Question

The charge $q$ on a plate of a condenser of capacity $C$ charged through a resistance $R$ by a steady voltage $V$ satisfies the differential equation $R\frac{dq}{dt}+\frac{q}{c}=V$.If $q=0$ at $t=0$ then the expression for the charge on the condenser is?

• A. $q=CV(1-cos\frac{-t}{RC})$
• B. $q=CV(1-sin\frac{-t}{RC})$
• C. $q=CV(1-e^{-t/RC}$
• D. $q=CV(1-cot\frac{-t}{RC})$

Answer 1

The correct option is C $q=CV(1-e^{-t/RC}$

We are given that $\frac{dq}{dt}+\frac{1}{RC}.q=\frac{V}{R}$.

This is a linear differential equation of the type $\frac{dy}{dx}+Py=Q$.

Its solution is $q.e{.}^{\int \left(1/RC\right)dt}=\int e^{\int \left(1/RC\right)dt}.\frac{V}{R}dt+k$

$\therefore q.e^{t/RC}=\int e^{t/RC}.\frac{V}{R}dt+k=\frac{V}{R}.\frac{e^{t/RC}}{\left(1/RC\right)}+k$

$=CV.e^{t/RC}+k$

By data when $t=0,q=0$ $\therefore k=-CV$

$\therefore q.e^{t/RC}=CV.e^{t/RC}-CV$

$\therefore q=CV-CV{e}^{-t/RC}=CV(1-e^{-1/RC})$

Further $i=\frac{dq}{dt}=CV.e^{-1/RC}.\frac{1}{RC}=\frac{V}{R}.e^{-t/RC}$