Question

An uncharged capacitor of capacitance $5\mu \text{F}$ is connected to a battery of EMF $50\text{V}$ at $t=0$ through a resistance of $2\text{M}\Omega$, then the time at which rate of energy stored is maximum is -

• A. $\left(10\ln 2\right)\text{s}$
• B. $(\ln 2\text{)}\text{s}$
• C. $\left(5\ln 2\right)\text{s}$
• D. $\left(0.5\ln 2\right)\text{s}$

Answer 1

The correct option is A $\left(10\ln 2\right)\text{s}$

The growth of charge on the capacitor during charging is given by $$q=q_0(1-e^{\frac{-t}{\tau }})$$
Energy stored in capacitor, $$E=\frac{q^2}{2C}$$
$\Rightarrow E=\frac{q_0^2}{2C}{(1-e^{-\frac{t}{\tau }})}^2$

$\Rightarrow E=\frac{(CV{)}^2}{2C}{(1-e^{-\frac{t}{\tau }})}^2\because [q_0=CV]$

$\Rightarrow E=\frac{C{V}^2}{2}{(1-e^{-\frac{t}{\tau }})}^2$

To get the rate of energy stored, differentiating the above equation with respect to time $t$,

$\Rightarrow \frac{dE}{dt}=\frac{C{V}^2}{2}\times 2\times (1-e^{-\frac{t}{\tau }})\times (0-e^{-\frac{t}{\tau }}\times -\frac{1}{\tau })$

$\Rightarrow \frac{dE}{dt}=\frac{C{V}^2}{\tau }(e^{-\frac{t}{\tau }}-e^{-\frac{2t}{\tau }})....\left(1\right)$

To get the time $t$ at which the rate of energy stored is maximum, differentiating the equation $\left(1\right)$ and equating to zero.

$\Rightarrow \frac{d^{2}E}{d{t}^2}=\frac{C{V}^2}{\tau }(e^{-\frac{t}{\tau }}\times (-\frac{1}{\tau })-e^{-\frac{2t}{\tau }}\times (-\frac{2}{\tau }))=0$

$\Rightarrow e^{(-\frac{t}{\tau })}=\frac{1}{2}$

Taking $\ln$ on both sides, we get,

$\Rightarrow t=\tau \ln 2$

As, $\tau =RC=(2\times {10}^6)\times (5\times {10}^{-6})=10\text{s}$

$\therefore t=\left(10\ln 2\right)\text{s}$

Hence, option $\left(A\right)$ is the correct answer.