Electrostatics Ques 158
- In the circuit shown in figure, the battery is an ideal one, with emf $V$. The capacitor is initially uncharged. The switch $S$ is closed at time $t=0$.
$(1998,8 \mathrm{M})$
(a) Find the charge $Q$ on the capacitor at time $t$.
(b) Find the current in $A B$ at time $t$. What is its limiting value as $t \rightarrow \infty$ ?
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Answer:
Correct Answer: 158.(a) $Q=\frac{C V}{2}\left(1-e^{-2 t / 3 R C}\right)$
(b) $ i_2=\frac{V}{2 R}-\frac{V}{6 R} e^{-2 t / 3 R C}, \frac{V}{2 R}$
Solution:
- Let at any time $t$ charge on capacitor $C$ be $Q$ and currents are as shown. Since, charge $Q$ will increase with time $t$.
Therefore,
$$ i_{1}=\frac{d Q}{d t} $$
(a) Applying Kirchhoff ’s second law in loop $M N A B M$
$$ \begin{aligned} & & =\left(i-i_{1}\right) R+i R \\ \text { or } & V & =2 i R-i_{1} R \end{aligned} $$
Similarly, applying Kirchhoff’s second law in loop MNSTM, we have
$$ V=i_{1} R+\frac{Q}{C}+i R $$
Eliminating $i$ from Eqs. (i) and (ii), we get
$$ V=3 i_{1} R+\frac{2 Q}{C} $$
$3 i_{1} R=V-\frac{2 Q}{C}$ or $i_{1}=\frac{1}{3 R} \quad V-\frac{2 Q}{C}$
or
$$ \frac{d Q}{d t}=\frac{1}{3 R} \quad V-\frac{2 Q}{C} \quad \text { or } \frac{d Q}{V-\frac{2 Q}{C}}=\frac{d t}{3 R} $$
or $\int_{0}^{Q} \frac{d Q}{V-\frac{2 Q}{C}}=\int_{0}^{t} \frac{d t}{3 R}$
This equation gives $Q=\frac{C V}{2}\left(1-e^{-2 t / 3 R C}\right)$
(b)
$$ i_{1}=\frac{d Q}{d t}=\frac{V}{3 R} e^{-2 t / 3 R C} $$
From Eq. (i) $i=\frac{V+i_{1} R}{2 R}=\frac{V+\frac{V}{3} e^{-2 t / 3 R C}}{2 R}$
$\therefore$ Current through $A B$
$$ \begin{aligned} i_{2} & =i-i_{1}=\frac{V+\frac{V}{3} e^{-2 t / 3 R C}}{2 R}-\frac{V}{3 R} e^{-2 t / 3 R C} \\ i_{2} & =\frac{V}{2 R}-\frac{V}{6 R} e^{-2 t / 3 R C} \\ i_{2} & =\frac{V}{2 R} \text { as } t \rightarrow \infty \end{aligned} $$
BETA
