Question

In the circuit shown in the figure, a voltage is applied between points $A$ and $B$ which changes with time as
$$E_0=\{\begin{array}{cc}kt& \text{for}0\le t\le t_0\\ k{t}_0& \text{for}t>t_0\end{array}$$ Plot the variation of potential difference $E$ as a function of time.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

For $0\le t\le t_0$, $E_0=\frac{q}{C}+Ri$

Where $q$ and $i$ are instantaneous values of charge on the capacitor and the current.

$kt=\frac{q}{C}+Ri$

Differentiating with respect to time $t$,

$k=\frac{1}{C}\frac{dq}{dt}+R\frac{di}{dt}$

$\Rightarrow k-\frac{i}{C}=R\frac{di}{dt}$

Thus we get, $\underset{0}{\overset{i}{\int }}\frac{di}{k-\frac{i}{C}}=\frac{1}{R}\underset{0}{\overset{i}{\int }}dt$

$\Rightarrow [\ln (k-\frac{i}{C}){]}_0^i=-\frac{t}{RC}$

$\Rightarrow \ln (1-\frac{i}{kC})=-\frac{t}{RC}$

$\Rightarrow i=kC(1-e^{-\frac{t}{RC}})$

$\therefore E=Ri=kcR(1-e^{-\frac{t}{RC}})$

Hence, voltage across $C$ and $D$ increases as per above equation till $t=t_0$

Let the current $\left(i\right)$ and voltage $\left(E\right)$ at $t=t_0$ be

$i_1=kC(1-e^{-\frac{t_0}{RC}})$ and $E_1=Ri=kcR(1-e^{-\frac{t_0}{RC}})$

For $t>t_0$ , the applied voltage remains constant at $k{t}_0$.

For simplicity, let's begin our time count from this instant itself.

$k{t}_0=\frac{q}{C}+iR$

Differentiating wrt ${}^{\prime }{t}^{\prime }$ on both sides,

$\frac{1}{C}\frac{dq}{dt}+R\frac{di}{dt}=0$

Thus we get, ${\int }_{i_1}^i\frac{di}{i}=-\frac{1}{RC}{\int }_0^{t}dt$

$\Rightarrow \ln (\frac{i}{i_1})=-\frac{t}{RC}$

$\therefore i=i_1{e}^{-\frac{t}{RC}}$ and $E=i_{1}R{e}^{-\frac{t}{RC}}$

Hence, $E$ decreases exponentially after $t=t_0$