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  4. In the circuit shown in the 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, <img class=img-fluid question-image alt=image src=https://cdn.tardigrade.in/img/question/physics/92ac5c9c54b9ed9ef6a93cc5d3774a4d-.png />

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Q. In the circuit shown in the 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 $K_{1}$ 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,
image

Alternating Current

Solution:

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