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Science practice

Physics - Electricity Questions with Answers

Electricity questions cover charge, current, potential difference, Ohm's law, resistance, series and parallel circuits, electric power and the heating effect of current, at NCERT class 10 level. RRB Group D, ALP and NTPC papers often ask for the unit of a quantity or a short numerical on V = IR, while SSC papers lean on units, devices and electrical energy. Tip: write the formula first, then substitute, and check that the final unit matches the quantity asked.

Formulas and key facts

  • Current I = Q / t; 1 ampere = 1 coulomb per second
  • Potential difference V = work done / charge; unit volt (1 V = 1 J/C)
  • Ohm's law: V = I × R, unit of resistance is ohm (Ω)
  • Resistance R = ρ × l / A; resistivity ρ depends on the material, not on length or area
  • Series: R = R₁ + R₂ + R₃; the same current flows through each resistor
  • Parallel: 1/R = 1/R₁ + 1/R₂ + 1/R₃; the same voltage across each resistor
  • Electric power P = V × I = I²R = V²/R, unit watt (W)
  • Electrical energy E = P × t; 1 kWh = 3.6 × 10⁶ J
  • Heating effect: H = I²Rt
  • An ammeter is connected in series; a voltmeter is connected in parallel
  • Fuse wire is connected in series and melts when the current is too high
  • Commercial unit of electrical energy is the kilowatt-hour (kWh), also called one unit

Worked examples

Q1. A potential difference of 12 V is applied across a resistor of 4 Ω. Find the current.

  1. Use Ohm's law: I = V / R.
  2. I = 12 / 4 = 3 A.

Answer: 3 A

Q2. A 1000 W heater runs for 3 hours every day for 30 days. If one unit of electricity costs ₹6, find the monthly cost.

  1. Power = 1000 W = 1 kW.
  2. Energy per day = 1 kW × 3 h = 3 kWh (3 units).
  3. Energy in 30 days = 3 × 30 = 90 units.
  4. Cost = 90 × ₹6 = ₹540.

Answer: ₹540

Q3. A 4 Ω resistor is in series with a parallel pair of 6 Ω and 3 Ω. The combination is connected to a 12 V supply. Find the total current and the voltage across the parallel pair.

  1. Parallel pair: 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6, so R = 2 Ω.
  2. Total resistance = 4 + 2 = 6 Ω.
  3. Total current I = 12 / 6 = 2 A.
  4. Voltage across the parallel pair = I × 2 Ω = 2 × 2 = 4 V (the other 8 V falls across the 4 Ω resistor).

Answer: Total current 2 A; voltage across the parallel pair 4 V

Shortcuts that save time

  • Parallel resistance is always smaller than the smallest resistor. If an option is larger than the smallest resistor, reject it immediately.
  • For two resistors in parallel use (product) / (sum), for example 6 × 3 / (6 + 3) = 2 Ω. This is faster and less error-prone than adding reciprocals.
  • If a wire is stretched to double its length (volume constant), its resistance becomes four times, since length doubles and area halves. Remember this instead of guessing between 2× and 4×.

Where this appears

RRB Group DGeneral Science section
RRB ALPGeneral Science section
RRB NTPCGeneral Awareness section
SSC CGLGeneral Awareness section

Physics - Electricity

Practice Set
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What is the equivalent resistance of a series combination of resistors?

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Physics - Electricity questions with answers

10 sample questions from this topic with the correct option and the solution.

  1. Q1. What is the full form of UPS?

    1. United Power Supply
    2. Uninterruptible Power Supply
    3. Universal Pulse Supply
    4. Unidirectional Power Supply

    Answer: (B) Uninterruptible Power Supply

    Solution
    UPS stands for Uninterruptible Power Supply. It is a device that provides backup power through a battery when the main electricity supply fails, so connected equipment such as computers keeps running without interruption.

  2. Q2. Two wires are of the same length and the same resistance, but the ratio of their cross-sectional areas is 1 ∶ 8. The ratio of their resistivities will be:

    1. 1 ∶ 8
    2. 3 ∶ 8
    3. 8 ∶ 1
    4. 8 ∶ 3

    Answer: (A) 1 ∶ 8

  3. Q3. The charge of two H + ions is ________.

    1. −1.6 × 10 -19 C
    2. +3.2 × 10 -10 C
    3. +3.2 × 10 -19 C
    4. + 1.6 × 10 -19 C

    Answer: (C) +3.2 × 10 -19 C

  4. Q4. To investigate the conduction of electric current, Ravi performed an experiment. He took different aqueous solutions or liquids (as electrolyte) and tried to pass electricity and connected the circuit with a bulb. In the presence of which of the following, will the bulb NOT glow?

    1. Octane (1)
    2. Copper sulphate (aq)
    3. Sodium chloride (aq)
    4. Calcium chloride (aq)

    Answer: (A) Octane (1)

  5. Q5. The charge flowing through a circuit in 10 minutes is 300 C. Find the current drawn by the circuit.

    1. 50 A
    2. 0.05 A
    3. 0.5 A
    4. 5 A

    Answer: (C) 0.5 A

  6. Q6. The correct relation between charge(Q) following through a conductor, time(t) for which the charge follows and the current (I) following through the conductor is:

    1. Q I = t
    2. I = Q 2 t
    3. I = Q t 2
    4. I = Qt

    Answer: (A) Q I = t

  7. Q7. The formula of current I = _______ /time (t).

    1. Charge (Q)
    2. Resistance (R)
    3. Volt (V)
    4. Power (P)

    Answer: (A) Charge (Q)

  8. Q8. The generation of heat in a conductor due to electric current is known as:

    1. Ohm’s Law
    2. electromagnetic induction
    3. electrolysis
    4. Joule’s heating

    Answer: (D) Joule’s heating

  9. Q9. The effective resistance of the parallel combination of two resistors is always _______

    1. less than the resistance of the individual resistors
    2. equal to the product of resistance of the individual resistors
    3. greater than the resistance of the individual resistors
    4. equal to the sum of the resistance of the resistors

    Answer: (A) less than the resistance of the individual resistors

  10. Q10. The electric dipole moment per unit volume is called ___________.

    1. Power
    2. Electrical Dipole
    3. Energy
    4. Polarization

    Answer: (D) Polarization