Electricity Important Questions.

Master Electricity with AI-powered practice. Our adaptive tutor identifies your gaps and provides step-by-step explanations for every topic in this chapter.

Practice questions covering all these NCERT topics — with AI explanations for each concept.

Practice these Electricity questions for Class 10 Science, each with the correct answer and a step-by-step explanation. Sign up free to practice all 48+ questions with adaptive difficulty.

1. Q1Easy

   Which of the following materials is the best conductor of electricity?

   • A.Nichrome
   • B.Copper
   • C.Constantan
   • D.Aluminum
   Answer: Copper

   Explanation: Copper has very low resistivity (in the range of 10^-8 to 10^-6 Ω·m) and is widely used in electrical circuits due to its high conductivity.

2. Q2Easy

   How is a voltmeter connected in a circuit to measure the potential difference between two points?

   • A.In series
   • B.In parallel
   • C.Across the battery
   • D.In any configuration
   Answer: In parallel

   Explanation: A voltmeter is connected in parallel to the component across which the potential difference is to be measured.

3. Q3Easy

   What is the current drawn from a 220V supply if a resistor of 4Ω is connected across it?

   • A.11A
   • B.22A
   • C.55A
   • D.220A
   Answer: 55A

   Explanation: Using Ohm’s Law, I = V/R = 220V / 4Ω = 55A.

4. Q4Easy

   Which of the following is the correct formula for calculating power in an electrical circuit?

   • A.P = IR
   • B.P = V²/R
   • C.P = IR²
   • D.P = VI²
   Answer: P = V²/R

   Explanation: P = V²/R is one of the valid formulas for calculating power in an electrical circuit.

5. Q5Medium

   When the length of a wire is doubled, how does the resistance of the wire change if the material and cross-sectional area remain constant?

   • A.Halves
   • B.Doubles
   • C.Remains the same
   • D.Quartered
   Answer: Doubles

   Explanation: Resistance is directly proportional to the length of the wire. Doubling the length doubles the resistance.

6. Q6Medium

   What happens to the current flowing through a wire when its cross-sectional area is increased while keeping the length and material constant?

   • A.Decreases
   • B.Increases
   • C.Remains the same
   • D.Becomes zero
   Answer: Increases

   Explanation: Current is inversely proportional to resistance. Increasing the cross-sectional area decreases resistance, thus increasing current.

7. Q7Medium

   Which of the following is not a valid expression for electrical power in a circuit?

   • A.I²R
   • B.VI
   • C.IR²
   • D.V²/R
   Answer: IR²

   Explanation: The valid expressions for electrical power are VI, I²R, and V²/R. IR² is not a standard expression for power.

8. Q8Medium

   An electric bulb is rated at 220V and 100W. If it is operated on 110V, what will be the power consumed?

   • A.100W
   • B.75W
   • C.50W
   • D.25W
   Answer: 25W

   Explanation: Using the power formula P = V²/R, if the voltage is halved, the power becomes one-fourth of the rated power. So, 100W/4 = 25W.

9. Q9Hard

   Two wires of the same material, length, and diameter are connected first in series and then in parallel across the same potential difference. What is the ratio of heat produced in series to parallel?

   • A.1:2
   • B.2:1
   • C.1:4
   • D.4:1
   Answer: 1:4

   Explanation: In series, total resistance is doubled, so heat produced is proportional to R (I²Rt). In parallel, total resistance is halved, so heat produced is proportional to R/4. The ratio of heat produced in series to parallel is 1:4.

10. Q10Hard

    What is the total resistance of three resistors connected in parallel with resistances 10Ω, 20Ω, and 30Ω?

    • A.10Ω
    • B.15Ω
    • C.5Ω
    • D.30Ω
    Answer: 5Ω

    Explanation: The formula for parallel resistors is 1/R_total = 1/R1 + 1/R2 + 1/R3. Substituting the values: 1/R_total = 1/10 + 1/20 + 1/30 = 0.2, thus R_total = 5Ω.

11. Q11Hard

    An electric heater of resistance 44Ω draws 5A from the mains for 2 hours. What is the rate at which heat is developed in the heater?

    • A.550W
    • B.1100W
    • C.2200W
    • D.440W
    Answer: 1100W

    Explanation: Using Joule's Law, P = I²R = (5A)² * 44Ω = 25 * 44 = 1100W.

12. Q12Hard

    What is the resistance of a wire if its resistivity is 1.6 × 10^-8 Ω·m, length is 1 meter, and cross-sectional area is 0.5 mm²?

    • A.0.16Ω
    • B.0.32Ω
    • C.1.6Ω
    • D.3.2Ω
    Answer: 0.32Ω

    Explanation: Using the formula R = ρL/A, where ρ = 1.6 × 10^-8 Ω·m, L = 1m, A = 0.5 × 10^-6 m². R = (1.6 × 10^-8 * 1) / 0.5 × 10^-6 = 0.32Ω.

Why practice Electricity on AskAide?