Chapter Exam

Chapter 11: Electricity — Online MCQ Test

Q: The SI unit of electric current is:

Answer: A. ampere — Electric current is the rate of flow of charge. Its SI unit is ampere (A).

Q: Electric current in a conductor is defined as the:

Answer: B. charge flowing per unit time — Electric current is given by I = Q/t, where Q is charge and t is time.

Q: The SI unit of potential difference is:

Answer: C. volt — Potential difference is measured in volts. One volt is one joule per coulomb.

Q: According to Ohm's law, the current through a conductor is directly proportional to:

Answer: B. potential difference — Ohm's law states that V = IR, so current is directly proportional to potential difference when resistance remains constant.

Q: A charge of 120 C flows through a wire in 60 s. What is the current in the wire?

Answer: C. 2 A — Using I = Q/t, current = 120/60 = 2 A.

Q: A current of 3 A flows through a resistor of 5 Ω. The potential difference across it is:

Answer: B. 15 V — By Ohm's law, V = IR = 3 × 5 = 15 V.

Q: A 4 Ω resistor and a 6 Ω resistor are connected in series across a 20 V battery. The potential difference across the 6 Ω resistor is:

Answer: C. 12 V — Total resistance = 10 Ω, so current = 20/10 = 2 A. Voltage across 6 Ω resistor = IR = 2 × 6 = 12 V.

Q: Resistors of 3 Ω and 6 Ω are connected in parallel across a 12 V battery. The total current drawn from the battery is:

Answer: C. 6 A — Current through 3 Ω is 12/3 = 4 A and through 6 Ω is 12/6 = 2 A. Total current = 6 A.

Q: Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel across a 6 V battery. The total power supplied by the battery is:

Answer: D. 36 W — For parallel combination, 1/R = 1/2 + 1/3 + 1/6 = 1, so R = 1 Ω. Power = V²/R = 36/1 = 36 W.

Q: Two 4 Ω resistors are connected in parallel, and this combination is connected in series with a 3 Ω resistor across a 10 V battery. The current through each 4 Ω resistor is:

Answer: B. 1 A — The parallel combination of two 4 Ω resistors is 2 Ω, so total resistance is 5 Ω and total current is 2 A. This current splits equally, giving 1 A through each 4 Ω resistor.

SCIENCE · Grade 10 · CBSE(NCERT)

Practice Chapter 11: Electricity with a free chapter-wise online MCQ test for CBSE(NCERT) Grade 10 SCIENCE. This chapter covers: Electric charge Electric current Potential difference Voltage Resistance Ohms law Resistivity Series circuit Parallel circuit Electric power Heating effect. AI-generated questions from basic to board-exam level, with instant results and explanations.

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Chapter 11: Electricity — Important Questions & Answers (FAQ)

Frequently asked questions from CBSE(NCERT) Grade 10 SCIENCE — Chapter 11: Electricity, with answers and explanations. These are sample questions; the exam has its own separate question set.

The SI unit of electric current is:

A. ampere ✓
B. volt
C. ohm
D. coulomb

Answer: A. ampere
Electric current is the rate of flow of charge. Its SI unit is ampere (A).

Electric current in a conductor is defined as the:

A. work done per unit charge
B. charge flowing per unit time ✓
C. resistance per unit length
D. heat produced per unit resistance

Answer: B. charge flowing per unit time
Electric current is given by I = Q/t, where Q is charge and t is time.

The SI unit of potential difference is:

A. joule
B. ampere
C. volt ✓
D. watt

Answer: C. volt
Potential difference is measured in volts. One volt is one joule per coulomb.

According to Ohm's law, the current through a conductor is directly proportional to:

A. resistance
B. potential difference ✓
C. resistivity
D. length of conductor

Answer: B. potential difference
Ohm's law states that V = IR, so current is directly proportional to potential difference when resistance remains constant.

A charge of 120 C flows through a wire in 60 s. What is the current in the wire?

A. 0.5 A
B. 1 A
C. 2 A ✓
D. 4 A

Answer: C. 2 A
Using I = Q/t, current = 120/60 = 2 A.

A current of 3 A flows through a resistor of 5 Ω. The potential difference across it is:

A. 8 V
B. 15 V ✓
C. 2 V
D. 1.67 V

Answer: B. 15 V
By Ohm's law, V = IR = 3 × 5 = 15 V.

A 4 Ω resistor and a 6 Ω resistor are connected in series across a 20 V battery. The potential difference across the 6 Ω resistor is:

A. 6 V
B. 8 V
C. 12 V ✓
D. 20 V

Answer: C. 12 V
Total resistance = 10 Ω, so current = 20/10 = 2 A. Voltage across 6 Ω resistor = IR = 2 × 6 = 12 V.

Resistors of 3 Ω and 6 Ω are connected in parallel across a 12 V battery. The total current drawn from the battery is:

A. 2 A
B. 4 A
C. 6 A ✓
D. 12 A

Answer: C. 6 A
Current through 3 Ω is 12/3 = 4 A and through 6 Ω is 12/6 = 2 A. Total current = 6 A.

Three resistors of 2 Ω, 3 Ω and 6 Ω are connected in parallel across a 6 V battery. The total power supplied by the battery is:

A. 6 W
B. 12 W
C. 24 W
D. 36 W ✓

Answer: D. 36 W
For parallel combination, 1/R = 1/2 + 1/3 + 1/6 = 1, so R = 1 Ω. Power = V²/R = 36/1 = 36 W.

Two 4 Ω resistors are connected in parallel, and this combination is connected in series with a 3 Ω resistor across a 10 V battery. The current through each 4 Ω resistor is:

A. 0.5 A
B. 1 A ✓
C. 2 A
D. 4 A

Answer: B. 1 A
The parallel combination of two 4 Ω resistors is 2 Ω, so total resistance is 5 Ω and total current is 2 A. This current splits equally, giving 1 A through each 4 Ω resistor.

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