A Level Physics: Alternating Currents — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Alternating Currents revision notes

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Section A

1. Define the root-mean-square value of an alternating current. [2]

2. Explain why the mean value of an alternating current over a complete cycle is not a useful measure. [2]

3. (Prerequisite, IGCSE/O Level background – not a new Topic 21 outcome) Explain why a transformer cannot operate on direct current. [2]

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Section B

4. A mains supply is quoted as 230 V, 50 Hz.

(a) State what the 230 V represents and calculate the peak voltage. [3]

(b) Calculate the period. [1]

(c) A 1150 W heater is connected. Calculate the r.m.s. current and the peak current. [3]

(d) Calculate the peak power dissipated, and state its relationship to the mean power. [2]

5. (Prerequisite, IGCSE/O Level background – not a new Topic 21 outcome) A transformer steps 230 V down to 12 V. The primary has 1150 turns.

(a) Calculate the number of secondary turns. [2]

(b) The transformer is 100% efficient and the secondary current is 3.0 A. Calculate the primary current. [2]

(c) Real transformers are not 100% efficient. State two causes of energy loss and how each is reduced. [4]

6. Electricity is transmitted at 400 kV rather than 230 V.

(a) Explain why, in terms of power loss. [3]

(b) A cable carries 500 MW. Calculate the current at 400 kV and at 40 kV, and comment on the effect on power loss. [3]

Section C

7. Sketch, on the same axes, the output waveform you would expect from half-wave and from full-wave rectification of a sinusoidal input, and state which arrangement of diodes produces full-wave rectification. [4]

8. A capacitor is connected across the load of a full-wave rectifier circuit to smooth the output.

(a) Explain how the capacitor smooths the output. [2]

(b) State two ways of reducing the ripple in the smoothed output, and explain why each works. [4]

9. A student claims that increasing the smoothing capacitance removes the ripple completely. Explain why this claim is wrong. [2]

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Answers

1. The value of the direct current [1] that would dissipate the same mean power in the same resistor [1].

2. The current spends equal times in each direction, so the mean over a full cycle is zero [1], which would wrongly imply no energy is transferred [1].

3. Direct current produces a constant magnetic flux in the core [1], so there is no rate of change of flux linkage in the secondary and no e.m.f. is induced [1].

4. (a) It is the r.m.s. value [1]. V₀ = V_rms × √2 = 230 × 1.414 [1] = 325 V [1].

(b) T = 1 ÷ 50 = 0.020 s [1].

(c) I_rms = P ÷ V_rms = 1150 ÷ 230 [1] = 5.0 A [1]. I₀ = 5.0 × √2 = 7.07 A [1].

(d) P₀ = I₀V₀ = 7.07 × 325 = 2300 W [1], which is twice the mean power [1].

5. (a) N_s = N_p × (V_s ÷ V_p) = 1150 × (12 ÷ 230) [1] = 60 turns [1].

(b) I_pV_p = I_sV_s, so I_p = (3.0 × 12) ÷ 230 [1] = 0.157 A [1].

(c) Any two, each with its remedy: Eddy currents in the core dissipate energy [1] — reduced by laminating the core with insulated layers [1]. Resistive (I²R) heating in the windings [1] — reduced by using thick, low-resistance copper wire [1]. (Also acceptable: hysteresis, reduced by using a soft iron core; flux leakage, reduced by better core design.)

6. (a) Power loss in the cables is I²R [1]. Transmitting at high voltage means a low current for the same power [1], and since loss depends on the square of the current, the loss is greatly reduced [1].

(b) At 400 kV: I = 500 × 10⁶ ÷ 400 × 10³ = 1250 A [1]. At 40 kV: I = 500 × 10⁶ ÷ 40 × 10³ = 12 500 A [1]. The current is ten times greater, so the power loss would be 100 times greater [1].

7. Half-wave: only the positive half-cycles of the input survive as unmodified humps, with a flat gap at zero where each negative half-cycle is blocked [1]. Full-wave: every half-cycle produces a hump — the negative half-cycles are inverted onto the positive side rather than blocked — giving a continuous train of humps at twice the half-wave pulse rate, with no flat gaps [1] [1]. A bridge arrangement of four diodes produces full-wave rectification [1].

8. (a) The capacitor charges up during each pulse and discharges gradually between pulses [1], reducing the fall in voltage that would otherwise occur between pulses [1].

(b) Increasing the capacitance — a larger capacitor stores more charge, so it discharges more slowly for the same load, increasing the time constant τ = RC and smoothing the ripple further [2]. Increasing the load resistance — a higher resistance discharges the capacitor more slowly for the same capacitance, again increasing τ = RC [2].

9. Increasing the capacitance increases the time constant RC, so the voltage falls more slowly between pulses and the ripple is reduced [1], but the capacitor still discharges to some extent between pulses, so the output is never a perfectly constant, ripple-free voltage [1].

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Where marks are usually lost

• Using peak values where r.m.s. is required, or vice versa.
• Saying no power is delivered because the mean current is zero.
• Forgetting mean power is half peak power.
• Describing half-wave vs full-wave rectification only in words when the question asks to sketch or otherwise distinguish them graphically – know the shape of each output waveform, not just a verbal description.
• Explaining transmission loss in terms of voltage rather than I²R.
• Confusing lamination (eddy currents) with a soft iron core (hysteresis).