OxfordAQA International AS and A-level Physics: Alternating currents and the transformer (9630) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow sections 3.10.5 (Alternating currents) and 3.10.6 (The operation of a transformer) of the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, including Required practical 7. Section 3.10 is International A-level only. All waves are sinusoidal unless stated.

Read the study guide first if you need the explanations, and keep the revision notes open for the formulae. See also the course hub, the printable checklist and the free diagnostics.

Questions

1. State what is meant by the root mean square value of an alternating current. [2]

2. A sinusoidal voltage has a peak value of 14.0 V. (a) Calculate its rms value. [2] (b) State its peak-to-peak value. [1]

3. A country’s mains supply is 110 V rms at a frequency of 60 Hz. (a) Calculate the peak voltage. [2] (b) Calculate the peak-to-peak voltage. [1] (c) Calculate the period of the supply. [1]

4. A 47 Ω resistor is connected to an alternating supply and the peak current is 0.38 A. (a) Calculate the rms current. [1] (b) Calculate the mean power dissipated in the resistor. [2] (c) Calculate the maximum instantaneous power and compare it with the mean power. [2]

5. An oscilloscope displays a sinusoidal signal. The Y-gain is 2.0 V per division and the time-base is 0.50 ms per division. The trace is 4.4 divisions from the lowest point to the highest point, and one complete cycle occupies 5.0 divisions horizontally. (a) Calculate the peak voltage. [2] (b) Calculate the rms voltage. [1] (c) Calculate the frequency of the signal. [2] (d) The time-base is changed to 1.0 ms per division. State how many divisions one cycle now occupies. [1]

6. An oscilloscope is used as a dc voltmeter to measure a cell. The Y-gain is 0.50 V per division. (a) With the cell connected, the trace sits 3.0 divisions above its zero position. Calculate the p.d. of the cell. [1] (b) The time-base is running. Describe the trace. [1] (c) State what you should do before connecting the cell so that the zero position is known. [1]

7. An ideal transformer changes 230 V rms to 9.0 V rms. The secondary coil has 36 turns. (a) Calculate the number of turns on the primary coil. [2] (b) The secondary delivers 1.8 A. Calculate the current in the primary coil. [2]

8. A transformer is supplied with 230 V and a current of 0.42 A. The secondary output is 24 V at 3.6 A. (a) Calculate the efficiency. [3] (b) Calculate the power wasted. [1] (c) State two causes of the wasted power. [2]

9. Explain why transformer cores are: (a) made of thin sheets insulated from each other rather than a solid block, [2] (b) made of soft iron rather than hard steel. [2]

10. A power of 8.0 MW is transmitted through a line of total resistance 2.5 Ω. (a) Calculate the line current when the power is transmitted at 33 kV. [2] (b) Calculate the power lost in the line and the percentage of the transmitted power this represents. [2] (c) The voltage is raised to 132 kV for the same power. Calculate the new current and power loss. [2] (d) Explain why a fourfold increase in voltage cuts the power loss by a factor of 16. [2]

11. In an investigation of a transformer’s efficiency, the primary p.d. is kept at 6.0 V alternating. The load on the secondary is changed and four readings are taken each time.

Ip / A	Vs / V	Is / A
0.20	11.4	0.095
0.36	11.2	0.17
0.62	10.9	0.28
1.10	10.2	0.48

(a) Calculate the efficiency for the first and the last rows. [4] (b) Describe the trend in the efficiency and explain it. [2] (c) State two ways of making the results more reliable. [2]

Answers

1. It is the steady direct current [1] that gives the same mean power in a resistor as the alternating current does [1]. Examiner insight: A definition that mentions only the method of squaring, averaging and square-rooting describes how to find the value, not what it means, and usually earns less than one that mentions equal power.

2. (a) Vrms = Vo/√2 = 14.0/√2 [1] = 9.90 V [1] (b) Peak-to-peak = 2 × 14.0 = 28.0 V [1] Examiner insight: Keep three significant figures to match the data in the question, and write “V” on the answer; an answer of 9.9 with no unit is easy to lose.

3. (a) Vo = √2 × 110 [1] = 156 V [1] (b) Peak-to-peak = 2 × 155.6 = 311 V [1] (c) T = 1/f = 1/60 = 16.7 ms [1] Examiner insight: Work (b) from the unrounded 155.6 V. Doubling a rounded 156 V gives 312 V, which is a small slip but avoidable.

4. (a) Irms = 0.38/√2 = 0.27 A [1] (b) P = Irms²R = 0.269² × 47 [1] = 3.4 W [1] (c) Maximum power = Io²R = 0.38² × 47 [1] = 6.8 W, which is twice the mean power [1] Examiner insight: In (b), using the peak current in Irms²R doubles the answer. The word “mean” in the question is the signal to use rms values.

