Question 174 of 180

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 $\left({C}_{eq}=\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}\right)$

A${C}_{1}E\left[1-\exp ⁡\left(-\dfrac{tR}{{C}_{1}}\right)\right]$B${C}_{eq}E\text{e}\text{x}\text{p}\left[-\dfrac{t}{R{C}_{eq}}\right]$C${C}_{eq}E\left[1-\exp ⁡\left(-\dfrac{t}{R{C}_{eq}}\right)\right]$D${C}_{2}E\left[1-\exp ⁡\left(-\dfrac{t}{R{C}_{2}}\right)\right]$

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Answer: (C) ${C}_{eq}E\left[1-\exp ⁡\left(-\dfrac{t}{R{C}_{eq}}\right)\right]$

In given circuit capacitors ${C}_{1}$ and ${C}_{2}$ are connected in series , so equivalent capacitance will be
$\dfrac{1}{{C}_{eq}}=\dfrac{1}{{C}_{1}}+\dfrac{1}{{C}_{2}} \Rightarrow {C}_{eq}=\dfrac{{C}_{1}{C}_{2}}{{C}_{1}+{C}_{2}}$
at $t=0$, the switch is closed and capacitors start charged, at time $t$
charge on the capacitor is given by
$q={q}_{0}\left[1-{e}^{-t/\tau }\right]$
where ${q}_{0}$ is the maximum charge on the capacitor at steady state and $\tau$ is the time constant.
and ${q}_{0}={C}_{eq}E$ and $\tau =R{C}_{eq}$
Then from the above equation
$q={C}_{eq}E\left[1-{e}^{-t/R{C}_{eq}}\right]$
$∴q={C}_{eq}E\left[1-exp\left(\dfrac{t}{R{C}_{eq}}\right)\right]$
Both capacitor will have same charge as they are connected in series