Question

The capacitor $C_1$ in the figure initially carries a charge $q_0$. When the switch $S_1$ and $S_2$ are closed, capacitor $C_1$ is connected to a resistor $R$ and a second capacitor $C_2$ which initially does not carry any charge. The current $i$, through resistor $R$ as a function of time $t$ is represented as:

• A. $i=\frac{q_0}{C_{1}R}{e}^{[\frac{-t(C_1+C_2)}{C_1{C}_{2}R}]}$
• B. $i=\frac{2q_0}{C_{1}R}{e}^{[\frac{-t}{C_{2}R}]}$
• C. $i=\frac{q_0}{C_1}{e}^{[\frac{-t(C_1+C_2)}{R}]}$
• D. $i=\frac{2q_0}{C_{2}R}{e}^{[\frac{-t}{C_{1}R}]}$

Answer 1

The correct option is A $i=\frac{q_0}{C_{1}R}{e}^{[\frac{-t(C_1+C_2)}{C_1{C}_{2}R}]}$

At steady state, capacitors will be in parallel because potential difference across them will be equal due to charge distribution and there will be no current through resistor $R$.

In parrallel combination, Charge will distribute in direct ratio of capacitance.

$q_1=\left(\frac{C_1}{C_1+C_2}\right)q_0$

$q_2=\left(\frac{C_2}{C_1+C_2}\right)q_0$

Initial EMF $E$ in the circuit is potential difference across capacitor $C_1$

$E_0=\frac{q_0}{C_1}$

Initial current (at $t=0$) in the circuit after both switches closed will be,

$i_0=\frac{E_0}{R}$

$\Rightarrow i_0=\frac{q_0}{C_{1}R}$

The capacitor $C_1$ starts discharging in given circuit as,$$i=i_0{e}^{\frac{-t}{\tau }}$$
Where, time constant $\tau =C_{eq}R$

$\tau =\left(\frac{C_1{C}_2}{C_1+C_2}\right)R$

Hence, at any time $t$ current in the circuit is given by:

$i=\frac{q_0}{C_{1}R}{e}^{[\frac{-t(C_1+C_2)}{C_1{C}_{2}R}]}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(b\right)$ is correct.