A transformer with efficiency `80%` works at `4 kW` and `100 V`. If the secondary voltage is `200 V`, then the primary and secondary currents are respectively

Options

A. 16A, 40A
B. 40A, 16A
C. 30A, 45A
D. 50A, 30A

Answer

B

Explanation

`P_(i) = 4 xx 10^(3) W`
`:. I_(P) = (P_i)/(v) = (4 xx 10^3)/(100) = 40 A` (Primary)
`eta = (P_o)/(P_i) :. 8/10 = (P_o)/(4 xx 10^3) :. P_(o) = 32 xx 10^(2)`
`:. I_(S) = (P_o)/(V_S) = (32 xx 10^(2))/(200) = 16 A` (Secondary).