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

To find the charge on each capacitor after 0.16 ms, we follow these steps:

1. Calculate Equivalent Capacitance ($C_{eq}$): The two capacitors, $C_1 = 4.0 \, \mu F$ and $C_2 = 4.0 \, \mu F$, are connected in parallel. $C_{eq} = C_1 + C_2 = 4.0 \, \mu F + 4.0 \, \mu F = 8.0 \, \mu F = 8.0 \times 10^{-6} \, F$

2. Calculate Maximum Charge ($Q_{max}$): The maximum charge stored on the equivalent capacitor is: $Q_{max} = C_{eq} \times V = (8.0 \times 10^{-6} \, F) \times (10.0 \, V) = 80 \times 10^{-6} \, C = 80 \, \mu C$

3. Calculate Time Constant ($\tau$): The time constant for an RC circuit is given by: $\tau = R \times C_{eq}$ Given Resistance $R = 20 \, \Omega$. $\tau = (20 \, \Omega) \times (8.0 \times 10^{-6} \, F) = 160 \times 10^{-6} \, s = 0.16 \times 10^{-3} \, s = 0.16 \, ms$

4. Calculate Charge on Equivalent Capacitor at time t: $Q(t) = Q_{max} (1 - e^{-t/\tau})$ Given time $t = 0.16 \, ms$. Since $t = \tau = 0.16 \, ms$, the ratio $t/\tau = 1$. $Q(t) = 80 \, \mu C (1 - e^{-1})$ Using the value $e^{-1} \approx 0.36788$: $Q(t) = 80 \, \mu C (1 - 0.36788)$ $Q(t) = 80 \, \mu C (0.63212)$ $Q(t) = 50.5696 \, \mu C$

5. Calculate Charge on Each Capacitor: Charge on each capacitor = $\frac{Q(t)}{2} = \frac{50.5696 \, \mu C}{2} = 25.2848 \, \mu C \approx 25.3 \, \mu C$

Therefore, the charge on each capacitor is approximately $25.3 \, \mu C$, which is closest to option C: $25.2 \, \mu C$.