Question

In the circuit shown here, $R$ 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,

• A. $I_{0}, \frac{I_{0}}{2}$
• B. $I_{0}, 2 I_{0}$
• C. $2 I_{0}, I_{0}$
• D. $2 I_{0}, \frac{I_{0}}{2}$

Answer 1

Answer: (A)

$I_{R}=\frac{V_{S}}{R}=I_{0}$
It remains unchanged, in the second case as, $I_{R}$ does not depend on $v$. In the first case,
$I_{L}=\frac{V_{S}}{X_{L}}=\frac{V_{S}}{2 \pi \cup L}=I_{0}$
When $\cup$ becomes $100 Hz$ or $50 Hz$ is doubled, current
through $I_{L}$ will be halved or $\frac{I_{0}}{2}$