Question

The charge on each of the capacitors 0.16 ms after the switch S is closed in figure is :

• A. 24 $\mu$C
• B. 26.8 $\mu$C
• C. 25.2 $\mu$C
• D. 40 $\mu$C

Answer 1

Answer: (C)

25.2 $\mu$C

Answer 2

The problem asks for the charge on each of the capacitors 0.16 ms after the switch S is closed.

1. Determine the equivalent capacitance ($C_{eq}$): The two capacitors, $C_1 = 4.0 \mu F$ and $C_2 = 4.0 \mu F$, are connected in parallel. For capacitors in parallel, the equivalent capacitance is the sum of individual capacitances: $C_{eq} = C_1 + C_2 = 4.0 \mu F + 4.0 \mu F = 8.0 \mu F$

2. Calculate the time constant ($\tau$) of the RC circuit: The circuit consists of a resistor $R = 20 \Omega$ and the equivalent capacitor $C_{eq}$. The time constant is given by: $\tau = R \times C_{eq}$ $\tau = 20 \, \Omega \times (8.0 \times 10^{-6} \, F)$ $\tau = 160 \times 10^{-6} \, s = 0.16 \times 10^{-3} \, s = 0.16 \, ms$

3. Determine the maximum charge ($Q_{max}$) on the equivalent capacitor: The maximum charge occurs when the capacitor is fully charged to the source voltage $V = 10.0 \, V$. $Q_{max} = C_{eq} \times V$ $Q_{max} = (8.0 \times 10^{-6} \, F) \times (10.0 \, V)$ $Q_{max} = 80 \times 10^{-6} \, C = 80 \, \mu C$

4. Calculate the total charge ($Q(t)$) on the equivalent capacitor at time $t = 0.16 \, ms$: The charge on a capacitor in an RC charging circuit at time $t$ is given by: $Q(t) = Q_{max} (1 - e^{-t/\tau})$ Given $t = 0.16 \, ms$ and we calculated $\tau = 0.16 \, ms$. So, $t/\tau = 0.16 \, ms / 0.16 \, ms = 1$. $Q(t) = Q_{max} (1 - e^{-1})$ Using the approximation $e^{-1} \approx 0.37$: $Q(t) = 80 \, \mu C (1 - 0.37)$ $Q(t) = 80 \, \mu C (0.63)$ $Q(t) = 50.4 \, \mu C$

5. Determine the charge on each capacitor: Since the two capacitors are identical ($C_1 = C_2$) and are connected in parallel, the total charge $Q(t)$ will be equally distributed between them. Charge on each capacitor ($Q_{each}$) $= \frac{Q(t)}{2}$ $Q_{each} = \frac{50.4 \, \mu C}{2}$ $Q_{each} = 25.2 \, \mu C$