Question

The current passing through the battery immediately after key $\left(K\right)$ is closed (Initially all the capacitors are uncharged and the value of $R=6\Omega$ and $C=4\mu \text{F}$) is

• A. $2\text{A}$
• B. $4\text{A}$
• C. $3\text{A}$
• D. $1\text{A}$

Answer 1

The correct option is D $1\text{A}$

When key $K$ is closed,
Inductor acts as open circuit
$L-$ no current will flow.
Capacitor acts as short circuit
$C-$ conducting wire



Hence, current in the circuit is
$i=\frac{E}{\frac{R}{3}+\frac{R}{2}}=\frac{6E}{5R}$
$i=\frac{6\times 5}{5\times 6}=1\text{A}$