Question

The switches given in the circuit $\left(\text{A}\right)$ and $\left(\text{B}\right)$ are closed at $t=0$ and reopened after a long time at $t=t_0$. Then

• A. The charge on $C$ just after $t=0$ is $EC$
• B. The current in $L$ just before $t=t_0$ is $\frac{E}{R}$
• C. The charge on $C$ long time after $t=0$ is $EC$
• D. The current in $L$ long time after $t=t_0$ is $\frac{E}{R}$

Answer 1

Answer: (B), (C)

The charge on $C$ long time after $t=0$ is $EC$

When a capacitor is connected to a battery for a long time, it behaves as an open circuit.

Hence, using $KVL$,

$E-\frac{q}{c}=0\Rightarrow q=EC$

It takes time to build up charge in the capacitor.

When an inductor is connected to a battery for a long time, it behaves as a short circuit.

Hence, using $KVL$,

$E-iR=0\Rightarrow i=\frac{E}{R}$

A long time after opening the switch, the current in the circuit
will be zero.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ and $\left(C\right)$ is the correct answer.

Why this question ? This question tests your knowledge of the basic understanding of the combination of a capacitor and of an inductor across a DC battery.