Question

Two identical capacitors $A$ and $B$, charged to the same potential of $5\text{V}$ each, are connected in two different circuits, as shown below, at the time instant $t=0$. If the charge on capacitors $A$ and $B$ at time $t=CR$, is $Q_A$ and $Q_B$ respectively, then $\left(\text{Here,}e\text{is the base of natural logarithm.}\right)$

• A. $Q_A=\frac{CV}{e},Q_B=\frac{CV}{2}$
• B. $Q_A=CV,Q_B=CV$
• C. $Q_A=CV,Q_B=\frac{CV}{e}$
• D. $Q_A=\frac{CV}{2},Q_B=\frac{CV}{e}$

Answer 1

The correct option is C $Q_A=CV,Q_B=\frac{CV}{e}$

In the first circuit, the diode is reverse biased, so no current flows through the circuit, at any time instant.

$\therefore Q_A=CV$

In the second circuit, current will flow as the diode is forward biased. So, it offers negligible resistance to the flow of current and thus can be replaced by a closed switch.

Now, the charge of the capacitor will leak through the resistance and decay exponentially with time.

During discharging of the capacitor, the potential difference across the capacitor, at any instant, is

$V^{\prime }=V_0{e}^{-\frac{t}{CR}}$

$\text{But}t=CR$

$\Rightarrow V^{\prime }=V{e}^{-1}=\frac{V}{e}$

$\therefore \text{Charge}{Q}_B=CV=\frac{CV}{e}$

Hence, $\left(C\right)$ is the correct answer.