C)

2 A

Worked solution:

2 R = ρ (4 L) / A_{B}

while

R = ρ L / A

. Comparing gives

A_{B} = 2 A

. Hence option (C) is correct.

Question 11 · MCQ · Understand · 1 mark

Temperature effect

Resistance and resistivity

At approximately constant geometry, a metallic wire is heated. Its resistance generally:

(A) Falls
(B) Rises
(C) Becomes zero
(D) Must remain identical

Show answer and solution

Answer: (B) Rises

Worked solution: Metals generally have a positive temperature coefficient of resistance. Hence option (B) is correct.

Question 12 · MCQ · Understand · 1 mark

Resistivity as a material property

Resistance and resistivity

At fixed material and temperature, a wire length doubles. What happens to resistivity

ρ

?

(A) Halves
(B) Doubles
(C) Quadruples
(D) Remains unchanged

Show answer and solution

Answer: (D) Remains unchanged

Worked solution: Resistivity is a material property at a fixed temperature. Resistance changes with length, but

ρ

does not. Hence option (D) is correct.

Question 13 · MCQ · Apply · 1 mark

Series current

Combination of resistors

4Ω

lamp and a

6Ω

resistor are in series. Current through the resistor is

0.30

A. Current through the lamp is:

(A)

0.18

A
(B)

0.30

A
(C)

0.50

A
(D)

0.75

Show answer and solution

Answer: (B)

0.30

Worked solution: There is one unbranched series path, so the same current flows through each component. Hence option (B) is correct.

Question 14 · MCQ · Apply · 1 mark

Maximum equivalent resistance

Combination of resistors

Five separate

2Ω

resistors must all be used. What is the largest achievable equivalent resistance?

(A)

0.4Ω

(B)

2Ω

(C)

5Ω

(D)

10Ω

Show answer and solution

Answer: (D)

10Ω

Worked solution: The largest equivalent resistance is obtained in series:

5 \times 2 = 10Ω

Question 15 · MCQ · Apply · 1 mark

Parallel branches

Combination of resistors

Three branches of

3Ω

6Ω

, and

6Ω

connect the same terminals X and Y. Their equivalent resistance is:

(A)

1.5Ω

(B)

2Ω

(C)

3Ω

(D)

15Ω

Show answer and solution

Answer: (A)

1.5Ω

Worked solution:

1/ R = 1/3 + 1/6 + 1/6 = 2/3

. Therefore

R = 1.5Ω

. Hence option (A) is correct.

Question 16 · MCQ · Apply · 1 mark

Four resistors in series

Combination of resistors

What is the maximum equivalent resistance using four

3Ω

resistors, all included?

(A)

3/4Ω

(B)

3Ω

(C)

9Ω

(D)

12Ω

Show answer and solution

Answer: (D)

12Ω

Worked solution: Connect all four resistors in series:

3 + 3 + 3 + 3 = 12Ω

Question 17 · MCQ · Apply · 1 mark

Series-parallel ratio

Combination of resistors

Four

3Ω

resistors give

R_{s}

in series and

R_{p}

in parallel. What is

R_{s} : R_{p}

?

(A)

4 : 1

(B)

8 : 1

(C)

16 : 1

(D)

1 : 16

Show answer and solution

Answer: (C)

16 : 1

Worked solution:

R_{s} = 12Ω

and

R_{p} = 3/4Ω

. Thus

R_{s} : R_{p} = 12 : (3/4) = 16 : 1

Question 18 · MCQ · Apply · 1 mark

Cells in series

Potential difference and meters

Three identical

1.5

V cells are connected in series aiding. Their net emf is:

(A)

1.5

V
(B)

3.0

V
(C)

4.5

V
(D)

0

Show answer and solution

Answer: (C)

4.5

Worked solution: In series aiding, emfs add:

1.5 + 1.5 + 1.5 = 4.5

V. Hence option (C) is correct.

Question 19 · MCQ · Apply · 1 mark

Heating at fixed voltage

Heating effect and electric power

An ideal fixed-voltage supply feeds a resistor. If its resistance becomes

3 R

while time stays

t

, the new heat compared with initial heat

H

is:

(A)

H /3

(B)

H

(C)

3 H

(D)

9 H

Show answer and solution

Answer: (A)

H /3

Worked solution: For fixed voltage and time,

H = V^{2} t / R

. Replacing

R

3 R

gives

H /3

Question 20 · MCQ · Apply · 1 mark

Heat in a series resistor

Heating effect and electric power

3Ω

and

9Ω

resistor are in series across

12

V for

4

s. Heat in the

9Ω

resistor is:

(A)

9

J
(B)

18

J
(C)

36

J
(D)

48

Show answer and solution

Answer: (C)

36

Worked solution: Total resistance is

12Ω

, so

I = 1

A. Heat in

9Ω

I^{2} R t = 1^{2} \times 9 \times 4 = 36

Question 21 · MCQ · Analyse · 1 mark

Greatest energy draw

Combination of resistors

Across the same

12

V supply for equal time, which draws the most electrical energy?

