These eleven questions cover induced voltage, generators and motors, transformer ratios, alternating current and transmission loss from induction and transformers. All data is invented for practice and is not taken from any exam paper. They go from easier to harder.
Attempt each question on paper first, with units in every line. Then open the answer. Keep a note of any slip in a mistake log and retest queue.
Questions
Q1. A magnet is pushed into a coil and the meter deflects. Name three changes that would each make the deflection larger.
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Push the magnet faster, use a stronger magnet, or use a coil with more turns. Each one makes the coil cut field lines more quickly, so the induced voltage is larger.
Q2. A bar magnet rests still inside a coil connected to a meter. The meter reads zero. Explain why.
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A voltage is induced only when the field through the coil is changing or field lines are being cut. The magnet is still, so nothing is changing and no voltage is induced.
Q3. A motor runs on 15 V and draws 1.2 A. It delivers 13.5 W of useful kinetic power. Find its efficiency.
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Input power = 15 × 1.2 = 18 W. Efficiency = 13.5 ÷ 18 = 0.75, so 75%. The wasted power is 4.5 W, mainly as thermal energy.
Q4. State the main energy transfer in (a) a generator, (b) a motor, and say what is always wasted in both.
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(a) Kinetic to electrical. (b) Electrical to kinetic. In both, some energy is wasted as thermal energy (from resistance in the coil and friction).
Q5. A transformer has 2300 primary turns and 100 secondary turns. The primary voltage is 230 V. Find the secondary voltage and say whether it is step-up or step-down.
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Vs = Vp × Ns ÷ Np = 230 × 100 ÷ 2300 = 23 000 ÷ 2300 = 10 V. The secondary has fewer turns, so it is step-down.
Q6. A 240 V supply is connected to a primary coil of 4800 turns. A secondary voltage of 9.0 V is required. How many turns are needed on the secondary?
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Ns = Np × Vs ÷ Vp = 4800 × 9.0 ÷ 240 = 43 200 ÷ 240 = 180 turns.
Q7. A transformer has a primary of 50 turns at 12 V and a secondary of 400 turns. Find the secondary voltage. Is it step-up or step-down?
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Vs = 12 × 400 ÷ 50 = 4800 ÷ 50 = 96 V. The secondary has more turns, so it is step-up. The voltage is multiplied by 8.
Q8. An ideal transformer supplies 2.0 A at 9.0 V on its secondary from a 240 V primary. Find the primary current.
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Power out = 9.0 × 2.0 = 18 W. For an ideal transformer, power in = 18 W. Ip = 18 ÷ 240 = 0.075 A.
Q9. A real transformer takes 2.5 A from a 240 V supply and delivers 570 W. Find its efficiency and the power wasted.
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Power in = 240 × 2.5 = 600 W. Efficiency = 570 ÷ 600 = 0.95, so 95%. Power wasted = 600 − 570 = 30 W, as thermal energy in the coils and core.
Q10. The trace of an alternating supply shows a period of 25 ms. Find its frequency. Then explain why a transformer needs this kind of supply and not a steady one.
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T = 25 ms = 0.025 s. f = 1 ÷ 0.025 = 40 Hz. A transformer needs a changing magnetic field in its core to induce a voltage in the secondary. An alternating current gives a continually changing field, while a steady current does not.
Q11. A fictional power station sends 120 kW through cables of total resistance 5.0 Ω. Find the power lost and the percentage of the power sent if the transmission voltage is (a) 6000 V, (b) 1200 V.
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(a) I = 120 000 ÷ 6000 = 20 A. Loss = 20² × 5.0 = 400 × 5.0 = 2000 W (2.0 kW). Percentage = 2000 ÷ 120 000 × 100 = 1.7% (to 2 s.f.).
(b) I = 120 000 ÷ 1200 = 100 A. Loss = 100² × 5.0 = 10 000 × 5.0 = 50 000 W (50 kW). Percentage = 50 000 ÷ 120 000 × 100 = 42% (to 2 s.f.).
The voltage is 5 times smaller in (b), so the loss is 5² = 25 times larger: 50 000 ÷ 2000 = 25. That matches.
If you got these wrong
| What went wrong | Where to look |
|---|---|
| Said a voltage appears with no change, or could not name what makes it larger | Explain a change needed for induced voltage (Q1, Q2) |
| Mixed up generator and motor, or forgot the wasted energy | Compare generator and motor energy transfers (Q3, Q4) |
| Turned the turns ratio upside down, or forgot power in equals power out | Use a transformer ratio under stated assumptions (Q5 to Q9) |
| Period and frequency mixed up, or no reason why a.c. is needed | Explain why alternating current is relevant (Q10) |
| Used V² ÷ R for the loss, or skipped finding the current first | Calculate a transmission loss in a fictional model (Q11) |
Correct answers with a muddled explanation are worth a second look too. Return to the topic overview to see how the skills connect.
Practice shows what goes wrong, and a teacher can show why it happens to you in particular. That is what we offer in online one-to-one Physics tuition.