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OCR A Physics · Module 4: Electrons, waves and photons

Energy, power & resistance: practice questions with answers

6 exam-style questions on energy, power & resistance. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.

  1. Rearranging equations2 marks

    A kettle is rated at 2.4 kW on a 230 V supply. Calculate the resistance of its heating element.

    R = Ω

    Show the answer and working

    Answer R = 22 Ω

    P = V²/R, so R = V² ÷ P = 230² ÷ 2400 = 22 Ω.

    • Got 10.4 Ω? This is the current in amperes, not the resistance. Use P = V²/R, so R = 230² ÷ 2400 = 22 Ω.
    • Got 0.0958 Ω? You divided voltage by power. R = V² ÷ P = 52 900 ÷ 2400 = 22 Ω.
  2. Using the right equation3 marks

    A constantan wire has length 1.50 m, diameter 0.50 mm and resistivity 4.9 × 10⁻⁷ Ω m. Calculate its resistance.

    R = Ω

    Show the answer and working

    Answer R = 3.74 Ω

    A = π(0.25 × 10⁻³)² = 1.96 × 10⁻⁷ m². R = ρL ÷ A = 4.9 × 10⁻⁷ × 1.50 ÷ 1.96 × 10⁻⁷ = 3.74 Ω.

    • Got 0.936 Ω? You used the diameter as the radius. A = π(0.25 × 10⁻³)² = 1.96 × 10⁻⁷ m², so R = ρL ÷ A = 3.74 Ω.
    • Got 2.94 Ω? You used d² for the area. A = πd²/4 = 1.96 × 10⁻⁷ m², so R = ρL ÷ A = 3.74 Ω.
  3. Unit prefixes2 marks

    A 3.0 kW heater runs for 4.5 hours. Calculate the energy transferred, in MJ.

    E = MJ

    Show the answer and working

    Answer E = 48.6 MJ

    E = Pt = 3000 W × (4.5 × 3600 s) = 4.86 × 10⁷ J = 48.6 MJ.

    • Got 13.5 MJ? This is the energy in kWh. 1 kWh = 3.6 MJ, so E = 13.5 × 3.6 = 48.6 MJ.
    • Got 810 MJ? You multiplied by 60 only. 1 kWh = 3600 kJ = 3.6 MJ, so E = 13.5 × 3.6 = 48.6 MJ.
  4. Using the right equation3 marks

    A 2.0 kW load is supplied at 500 V through cables of total resistance 0.80 Ω. Calculate the power wasted in the cables.

    P = W

    Show the answer and working

    Answer P = 12.8 W

    Current I = P ÷ V = 2000 ÷ 500 = 4.0 A. Power lost = I²R = 4.0² × 0.80 = 12.8 W.

    • Got 3.13 × 10⁵ W? You used the supply voltage across the cable. The p.d. across the cable is much smaller. Find I = P ÷ V = 4.0 A, then P = I²R = 12.8 W.
    • Got 3.2 W? You calculated IR, which is a voltage. Power loss is I²R = 4.0² × 0.80 = 12.8 W.
  5. Using the right equation1 mark

    What happens to the resistance of an NTC thermistor when its temperature rises?

    • A It decreases because more charge carriers are released.
    • B It increases because ions vibrate more.
    • C It stays the same.
    Show the answer and working

    Answer A: It decreases because more charge carriers are released.

    Heating releases more charge carriers, so the number density rises and resistance falls.

    • Why not B? That describes a metal conductor; in an NTC thermistor more carriers are released as it warms.
    • Why not C? The resistance changes significantly with temperature, which is why thermistors are used as sensors.
  6. Reading graphs1 mark

    The I-V graph of a filament lamp curves so that the gradient decreases as V increases. What does this show?

    • A Its resistance decreases as the filament heats up.
    • B Its resistance increases as the filament heats up.
    • C The lamp obeys Ohm's law.
    Show the answer and working

    Answer B: Its resistance increases as the filament heats up.

    R = V/I rises because the hotter metal ions vibrate more and impede the electrons.

    • Why not A? A falling gradient on an I against V graph means V/I is increasing, so resistance rises.
    • Why not C? Ohm's law gives a straight line through the origin, not a curve.