A time varying voltage is applied to the clamps $A and B$ such that voltage across the capacitor plates is as shown in the figure. Plot the time dependence of voltage across the terminals of the resistance $E and D$. If the given graph is between $V_{AB}$ and time. Plot the graph between $V_{ED}$ and time.

From O to A $V=at$

Here $a$ is a positive constant.

$\therefore at=\frac{q}{C}+iR$

Differentiating w.r.t. time we have 

$a=\frac{1}{C}\left(\frac{dq}{dt}\right)+\left(\frac{di}{dt}\right)R$

or, $\left(\frac{di}{dt}\right)R=a-\frac{i}{C} \left(\text{ as }i=\frac{dq}{dt}\right)$

${\int }_0^i\frac{di}{a-i/C}={\int }_0^t\frac{dt}{R}$

$\therefore i=aC\left(1-e^{-t/CR}\right)$

i.e. current in the circuit increases exponentially. 

$\therefore V_{ED}=iR=aCR(1-e^{-t/CR})$

or, $V_{ED}$ also increases exponentially. 

From A onwards When $V=\mathrm{constant} \left(\mathrm{say}{V}_0\right)$

$V_0=at$

$t=\frac{V_0}{a}$

$V_{ED}=aCR\left(1-e^{-V_0/aCR}\right)$

After this $V_{ED}$ will decrease exponentially. Hence, a rough graph is as shown in figure.