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In the circuit shown in figure, R is a pure resistor, L is an inductor of negligible resistance (as compared to R), S is a 100 V 50 Hz ac source of negligible resistance. With either key K1 alone or K2 alone closed, the current is I0. If the source is changed to 100 V, 100 Hz the current with K1 alone closed and with K2 alone closed will be, respectively,

a
I0,I02
b
I0,2I0
c
2I0,I0
d
2I0,I02

detailed solution

Correct option is A

Current remains unchanged in R. However, it becomes half in L,because reactance is doubled on doubling the frequency.

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