Secondary 4 Pure Physics Electricity Magnetism Quiz

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Secondary 4 Pure Physics Quiz - Electricity Magnetism

Name: ___________________________
Class: ___________________________
Date: ___________________________
Score: _______ / 40

Duration: 60 minutes
Total Marks: 40

Instructions:

• Answer all 20 questions.
• Section A: Short structured questions (1–8)
• Section B: Calculation and data questions (9–14)
• Section C: Extended response and interpretation (15–20)
• Show all working clearly. Use SI units.
• Write answers in the spaces provided.

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Section A: Short Structured Questions (1–8) [8 marks]

1. State the function of the neutral wire in a household alternating-current circuit. [1]

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2. State one advantage of using a circuit breaker rather than a fuse to protect an appliance. [1]

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3. A positively charged rod is moved quickly away from a coil connected to a galvanometer. State what is observed on the galvanometer. [1]

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4. Define the turns ratio of a transformer. [1]

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5. State the energy transformation that occurs in the secondary coil of a transformer during normal operation. [1]

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6. State one safety feature of the earth wire in a household appliance. [1]

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7. A magnet is pushed into a stationary coil. State the law that predicts the direction of the induced current. [1]

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8. State the relationship between primary and secondary currents in an ideal transformer. [1]

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Section B: Calculation and Data Questions (9–14) [12 marks]

9. A transformer has a primary voltage of 240 V and a secondary voltage of 12 V. The secondary current is 2.0 A and the transformer is 100% efficient. Calculate the current in the primary coil. [2]

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10. Transformer X has 500 turns on the primary coil and 100 turns on the secondary coil. The primary current is 0.50 A and the transformer is ideal. Calculate the secondary current. [2]

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11. A transformer is 80% efficient. The secondary coil delivers 60 V at 3.0 A. The primary voltage is 240 V. Calculate the primary current. [2]

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12. The diagram shows a simple DC circuit with two resistors.

Image pending generation: diagram for Q12.

Calculate (a) the total resistance, [1] and (b) the current through the circuit. [1]

13. A coil of wire has a resistance of 4.0 Ω and is connected to a 12 V supply. Calculate the power dissipated in the coil. [2]

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14. A step-up transformer has 200 primary turns and 800 secondary turns. If the primary voltage is 50 V, calculate the secondary voltage assuming ideal operation. [2]

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Section C: Extended Response and Interpretation (15–20) [20 marks]

15. (a) State Faraday’s law of electromagnetic induction. [1]
(b) A magnet is dropped through a coil connected to a datalogger. Explain why the induced electromotive force (e.m.f.) is zero when the magnet is at the centre of the coil. [2]

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16. A household appliance has a metal casing and is connected to a 230 V supply with a live, neutral, and earth wire.
(a) State the function of the earth wire. [1]
(b) Explain how the earth wire protects a person from electric shock if the live wire touches the metal casing. [3]

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17. The graph shows the output voltage of a transformer primary and secondary over time.

Image pending generation: graph for Q17.

(a) Determine the turns ratio of the transformer. [2]
(b) State whether this is a step-up or step-down transformer and give a reason. [2]

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18. A student builds an electromagnet by wrapping 100 turns of insulated wire around a soft-iron core and connects it to a 6.0 V battery with total circuit resistance 2.0 Ω.
(a) Calculate the current in the wire. [2]
(b) State two ways to increase the strength of the magnetic field of the electromagnet. [2]

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19. A transformer is used to charge a laptop. The primary is connected to 230 V mains and the secondary provides 19 V at 3.4 A. The transformer has an efficiency of 90%.
(a) Calculate the useful output power. [2]
(b) Calculate the input power drawn from the mains. [2]
(c) Calculate the primary current. [2]

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20. (a) Explain how electromagnetic induction is used in a bicycle dynamo to light a lamp. [3]
(b) State one factor that affects the magnitude of the induced e.m.f. in the dynamo and explain why. [2]

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Secondary 4 Pure Physics Quiz - Electricity Magnetism: Answer Key

Total Marks: 40
Topic: Electricity & Magnetism

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Section A: Short Structured Questions

1. [1 mark]
Answer: The neutral wire provides the return path for current to the supply and is at approximately zero potential.
Teaching note: In an AC household circuit, current flows out through the live wire and returns through the neutral wire, completing the circuit.

2. [1 mark]
Answer: A circuit breaker can be reset and reused after tripping; a fuse must be replaced. (Also accept: faster response to overcurrent.)
Teaching note: Fuses melt and are single-use; circuit breakers trip electronically and can be switched on again.

3. [1 mark]
Answer: The galvanometer needle deflects momentarily in the opposite direction to when the rod was brought near.
Teaching note: Removing the charged rod changes magnetic flux; by Lenz’s law the induced current opposes the change, causing opposite deflection.

