First slide

Electrical machines and devices

Question

The coils of a step down transformer have 500 and 5000 turns. In the primary coil an ac of 4 ampere at 2200 volts is sent. The value of the current and potential difference in the secondary coil will be

Moderate

A

20 A, 220 V
B

0.4 A, 22000 V
C

40 A, 220 V
D

40 A, 22000 V

Solution

$\frac{N_{p}}{N_{s}} = \frac{V_{p}}{V_{s}} = \frac{i_{s}}{i_{p}}$ . The transformer is step-down type, so primary coil will have more turns. Hence $\frac{5000}{500} = \frac{2200}{V_{s}} = \frac{i_{s}}{4} \Rightarrow V_{s} = 220 V, i_{s} = 40 amp$

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