(A) One

4Ω

resistor
(B) Two

4Ω

resistors in series
(C) Two

4Ω

resistors in parallel
(D) All are equal

Show answer and solution

Answer: (C) Two

4Ω

resistors in parallel

Worked solution: Equivalent resistances are

4Ω

8Ω

, and

2Ω

. Since

W = V^{2} t / R

, the smallest resistance draws the most energy.

Question 22 · MCQ · Apply · 1 mark

Voltage from power

Heating effect and electric power

5Ω

resistor converts

45

J to heat in

1

s. Voltage across it is:

(A)

5

V
(B)

10

V
(C)

15

V
(D)

25

Show answer and solution

Answer: (C)

15

Worked solution: Power is

45

W. Using

P = V^{2} / R

45 = V^{2} /5

, so

V^{2} = 225

and

V = 15

Question 23 · MCQ · Apply · 1 mark

Fuse rating

Household circuits and safety

900

W device operates at

230

V. Among idealised ratings

2

4

6

A and

10

A, choose the smallest rating above normal current.

(A)

2

A
(B)

4

A
(C)

6

A
(D)

10

Show answer and solution

Answer: (B)

4

Worked solution: Normal current is

I = P / V = 900/230 \approx 3.91

A. The smallest listed rating above this is

4

Question 24 · MCQ · Understand · 1 mark

High-power appliance current

Household circuits and safety

At the same household voltage, which appliance draws greater current?

(A)

150

W fan
(B)

1200

W heater
(C) Both equal
(D) Impossible to tell

Show answer and solution

Answer: (B)

1200

W heater

Worked solution: At fixed voltage,

I = P / V

. The higher rated power draws greater current. Hence option (B) is correct.

Question 25 · MCQ · Apply · 1 mark

Power change with current

Heating effect and electric power

Through a fixed resistor, current becomes

3 I

. What is the percentage increase in power?

(A)

200%

(B)

300%

(C)

800%

(D)

900%

Show answer and solution

Answer: (C)

800%

Worked solution:

P_{2} = (3 I)^{2} R = 9 P_{1}

. The increase is

8 P_{1}

, which is

800%

Question 26 · A-R · Apply · 1 mark

Length and resistance

Resistance and resistivity

Assertion (A): At constant cross-section, material and temperature, tripling a wire's length triples its resistance.
Reason (R):

R = ρ L / A

.

(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Show answer and solution

Answer: (A) Both A and R are true, and R correctly explains A.

Worked solution: With all other factors fixed,

R

is directly proportional to

L

. The stated formula directly explains the assertion.

Question 27 · A-R · Analyse · 1 mark

Copper lead heat loss

Heating effect and electric power

Assertion (A): A copper lead typically wastes less energy than an identical high-resistivity alloy lead carrying the same current.
Reason (R): For fixed current and time,

H = I^{2} R t

.

(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Show answer and solution

Answer: (A) Both A and R are true, and R correctly explains A.

Worked solution: Copper has lower resistance for the same dimensions. At fixed current and time, lower

R

produces lower heat loss by

H = I^{2} R t

Question 28 · A-R · Analyse · 1 mark

Temperature and short circuit

Household circuits and safety

Show answer and solution

Answer: (D) A is false, but R is true.

Worked solution: A short circuit is an unintended low-resistance path. Heating a metal wire does not itself create that path, although its resistance generally increases.

Question 29 · A-R · Understand · 1 mark

Nichrome heating elements

Heating effect and electric power

Show answer and solution

Answer: (A) Both A and R are true, and R correctly explains A.

Worked solution: Nichrome's high resistivity helps generate heat and its high melting point helps it withstand the temperature. The reason explains the assertion.

Question 30 · A-R · Understand · 1 mark

Heating element survival

Heating effect and electric power

Show answer and solution

Answer: (A) Both A and R are true, and R correctly explains A.

Worked solution: An element must not melt during use. A suitable high melting point directly supports this requirement.

Question 31 · A-R · Apply · 1 mark

Bulb rating comparison

Heating effect and electric power

Assertion (A): A

60

W and a

100

W incandescent bulb connected in parallel across their common rated voltage: the

100

W bulb consumes more power.
Reason (R): Their rated powers at that voltage are

100

W and

60

W respectively.

(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Show answer and solution

Answer: (A) Both A and R are true, and R correctly explains A.

