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

Electrical circuits: practice questions with answers

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

  1. Using the right equation2 marks

    A 12 Ω resistor and an 18 Ω resistor are connected in parallel. This combination is in series with a 5.0 Ω resistor. Calculate the total resistance.

    R = Ω

    Show the answer and working

    Answer R = 12.2 Ω

    Parallel: 1/R = 1/12 + 1/18 gives 7.2 Ω. Series: 7.2 + 5.0 = 12.2 Ω.

    • Got 35 Ω? You added all three as if in series. The parallel pair gives 1/R = 1/12 + 1/18, so R = 7.2 Ω, and the total is 7.2 + 5.0 = 12.2 Ω.
    • Got 7.2 Ω? You left out the series 5.0 Ω resistor. The total is 7.2 + 5.0 = 12.2 Ω.
  2. Using the right equation3 marks

    A cell of e.m.f. 1.50 V drives a current of 0.30 A through an external resistor of 4.2 Ω. Calculate the internal resistance of the cell.

    r = Ω

    Show the answer and working

    Answer r = 0.8 Ω

    E = I(R + r), so R + r = 1.50 ÷ 0.30 = 5.0 Ω and r = 5.0 - 4.2 = 0.8 Ω.

    • Got 5 Ω? This is the total circuit resistance. Subtract the external resistance: r = 5.0 - 4.2 = 0.8 Ω.
    • Got 0.24 Ω? This is the lost volts (0.24 V). Divide by the current to get resistance: r = 0.24 ÷ 0.30 = 0.8 Ω.
  3. Using the right equation2 marks

    A potential divider has a 4.7 kΩ resistor in series with a 2.2 kΩ resistor across a 9.0 V supply. Calculate the p.d. across the 2.2 kΩ resistor.

    V = V

    Show the answer and working

    Answer V = 2.87 V

    Vout = V × R2 ÷ (R1 + R2) = 9.0 × 2.2 ÷ 6.9 = 2.87 V.

    • Got 6.13 V? You found the p.d. across the 4.7 kΩ resistor. The 2.2 kΩ resistor gets 2.2 ÷ 6.9 of the supply: 2.87 V.
    • Got 4.21 V? You left out the 2.2 kΩ in the total resistance. The divider is V × 2.2 ÷ (4.7 + 2.2) = 2.87 V.
  4. Using the right equation3 marks

    A 12 V cell and a 6.0 V cell are connected in series but with opposite polarity, so their e.m.f.s oppose. The total resistance of the circuit is 9.0 Ω. Calculate the current, in mA.

    I = mA

    Show the answer and working

    Answer I = 667 mA

    Net e.m.f. = 12 - 6.0 = 6.0 V. I = 6.0 ÷ 9.0 = 0.667 A = 667 mA.

    • Got 2000 mA? You added the e.m.f.s. They oppose, so the net e.m.f. is 12 - 6.0 = 6.0 V and I = 6.0 ÷ 9.0 = 667 mA.
    • Got 1330 mA? You ignored the opposing cell. The net e.m.f. is 12 - 6.0 = 6.0 V, so I = V ÷ R = 667 mA.
  5. Using the right equation1 mark

    A voltmeter is used to measure the p.d. across a component. How should it be connected and what should its resistance be?

    • A In series, with a very low resistance.
    • B In parallel, with a very low resistance.
    • C In parallel, with a very high resistance.
    Show the answer and working

    Answer C: In parallel, with a very high resistance.

    It goes in parallel, and a high resistance means it draws almost no current.

    • Why not A? A voltmeter in series would alter the current and a low resistance would waste current.
    • Why not B? A low resistance would draw significant current and change the p.d. being measured.
  6. Reading graphs1 mark

    A cell has internal resistance. What happens to its terminal p.d. as the current drawn from it increases?

    • A It decreases.
    • B It increases.
    • C It stays equal to the e.m.f.
    Show the answer and working

    Answer A: It decreases.

    V = E - Ir, so the terminal p.d. falls linearly as I increases.

    • Why not B? More current means more lost volts (Ir) so the terminal p.d. cannot rise.
    • Why not C? The terminal p.d. only equals the e.m.f. when no current is drawn.