Question 3: Electrostatics & Capacitance

Question

In the given circuit, find charge on capacitor after 1s of opening the switch at $t=\infty$.

Accepted Answer

At $t ightarrow \infty$, capacitor
Works as a open circuit as shown in the Figure.
Therefore, current flow in the circuit,
$\begin{aligned}
I_1 & =\frac{9}{(12+15) 10^3}=\frac{1}{3} 	imes 10^{-3} \mathrm{~A} \
	herefore \quad V_0 & =I_1 	imes 15 	imes 10^3 \
& =\frac{1}{3} 	imes 10^{-3} 	imes 15 	imes 10^3=5 \mathrm{~V}
\end{aligned}$
$	herefore$ Charge on capacitor of $5 \mu \mathrm{F}$,
$\begin{aligned}
q_0 & =C V_0=5 	imes 10^{-6} 	imes 5 \
& =25 	imes 10^{-6} \mathrm{C}=25 \mu \mathrm{C}
\end{aligned}$
When switch is open, then capacitor starts discharging across $(5+15=20 \mathrm{k} \Omega)$ and instantaneous charge on capacitor is given by
$\begin{aligned}
q & =q_0 e^{-t / R C} \
& =25 e^{-\frac{1}{20 	imes 10^3 	imes 5 	imes 10^{-6}}} \quad[\because t=1 \mathrm{~s}] \
& =25 e^{-10} \mu \mathrm{C}
\end{aligned}$

Answer

$20~ e^{-10} \mu \mathrm{C}$

Answer

$25 ~e^{-10} \mu \mathrm{C}$

Answer

$30 ~e^{-10} \mu \mathrm{C}$

Answer

$35 ~e^{-10} \mu \mathrm{C}$

Electrostatics & Capacitance

In the given circuit, find charge on capacitor after 1s of opening the switch at t=\infty.

Choose the correct answer: