A transformer has an efficiency of 80%. It works at 4 kW
and 100 V. If secondary voltage is 240 V, the current in primary
coil is
(a) 0.4 A (b) 4 A (c) 10 A (d) 40 A
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Answer:
D) 40A
Explanation:
Efficiency of the transformer n = 80% (Given)
Power = P = 4kW (Given)
Voltage across primary Ep = 100v (Given)
Voltage across secondary Es = 240v (Given)
On conversion, Ep = Ip = 4kW= 4000W
Thus,
The current across primary coil will be found by = EpIp/Ep
= 4000/100
= 40A
Therefore, the current in primary coil is = 40A
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