An ideal inductor takes a current of `10 A` when connected to a `125 V, 50 Hz AC` supply, A pure resistor across the same source takes `12.5 A`. If the two are connected in series across a `100 sqrt(2) V, 40 Hz` supply, the current through the circuit will be

Options

A. `10 A`
B. `12.5 A`
C. `20 A`
D. `25 A`

Answer

1

Explanation

For `50 Hz` and `125 V` supply
`X_(L) = omega L = (V)/(i_(L)) implies L = (1)/(8pi), R = (V)/(i_(R)) = 10 Omega`
For `40 Hz`, `100 sqrt(2) V` supply
`i= (V)/(sqrt(R^(2) + X_(L)^(2))) = (1)/(sqrt(R^(2) + 4 pi^(2) f^(2) L^(2)))`