2018 A/L Physics Past Paper · Question 08 (English Medium)

Secondary voltage of a transformer

A certain transformer has 900 turns in the primary coil and 30 turns in the secondary coil. When 240 V alternating voltage is applied across the primary coil, the voltage across the secondary coil is

  1. (1) 0 V
  2. (2) 8 V
  3. (3) 12 V
  4. (4) 72 V
  5. (5) 7.2 kV
Answer and explanation

Answer: (2) 8 V

Given information

The primary coil has Np=900N_p = 900 turns.

The secondary coil has Ns=30N_s = 30 turns.

The primary has Vp=240V_p = 240 V a.c. across it.

What the question asks

We want: the voltage VsV_s across the secondary coil.

Relevant theory

Accordingly, the transformer is an electrical device which is used to transform an alternating voltage to another either at a higher value or at a lower value without any change of its frequency.

Required equations

VsVp=NsNp\frac{V_s}{V_p} = \frac{N_s}{N_p}

Vs=Vp NsNpV_s = V_p\,\frac{N_s}{N_p}

Applying it to this question

Vs=240×30900V_s = 240 \times \dfrac{30}{900}

Reasoning and calculation

Vs=240×130V_s = 240 \times \dfrac{1}{30}

Vs=24030V_s = \dfrac{240}{30}

88

VsV_s

Checks

Vp∗Ns/NpVp * Ns / Np

The turns ratio has no unit.

So VsV_s has the same unit as VpV_p, the volt.

The secondary has fewer turns than the primary. So its voltage must be lower.

88 V is less than 240240 V. That makes sense.

Conclusion

Vs=8 VV_s = 8\ \text{V}

Vs=240×30900=8 VV_s = 240 \times \frac{30}{900} = 8\ \text{V}

The answer is option (2).

The coil with fewer turns has the lower voltage.

Check your answer against this first.

Exam tip

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