Worked solution: In parallel each bulb receives its rated voltage. Their stated ratings therefore show that the

100

W bulb consumes more power.

Question 32 · A-R · Apply · 1 mark

Measured V and I

Potential difference and meters

Assertion (A): Measuring

V

across and

I

through an ohmic resistor permits calculation of its resistance and power.
Reason (R): For this resistor,

R = V / I

and

P = V I

.

(a) Both A and R are true, and R correctly explains A.
(b) Both A and R are true, but R does not explain A.
(c) A is true, but R is false.
(d) A is false, but R is true.

Show answer and solution

Answer: (A) Both A and R are true, and R correctly explains A.

Worked solution: The formulas use the simultaneously measured potential difference and current. The reason directly explains the assertion.

Question 33 · Case Study · Apply · 5 marks

Workstation circuit

Combination of resistors

12

V workstation supply feeds a

2Ω

resistor in series with a parallel pair of

6Ω

and

3Ω

6Ω

and

3Ω

branches. [1 mark]
(v) If the

6Ω

branch is disconnected, find the new total current. [1 mark]

Show answer and solution

Answer: (i)

2Ω

; (ii)

4Ω

3

A; (iii)

6

V; (iv)

1

A and

2

A; (v)

2.4

Worked solution: (i)

1/ R_{p} = 1/6 + 1/3 = 1/2

, so

R_{p} = 2Ω

. (ii)

R_{t o t a l} = 2 + 2 = 4Ω

, so

I = 12/4 = 3

A. (iii) The series

2Ω

drops

I R = 6

V, so the parallel pair has

6

V. (iv)

I_{6} = 6/6 = 1

A and

I_{3} = 6/3 = 2

A. (v) With only

3Ω

branch, total resistance is

2 + 3 = 5Ω

, so current is

12/5 = 2.4

Question 34 · Case Study · Analyse · 5 marks

Testing a solar lamp

Resistance and resistivity

An ohmic lamp test at nearly constant temperature gives

(V, I)

readings

(2 V, 0.1 A)

(4 V, 0.2 A)

, and

(6 V, 0.3 A)

V

becomes

8

V at constant resistance, find current. [1 mark]
(iv) At

6

V, find power. [1 mark]
(v) If both

V

and

R

double, what happens to current? [1 mark]

Show answer and solution

Answer: (i)

20Ω

; (ii)

0.05

A/V; (iii)

0.4

A; (iv)

1.8

W; (v) Current stays unchanged.

Worked solution: (i)

R = V / I = 2/0.1 = 20Ω

. (ii) Slope

Δ I /Δ V = 0.1/2 = 0.05

A/V, which is

1/ R

. (iii)

I = 8/20 = 0.4

A. (iv)

P = V I = 6 \times 0.3 = 1.8

W. (v)

I^{'} = 2 V / (2 R) = V / R

, so it remains the same.

Question 35 · VSA · Apply · 2 marks

Charge transferred

Electric current and charge

0.5

A current flows for

30

s. Find the charge transferred.

Show answer and solution

Answer:

15

Worked solution: Use

Q = I t = 0.5 \times 30 = 15

Question 36 · VSA · Understand · 2 marks

Unit of resistivity

Resistance and resistivity

Give the SI unit of resistivity.

Show answer and solution

Answer:

Ω

Worked solution: From

R = ρ L / A

, we get

ρ = R A / L

. Its SI unit is ohm metre, written

Ω

Question 37 · VSA · Understand · 2 marks

Current relation in series

Combination of resistors

In a series circuit, two lamps carry currents

I_{1}

and

I_{2}

. State their relation.

Show answer and solution

Answer:

I_{1} = I_{2}

Worked solution: A series circuit has one unbranched path, so the same current passes through every component.

Question 38 · SA · Apply · 3 marks

Measuring resistance

Potential difference and meters

Describe how to measure the resistance of a

5Ω

resistor using a DC supply, ammeter and voltmeter. With readings

V = 10

V and

I = 2

A, verify the resistance.

Show answer and solution

Answer: Connect the ammeter in series and the voltmeter across the resistor. The resistance is

5Ω

R = V / I = 10/2 = 5Ω

Question 39 · SA · Apply · 3 marks

Stretched wire

Resistance and resistivity

A wire of resistance

8Ω

is stretched to twice its length while volume stays constant. Find new resistance, assuming unchanged resistivity and temperature.

Show answer and solution

Answer:

32Ω

Worked solution: Constant volume gives

A L = A^{'} L^{'}

, so when length doubles, area becomes

A /2

. Thus

R^{'} = ρ (2 L) / (A /2) = 4 R = 32Ω

Question 40 · SA · Apply · 3 marks

Power and heat

Heating effect and electric power

10Ω

resistor receives

20

V for

30

s. Find current, power and heat produced.

