Question

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 $100V,50Hz$ 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 $100V,100Hz$ the current with $K_1$ alone closed and with $K_2$ alone closed will be, respectively,

• A. $I_0,\frac{I_0}{2}$
• B. $I_0,2I_0$
• C. $2I_0,I_0$
• D. $2I_0,\frac{I_0}{2}$

Answer 1

Answer: (A)

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