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In the circuit shown in Fig, R is a pure resistor, L is an inductor of negligible resistance (as compared to R ), S is a 100 \mathrm{~V}, 50 \mathrm{~Hz} ac source of negligible resistance. With either key K_1 alone or K_2 alone closed, the current is I_0. If the source is changed to 100 \mathrm{~V}, 100 \mathrm{~Hz} the current with K_1 alone closed and with K_2 alone closed will be respectively

Option: 1

I_0, \frac{I_0}{2}


Option: 2

I_0, 2 I_0


Option: 3

2 I_0, I_0


Option: 4

2 I_0, \frac{I_0}{2}


Answers (1)

best_answer

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

Posted by

Irshad Anwar

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