Question

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

Answer 1

Hola User

a) 0.4 A

Thanks

Answer 2

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