Show answer and solution

Answer: Current

2

A; power

40

W; heat

1200

Worked solution:

I = V / R = 20/10 = 2

P = V I = 20 \times 2 = 40

W. Heat

H = P t = 40 \times 30 = 1200

Question 41 · LA · Apply · 5 marks

Series-parallel supply

Combination of resistors

24

V ideal source is connected to a

4Ω

resistor in series with a parallel pair of

12Ω

and

6Ω

. Calculate (i) total resistance, (ii) source current, (iii) branch currents, (iv) total power and (v) verify that branch currents add correctly.

Show answer and solution

Answer:

R_{t o t a l} = 8Ω

; source current

3

A; branch currents

1

A and

2

A; total power

72

Worked solution:

1/ R_{p} = 1/12 + 1/6 = 3/12

, so

R_{p} = 4Ω

and

R_{t o t a l} = 8Ω

. Source current is

24/8 = 3

A. The series

4Ω

drops

12

V, so branch currents are

12/12 = 1

A and

12/6 = 2

A. Total power is

V I = 24 \times 3 = 72

W. Finally,

1 + 2 = 3

A, matching source current.

Question 42 · LA · Analyse · 5 marks

Changing an element

Heating effect and electric power

A device contains a

6Ω

element on a fixed

12

V supply for

5

minutes. It is replaced with a

12Ω

element while voltage and duration remain fixed. Calculate original and new current, power and electrical energy; state the energy ratio.

Show answer and solution

Answer: Original:

2

24

7200

J. New:

1

12

3600

J. Energy ratio

1 : 2

Worked solution: Time is

300

s. Initially

I_{1} = 12/6 = 2

P_{1} = 1 2^{2} /6 = 24

W and

E_{1} = 24 \times 300 = 7200

J. With

12Ω

I_{2} = 1

P_{2} = 1 2^{2} /12 = 12

W and

E_{2} = 12 \times 300 = 3600

J. Thus

E_{2} / E_{1} = 1/2

Question 43 · LA · Analyse · 5 marks

Comparing two conductors

Resistance and resistivity

Two ohmic conductors give points on an I-versus-V graph: conductor A

(3 V, 0.30 A)

and conductor B

(3 V, 0.15 A)

. (i) Calculate both resistances and slopes. (ii) At common

6

V, find currents and their ratio. (iii) Explain which graph is steeper.

Show answer and solution

Answer:

R_{A} = 10Ω

R_{B} = 20Ω

; slopes

0.10

A/V and

0.05

A/V; currents

0.60

A and

0.30

A, ratio

2 : 1

; A is steeper.

Worked solution:

R_{A} = 3/0.30 = 10Ω

and

R_{B} = 3/0.15 = 20Ω

. Their I-V slopes are

0.10

A/V and

0.05

A/V. At

6

V, currents are

6/10 = 0.60

A and

6/20 = 0.30

A, giving ratio

2 : 1

. A is steeper because I-V slope equals

1/ R

Electricity Competency Based Questions

Electron flow from current

Charge and electrons

Electrical energy

Meter connections

Voltmeter position

I-V graph slope

Wire diameter and resistance

Equal resistance wires

Least resistance

Cross-sectional area

Temperature effect

Resistivity as a material property

Series current

Maximum equivalent resistance

Parallel branches

Four resistors in series

Series-parallel ratio

Cells in series

Heating at fixed voltage

Heat in a series resistor

Greatest energy draw

Voltage from power

Fuse rating

High-power appliance current

Power change with current

Length and resistance

Copper lead heat loss

Temperature and short circuit

Nichrome heating elements

Heating element survival

Bulb rating comparison

Measured V and I

Workstation circuit

Testing a solar lamp

Charge transferred

Unit of resistivity

Current relation in series

Measuring resistance

Stretched wire

Power and heat

Series-parallel supply

Changing an element

Comparing two conductors

More Science competency questions

Electricity Competency Based Questions

Electron flow from current

Charge and electrons

Electrical energy

Meter connections

Voltmeter position

I-V graph slope

Wire diameter and resistance

Equal resistance wires

Least resistance

Cross-sectional area

Temperature effect

Resistivity as a material property

Series current

Maximum equivalent resistance

Parallel branches

Four resistors in series

Series-parallel ratio

Cells in series

Heating at fixed voltage

Heat in a series resistor

Greatest energy draw

Voltage from power

Fuse rating

High-power appliance current

Power change with current

Length and resistance

Copper lead heat loss

Temperature and short circuit

Nichrome heating elements

Heating element survival

Bulb rating comparison

Measured V and I

Workstation circuit

Testing a solar lamp