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 $100V,50Hz$ 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)

$I_R=\frac{V_S}{R}=I_0$
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 $100Hz$ or $50Hz$ is doubled, current
through $I_L$ will be halved or $\frac{I_0}{2}$