4. [1 mark]
Answer: Turns ratio = number of turns on primary coil ÷ number of turns on secondary coil = $N_p/N_s$.
Teaching note: This ratio determines voltage and current changes in a transformer.

5. [1 mark]
Answer: Electrical energy in the primary is transferred magnetically and induced as electrical energy in the secondary (electrical → magnetic → electrical).
Teaching note: No direct electrical connection exists between coils.

6. [1 mark]
Answer: The earth wire connects the metal casing to ground so it cannot become live.
Teaching note: Prevents dangerous potential on exposed metal.

7. [1 mark]
Answer: Lenz’s law.
Teaching note: Lenz’s law states induced current direction opposes the change producing it.

8. [1 mark]
Answer: $I_p / I_s = N_s / N_p$ (or $I_s / I_p = N_p / N_s$); currents are inversely proportional to turns.
Teaching note: Ideal transformer conserves power: $V_p I_p = V_s I_s$.

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Section B: Calculation and Data Questions

9. [2 marks]
Given: $V_p = 240$ V, $V_s = 12$ V, $I_s = 2.0$ A, $\eta = 100\% = 1$.
Step 1: $V_p I_p = V_s I_s$
Step 2: $I_p = \frac{V_s I_s}{V_p} = \frac{12 \times 2.0}{240} = 0.10$ A
Answer: 0.10 A
Marking: 1 mark substitution, 1 mark correct answer with unit.

10. [2 marks]
Given: $N_p = 500$, $N_s = 100$, $I_p = 0.50$ A, ideal.
Step 1: $\frac{I_s}{I_p} = \frac{N_p}{N_s}$
Step 2: $I_s = 0.50 \times \frac{500}{100} = 2.5$ A
Answer: 2.5 A
Marking: 1 mark for ratio use, 1 mark final answer.

11. [2 marks]
Given: $\eta = 0.80$, $V_s = 60$ V, $I_s = 3.0$ A, $V_p = 240$ V.
Step 1: Output power $P_s = V_s I_s = 60 \times 3.0 = 180$ W
Step 2: Input power $P_p = P_s / \eta = 180 / 0.80 = 225$ W
Step 3: $I_p = P_p / V_p = 225 / 240 = 0.9375 \approx 0.94$ A
Answer: 0.94 A
Marking: 1 mark working, 1 mark answer.

12. [2 marks]
(a) [1] $R_{total} = R_1 + R_2 = 6.0 + 3.0 = 9.0\ \Omega$
(b) [1] $I = V / R = 9.0 / 9.0 = 1.0$ A
Answer: (a) 9.0 Ω, (b) 1.0 A

13. [2 marks]
Given: $R = 4.0\ \Omega$, $V = 12$ V.
Step 1: $P = V^2 / R = 12^2 / 4.0 = 144 / 4.0 = 36$ W
Answer: 36 W
Marking: 1 mark formula, 1 mark answer.

14. [2 marks]
Given: $N_p = 200$, $N_s = 800$, $V_p = 50$ V.
Step 1: $\frac{V_s}{V_p} = \frac{N_s}{N_p}$
Step 2: $V_s = 50 \times \frac{800}{200} = 200$ V
Answer: 200 V
Marking: 1 mark ratio, 1 mark answer.

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Section C: Extended Response and Interpretation

15. [3 marks]
(a) [1] Faraday’s law: induced e.m.f. is proportional to the rate of change of magnetic flux linkage.
(b) [2] At the centre, the rate of change of flux is zero because the magnet’s field through the coil is symmetric and not changing at that instant; hence e.m.f. = 0.
Marking: 1 mark law, 2 marks explanation with flux change zero.

16. [4 marks]
(a) [1] Earth wire connects casing to ground at zero potential.
(b) [3] If live touches casing, current flows through earth wire to ground (low resistance path). This large current trips breaker/fuse, cutting supply. Person touching casing not shocked as casing near 0 V.
Marking: 1 + 3 with reasoning.

17. [4 marks]
(a) [2] Turns ratio $N_p/N_s = V_p/V_s = 240/24 = 10$.
(b) [2] Step-down; secondary voltage lower than primary.
Marking: 2 + 2.

18. [4 marks]
(a) [2] $I = V/R = 6.0 / 2.0 = 3.0$ A
(b) [2] Increase turns, increase current, use soft-iron core (any two).
Marking: 2 + 2.

19. [6 marks]
(a) [2] $P_{out} = 19 \times 3.4 = 64.6$ W
(b) [2] $P_{in} = 64.6 / 0.90 = 71.8$ W
(c) [2] $I_p = 71.8 / 230 = 0.312$ A
Marking: 2 each step.

20. [5 marks]
(a) [3] Dynamo has coil rotating in magnetic field (or magnet near coil). Rotation changes flux, inducing e.m.f. in coil, current lights lamp.
(b) [2] Factor: speed of rotation; faster rotation → greater rate of flux change → larger e.m.f. (also accept number of turns, field strength).
Marking: 3 + 2.