Chapter 11: Electricity — Online MCQ Test
SCIENCE · CLASS 10 · Andhra State Board
Practice Chapter 11: Electricity with a free chapter-wise online MCQ test.
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.
Chapter 11: Electricity — Important Questions & Answers
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 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.
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.
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.
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.
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.
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.
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.
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.
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.
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.