Question 706: Electrostatics & Capacitance

Question

In the given circuit, the switch S is closed at time $$t=0$$. The charge Q on the capacitor at any instant t is given by $$Q(t)=Q_0(1-e^{-\alpha t})$$. Find the value of Q$$_0$$ and $$\alpha$$ in terms of given parameters shown in the circuit.

Accepted Answer

Given, $$Q=Q_{0}\left(1-e^{-\alpha t}ight)$$
Where, Q$$_0$$ is the steady state charge stored in 
the capacitor.
Q$$_0$$ = C[PD across capacitor in steady state]
$$\mathrm{Q}_{0}=C\left(\frac{\mathrm{V}}{\mathrm{R}_{1}+\mathrm{R}_{2}}ight) \mathrm{R}_{2}$$
$$	herefore \quad \mathrm{Q}_{0}=\frac{\mathrm{CVR}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}$$
From, $$Q = {Q_0}(1 - {e^{ - \alpha t}})$$
$$\alpha$$ is $$\frac{1}{	au}$$ or $$\frac{1}{\mathrm{CR}_{e q}}$$
Here, $$\mathrm{R}_{e q}$$ is equivalent resistance across capacitor after short circuiting the battery.
$$\begin{aligned}
	herefore \quad R_{e q} & =\frac{\mathrm{R}_{1} R_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}} \
\alpha & =\frac{1}{C\left(\frac{R_{1} R_{2}}{R_{1}+R_{2}}ight)}=\frac{R_{1}+R_{2}}{C_{1} R_{2}}
\end{aligned}$$

Answer

$$\mathrm{Q_0=\frac{2CVR_2}{R_1+R_2};\alpha=\frac{(R_1+R_2)}{CR_1R_2}}$$

Answer

$$\mathrm{Q_0=\frac{CVR_2}{R_1+R_2};\alpha=\frac{(R_1+R_2)}{3CR_1R_2}}$$

Answer

$$\mathrm{Q_0=\frac{CVR_2}{R_1+R_2};\alpha=\frac{(R_1+R_2)}{CR_1R_2}}$$

Answer

$$\mathrm{Q_0=\frac{2CVR_2}{R_1+R_2};\alpha=\frac{(R_1+R_2)}{3CR_1R_2}}$$

Electrostatics & Capacitance

In the given circuit, the switch S is closed at time $t=0. The charge Q on the capacitor at any instant t is given by Q(t)=Q_0(1-e^{-\alpha t}). Find the value of Q_0 and \alpha$ in terms of given parameters shown in the circuit.

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