Circuit Transient Analysis with Multiple Capacitors and Switches

To determine the correct statements, we analyze the circuit under the given initial conditions and operations.

1. Analysis for Option (A):

At $t = 0$, key $S_1$ is closed while switch $S_2$ remains open. Initially, no charge exists on any capacitor, so all capacitors act as short circuits (zero potential drop across them).

The active loop consists of the $5\text{ V}$ battery, the resistors, and uncharged capacitors $C_1, C_4, C_3$.

The total resistance in this closed loop is: $R_{\text{total}} = 70\,\Omega + 30\,\Omega + 100\,\Omega = 200\,\Omega$

Using Ohm's law, the instantaneous current through the circuit at $t = 0$ is: $I(0) = \frac{E}{R_{\text{total}}} = \frac{5\text{ V}}{200\,\Omega} = 0.025\text{ A} = 25\text{ mA}$

Therefore, Option (A) is correct.

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2. Analysis for Option (B):

When $S_1$ is kept closed for a long time, the capacitors become fully charged and act as open circuits, stopping current flow through all resistors.

Capacitors $C_1, C_4,$ and $C_3$ are in series across the $5\text{ V}$ battery. The equivalent capacitance $C_{\text{eq}}$ of the series combination is given by: $\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_4} + \frac{1}{C_3} = \frac{1}{10\,\mu\text{F}} + \frac{1}{80\,\mu\text{F}} + \frac{1}{80\,\mu\text{F}} = \frac{1}{10} + \frac{1}{40} = \frac{5}{40} = \frac{1}{8\,\mu\text{F}}$

$C_{\text{eq}} = 8\,\mu\text{F}$

The total charge $Q$ supplied by the $5\text{ V}$ battery to the series circuit is: $Q = C_{\text{eq}} \times V = 8\,\mu\text{F} \times 5\text{ V} = 40\,\mu\text{C}$

Since the capacitors are in series, each capacitor holds the same charge $Q = 40\,\mu\text{C}$.

The potential difference across capacitor $C_1$ is: $V_{C1} = \frac{Q}{C_1} = \frac{40\,\mu\text{C}}{10\,\mu\text{F}} = 4\text{ V}$

Therefore, Option (B) is correct.

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3. Analysis for Option (D):

In the steady state (with $S_1$ closed and $S_2$ open), no current flows through the $70\,\Omega$ resistor connected in series with $C_1$. Therefore, the potential difference between node $\text{P}$ and node $\text{Q}$ is equal to the voltage across capacitor $C_1$: $V_{\text{P}} - V_{\text{Q}} = V_{C1} = 4\text{ V}$

Since $V_{\text{P}} - V_{\text{Q}} = 4\text{ V} \neq 10\text{ V}$, Option (D) is incorrect.

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4. Analysis for Option (C):

Before switch $S_2$ is closed, the voltages across the capacitors in steady state are:

• $V_{C1} = 4\text{ V}$ (top plate positive)
• $V_{C4} = \frac{Q}{C_4} = \frac{40\,\mu\text{C}}{80\,\mu\text{F}} = 0.5\text{ V}$ (left plate positive)
• $V_{C3} = \frac{Q}{C_3} = \frac{40\,\mu\text{C}}{80\,\mu\text{F}} = 0.5\text{ V}$ (top plate positive)
• $V_{C2} = 0\text{ V}$ (uncharged)

Immediately after switch $S_2$ is closed at $t = 0^+$, the capacitors retain their instantaneous voltages. We set node $\text{Q}$ as the reference potential ($V_{\text{Q}} = 0\text{ V}$).

Applying Kirchhoff's Current Law (KCL) at node $\text{P}$:

1. Left branch (through $S_1$): $I_{\text{left}} = \frac{V_{\text{P}} - 4}{70}$

2. Middle branch (through $S_2$): The $10\text{ V}$ battery has its positive terminal connected towards $\text{Q}$ and negative terminal towards $\text{P}$. $I_{\text{middle}} = \frac{V_{\text{P}} - (-10)}{30} = \frac{V_{\text{P}} + 10}{30}$

3. Right branch: Total drop in fixed potentials = $V_{C4} + 5\text{ V} + V_{C3} = 0.5\text{ V} + 5\text{ V} + 0.5\text{ V} = 6\text{ V}$. Total resistance in this branch = $30\,\Omega + 100\,\Omega = 130\,\Omega$. $I_{\text{right}} = \frac{V_{\text{P}} - 6}{130}$

Equating the sum of currents leaving node $\text{P}$ to zero: $\frac{V_{\text{P}} - 4}{70} + \frac{V_{\text{P}} + 10}{30} + \frac{V_{\text{P}} - 6}{130} = 0$

Multiplying the equation by $2730$ (the LCM of 70, 30, and 130): $39(V_{\text{P}} - 4) + 91(V_{\text{P}} + 10) + 21(V_{\text{P}} - 6) = 0$ $39V_{\text{P}} - 156 + 91V_{\text{P}} + 910 + 21V_{\text{P}} - 126 = 0$ $151 V_{\text{P}} + 628 = 0 \implies V_{\text{P}} = -\frac{628}{151}\text{ V} \approx -4.159\text{ V}$

The immediate current through the $30\,\Omega$ resistor in the middle branch is: $I_{\text{middle}} = \frac{V_{\text{P}} + 10}{30} = \frac{-4.159 + 10}{30} = \frac{5.841}{30} \approx 0.1947\text{ A}$

Rounding off to the $1^{\text{st}}$ decimal place: $I_{\text{middle}} \approx 0.2\text{ A}$

Therefore, Option (C) is correct.

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Correct Answer:

A, B, C