Question

In the following circuit, the switch $S$ is closed at $t=0$. The charge on the capacitor $C_1$ as a function of time will be given by $\left(C_{eq}=\frac{C_1{C}_2}{C_1+C_2}\right)$

• A. $C_{1}E[1-exp(-tR/C_1)]$
• B. $C_{2}E[1-exp(-tR/C_2)]$
• C. $C_{eq}E[1-exp(-tR/C_{eq})]$
• D. $C_{eq}exp(-tR/C_{eq})$

Answer 1

Answer: (C)

Since the two capacitors are joined in series, charge that passes through each will be same.

And equivalent capacitor is given by $C_{eq}=\left[\frac{C_1{C}_2}{C_1+C_2}\right]$.
We know that growth of charge with time is given by $Q=Q_0\left[1-\exp \left(-\frac{t}{C_{eq}R}\right)\right]$.
Charge on the capacitor $C_1$ is given by $Q_0=C_{eq}E.$ Therefore,
$Q=E{C}_{eq}\left[1-\exp \left(-\frac{t}{C_{eq}R}\right)\right]$