Question

A capacitor of capacitance $10\mu F$ is charged by connecting through a resistance of $20\Omega$ and a battery of $20V$. What is the energy supplied by the battery

• A. $2mJ$
• B. More than $2mJ$
• C. Cannot be predicted
• D. Less than $2mJ$

Answer 1

Answer: (B)

Given,

$C=10\mu F$

$R=20\Omega$

$V=20V$

We know, current in an RC circuit is given by
$i=\frac{V}{R}{e}^{\left(\frac{-t}{RC}\right)}$

Power dissipated by the resistor is given by
$P_R=i^{2}R={\left(\frac{V}{R}{e}^{\left(\frac{-t}{RC}\right)}\right)}^{2}R=\frac{V^2}{R}{e}^{\left(\frac{-2t}{RC}\right)}$

Energy consumed by the resistor $E_R={\int }_0^{\infty }{P}_{R}dt$

$E_R={\int }_0^{\infty }\frac{V^2}{R}{e}^{\left(\frac{-2t}{RC}\right)}dt$

$E_R=\frac{V^2}{R}{\int }_0^{\infty }{e}^{\left(\frac{-2t}{RC}\right)}dt$

$E_R=\frac{V^{2}RC}{R2}|e^0-e^{\infty }|=\frac{1}{2}C{V}^2$

$E_R=\frac{1}{2}\times 10\times {10}^{-6}\times {20}^2=2\times {10}^{-3}J=2mJ$

Energy consumed by resistor alone is equal to $2mJ$

$\therefore$ Energy supplied by battery must be $>2mJ$

Hence, the correct answer is OPTION C.