Switching Circuit from AC Source to DC Battery with Capacitor

To determine the correct statements, we analyze the circuit during each phase of the operation.

Phase 1: From $t = 0$ to $t = \frac{7\pi}{6\omega}$ (A is connected to B)

During this time, the capacitor of capacitance $C = 20\,\mu\text{F}$ is connected solely to the AC current source. The current supplied to the capacitor is given by: $I(t) = I_0 \cos(\omega t)$ with $I_0 = 1\text{ A}$ and $\omega = 500\text{ rad s}^{-1}$.

The arrow indicates that for $I(t) > 0$, current flows clockwise in the left loop, which deposits positive charge on the upper plate of the capacitor. Assuming the capacitor is initially uncharged ($q(0) = 0$), the charge on the upper plate of the capacitor at any time $t$ is: $q(t) = \int_0^t I(t')\,dt' = \int_0^t I_0 \cos(\omega t')\,dt' = \frac{I_0}{\omega} \sin(\omega t)$

1. Maximum charge on the capacitor: The maximum charge on the capacitor is: $q_{\max} = \frac{I_0}{\omega} = \frac{1\text{ A}}{500\text{ rad s}^{-1}} = 2 \times 10^{-3}\text{ C}$ Therefore, statement (A) is incorrect.

2. Direction of current just before $t = \frac{7\pi}{6\omega}$: The value of the current just before switching is: $I\left(\frac{7\pi}{6\omega}\right) = I_0 \cos\left(\frac{7\pi}{6}\right) = 1 \cdot \left(-\frac{\sqrt{3}}{2}\right)\text{ A} = -\frac{\sqrt{3}}{2}\text{ A}$ Since the current is negative, its direction is opposite to the initial reference direction (clockwise), which means the current is flowing counter-clockwise. Therefore, statement (B) is incorrect.

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Phase 2: After $t = \frac{7\pi}{6\omega}$ (A is connected to D)

At the instant of switching ($t = \frac{7\pi}{6\omega}$):

• The charge on the upper plate of the capacitor is: $q_i = q\left(\frac{7\pi}{6\omega}\right) = \frac{I_0}{\omega} \sin\left(\frac{7\pi}{6}\right) = \frac{1}{500} \left(-\frac{1}{2}\right) = -10^{-3}\text{ C} = -1\text{ mC}$

• The potential difference across the capacitor plates (upper plate relative to the lower plate) is: $V_{\text{upper}} - V_{\text{lower}} = \frac{q_i}{C} = \frac{-10^{-3}\text{ C}}{20 \times 10^{-6}\text{ F}} = -50\text{ V}$

1. Current in resistor $R$ immediately after switching: When terminal $\text{A}$ is connected to $\text{D}$, the capacitor is placed in series with the ideal $50\text{ V}$ battery and the resistor $R = 10\,\Omega$. Taking the lower plate of the capacitor as the reference potential ($V_{\text{lower}} = 0\text{ V}$):

   • The upper plate has potential $V_{\text{upper}} = -50\text{ V}$, so terminal $\text{D}$ is at $V_D = -50\text{ V}$.
   • The battery's positive terminal is connected to $\text{D}$, meaning its negative terminal is at a potential of: $V_{\text{neg}} = V_D - 50\text{ V} = -50\text{ V} - 50\text{ V} = -100\text{ V}$
   • The resistor $R$ is connected between the lower plate ($0\text{ V}$) and the negative terminal of the battery ($-100\text{ V}$).

   Thus, the initial potential drop across the resistor $R$ is: $V_R = 0 - (-100\text{ V}) = 100\text{ V}$ The current flowing through $R$ immediately after the connection is: $I = \frac{V_R}{R} = \frac{100\text{ V}}{10\,\Omega} = 10\text{ A}$ Therefore, statement (C) is correct.

2. Total charge $Q$ flowing from the battery: In the final steady state, the capacitor is fully charged by the $50\text{ V}$ DC source, such that its upper plate is at a potential of $+50\text{ V}$ relative to the lower plate: $q_f = C \cdot V = (20 \times 10^{-6}\text{ F}) \times (50\text{ V}) = +10^{-3}\text{ C} = +1\text{ mC}$

   The total charge that leaves the positive terminal of the battery to reach the upper plate of the capacitor is: $Q = q_f - q_i = (+1 \times 10^{-3}\text{ C}) - (-1 \times 10^{-3}\text{ C}) = 2 \times 10^{-3}\text{ C}$ Therefore, statement (D) is correct.

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Conclusion:

The correct statements are (C) and (D).