5. (a) Peak-to-peak = 4.4 × 2.0 = 8.8 V [1]; peak = 8.8/2 = 4.4 V [1] (b) Vrms = 4.4/√2 = 3.1 V [1] (c) T = 5.0 × 0.50 ms = 2.5 ms [1]; f = 1/0.0025 = 400 Hz [1] (d) 2.5 ms ÷ 1.0 ms per division = 2.5 divisions [1] Examiner insight: The 4.4 divisions is a full height, not the peak. Halving it before multiplying by the Y-gain, and showing that step, protects the working if a later slip occurs.

6. (a) V = 3.0 × 0.50 = 1.5 V [1] (b) A horizontal straight line 3.0 divisions above the zero position, because the voltage is constant [1] (c) Connect the input to zero (or ground) and use the Y-shift to place the trace on a marked graticule line [1] Examiner insight: Saying “the trace is a flat line” is not enough on its own; link it to the constant p.d. and give its position.

7. (a) Np/Ns = Vp/Vs, so Np = 36 × 230/9.0 [1] = 920 turns [1] (b) For an ideal transformer IpVp = IsVs [1]; Ip = 9.0 × 1.8/230 = 0.070 A [1] Examiner insight: A step-down transformer must give a smaller current in the primary than the secondary. Checking that your answer is less than 1.8 A catches an inverted ratio.

8. (a) Input power = 230 × 0.42 = 96.6 W [1]; output power = 24 × 3.6 = 86.4 W [1]; efficiency = 86.4/96.6 = 0.89 (89%) [1] (b) 96.6 − 86.4 = 10 W [1] (c) Heating in the resistance of the coils [1]; eddy currents in the core [1] (hysteresis and flux leakage are also valid causes) Examiner insight: “Heat” is the result, not the cause. Name where the heating arises, such as coil resistance or eddy currents.

9. (a) The changing flux induces eddy currents in the core [1]; thin insulated sheets limit the current paths, so the currents and the heating are smaller [1] (b) Each cycle the core is magnetised and demagnetised, which takes energy [1]; soft iron is easily magnetised and demagnetised, so less energy is wasted [1] Examiner insight: Answers that say the sheets “stop” eddy currents overstate it. Say that they reduce them.

10. (a) I = P/V = 8.0 × 10⁶/33 000 [1] = 242 A [1] (b) P(loss) = I²R = 242² × 2.5 = 147 kW [1]; 147/8000 = 1.8% [1] (c) I = 8.0 × 10⁶/132 000 = 61 A [1]; loss = 60.6² × 2.5 = 9.2 kW [1] (d) The current is inversely proportional to the voltage for a given power, so four times the voltage gives one quarter the current [1]; the loss is proportional to I², so it falls by 4² = 16 [1] Examiner insight: A common slip is using 33 kV in V²/R. The line loss uses the current in the line, or the p.d. across the line, not the transmission voltage.

11. (a) Row 1: Pout = 11.4 × 0.095 = 1.08 W and Pin = 6.0 × 0.20 = 1.20 W [1]; efficiency = 90% [1]. Row 4: Pout = 10.2 × 0.48 = 4.90 W and Pin = 6.0 × 1.10 = 6.60 W [1]; efficiency = 74% [1] (b) The efficiency falls as the load current rises [1]; the larger currents cause more I²R heating in the coils, so a larger fraction of the input power is wasted [1] (c) Repeat each set of readings and take means [1]; allow the coils to cool between readings, since a hot coil has more resistance [1]. Checking that the meters are on their ac ranges is also valid. Examiner insight: Show both power values before dividing. If a slip happens in the division, the working still shows which quantities you used.

Where marks are usually lost

• Using Vo/√2 as if it applied to every waveform, or applying it to the wrong value.
• Mixing up peak and peak-to-peak when reading an oscilloscope.
• Reading divisions without multiplying by the Y-gain or the time-base setting.
• Using peak values in Irms²R and so doubling the mean power.
• Dividing output by input the wrong way round, so efficiency exceeds 100%.
• Inverting the turns ratio, giving a step-up result for a step-down transformer.
• Writing “heat” as a loss without naming eddy currents, coil resistance or hysteresis.
• Using the transmission voltage in I²R instead of the line current.
• Giving a practical improvement such as “be more careful” with no specific action.

Next steps

• Condense the topic with the revision notes.
• Re-read the explanations in the study guide.
• Use the course hub and the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), specification Version 4.4, for exams from May/June 2020 onwards, published by OxfordAQA. These questions cover sections 3.10.5 and 3.10.6 and Required practical 7.