Question

80V C=4 C=3 C=6 C=10 30V

• A. Charge on C=4 is 150 μC, C=3 is 25 μC, C=6 is 25 μC, C=10 is 125 μC
• B. Charge on C=4 is 125 μC, C=3 is 25 μC, C=6 is 25 μC, C=10 is 150 μC
• C. Charge on C=4 is 150 μC, C=3 is 150 μC, C=6 is 150 μC, C=10 is 150 μC
• D. Charge on C=4 is 37.5 μC, C=3 is 12.5 μC, C=6 is 12.5 μC, C=10 is 12.5 μC

Answer 1

Answer: (A)

Charge on C=4 is 150 μC, C=3 is 25 μC, C=6 is 25 μC, C=10 is 125 μC

Answer 2

The circuit is analyzed by assigning potentials to nodes. The common negative terminal of the voltage sources is set to 0V. The positive terminals are at 80V and 30V respectively. A junction $X$ connects the 4µF capacitor to the parallel combination of capacitors. Charge conservation at node $X$ ($Q_1 = Q_p$) is used to find the potential $V_X$. The charges on individual capacitors are then calculated using $Q = CV$ and the principle of charge distribution in series and parallel combinations.

The equivalent capacitance of the parallel combination of $C=3$ and $C=6$ in series is $C_{23} = \frac{3 \times 6}{3 + 6} = 2$. This is in parallel with $C=10$, so the total parallel capacitance is $C_p = 2 + 10 = 12$. Let $V_X$ be the potential at the junction. Charge on $C=4$: $Q_1 = 4(80 - V_X)$. Charge on $C_p$: $Q_p = 12(V_X - 30)$. By charge conservation, $Q_1 = Q_p$: $4(80 - V_X) = 12(V_X - 30)$ $320 - 4V_X = 12V_X - 360$ $680 = 16V_X$ $V_X = 42.5V$.

Charges: $Q(C=4) = 4(80 - 42.5) = 4(37.5) = 150 \mu C$. The voltage across the parallel combination is $V_p = V_X - 30 = 42.5 - 30 = 12.5V$. $Q(C=10) = C_p \times V_p = 10 \times 12.5 = 125 \mu C$. The charge on the series combination $C_{23}$ is $Q_{23} = C_{23} \times V_p = 2 \times 12.5 = 25 \mu C$. Since $C=3$ and $C=6$ are in series, $Q(C=3) = Q(C=6) = Q_{23} = 25 \mu C$.