Question

In the circuit shown, `R` is a pure resistor, `L` is an inductor of negligible resistance (as compared to `R`) and `S` is a`100V,50 Hz AC` source of negligible resistance. With eigther key `k_(1)` alone or `k_(2)` alone closed, the current is `I_(0)`. if the source is changed to `100 V, 100 Hz`, the current with `k_(1)` alone closed and with `k_(2)` alone closed will be respectively

A. `I, I//2`
B. `I, 2 I`
C. `2 I, I`
D. `2I, I//2`

Answer 1

Correct Answer - 1
In the second case induction reactance becomes 2 times thus current through `L` when `K_(2)` is closed becomes `(i)/(2)`. But current through `R` when `K_(1)` is closed does not change