Question

An uncharged capacitor of capacitance $100\mu \text{F}$ is connected to a battery of emf $20\text{V}$ at $t=0$ through a resistance $10\Omega$, then the maximum rate at which energy is stored in the capacitor is :

• A. $20\text{J/s}$
• B. $15\text{J/s}$
• C. $10\text{J/s}$
• D. $30\text{J/s}$

Answer 1

The correct option is C $10\text{J/s}$

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,

$\Rightarrow t=\tau \ln 2$

So, maximum rate of energy stored,

$\Rightarrow \frac{dE}{dt}\left(\text{max}\right)=\frac{C{V}^2}{\tau }(e^{-\frac{\tau \ln 2}{\tau }}-e^{-\frac{2\tau \ln 2}{\tau }})$

$\Rightarrow \frac{dE}{dt}\left(\text{max}\right)=\frac{C{V}^2}{\tau }(\frac{1}{2}-\frac{1}{4})$

As, $\tau =RC$

$\Rightarrow \frac{dE}{dt}\left(\text{max}\right)=\frac{C{V}^2}{4RC}$

$\Rightarrow \frac{dE}{dt}\left(\text{max}\right)=\frac{V^2}{4R}$

Substituting the values,

$\Rightarrow \frac{dE}{dt}\left(\text{max}\right)=\frac{{20}^2}{4\times 10}=10\text{J/s}$

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