Transformers and power transmission (P) – IGCSE Physics exam questions with mark scheme

1Grade 4–5Triple only5 marks

A transformer is used to operate a 12 V lamp from the 230 V mains supply. The primary coil has 1150 turns.

(a) State whether this is a step-up or a step-down transformer. Give a reason. [1]

(b) Calculate the number of turns on the secondary coil. [2]

(c) The current in the lamp is 2.5 A. Assuming the transformer is 100% efficient, calculate the current in the primary coil. [2]

(a) step-down because the output (secondary) voltage is less than the input (primary) voltage; allow 12 V is less than 230 V; both 'step-down' and reason needed(1)
(b) substitution: $\dfrac{230}{12} = \dfrac{1150}{N_s}$(1)
(b) 60 (turns)(1)
(c) substitution: $230 \times I_p = 12 \times 2.5$(1)
(c) 0.13 A; accept 0.130 A or 0.1304 A(1)
Answer: (a) step-down (b) 60 turns (c) 0.13 A
Examiner tip: Write the ratio with primary over secondary on both sides; mixing them up gives about 22 000 turns.

2Grade 6–7Triple only6 marks

A power station transmits 500 kW of electrical power through overhead cables. A step-up transformer at the power station increases the voltage to 25 kV for transmission. The total resistance of the cables is 4.0 Ω.

(a) Calculate the current in the cables. [2]

(b) Calculate the power wasted as heat in the cables. [2]

(a) substitution: $500\,000 = I \times 25\,000$ (P = IV)(1)
(a) 20 A(1)
(b) voltage across the cables $V = IR = 20 \times 4.0 = 80$ V; allow $P = I^2R = 20^2 \times 4.0$; allow their (a)(1)
(b) $P = 80 \times 20 =$ 1600 W; accept 1.6 kW(1)
(c) alternating current (in primary) produces a changing / alternating magnetic field in the (iron) core(1)
(c) changing field (passes through the secondary coil and) induces a voltage in the secondary; allow d.c. gives a constant field so no voltage is induced(1)
Answer: (a) 20 A (b) 1600 W (c) a.c. gives a changing field which induces a voltage in the secondary
Examiner tip: In (b) do not use 25 kV with the 4.0 Ω: 25 kV is the transmission voltage, not the voltage across the cables. Find the voltage drop across the cables with V = IR (80 V), then use P = IV.

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