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In the circuit shown, R is a pure resist...

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`

Text Solution

Verified by Experts

The correct Answer is:
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
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