Question

A capacitor of capacitance C is connected to a battery of emf ε at t = 0 through a resistance R. Find the maximum rate at which energy is stored in the capacitor. When does the rate have this maximum value?

Answer 1

The rate of growth of charge for the capacitor,
q = εC (1 − ${\mathrm{e}}^{\frac{-\mathrm{t}}{\mathrm{RC}}}$)

Let E be the energy stored inside the capacitor. Then,
$$\begin{gathered} E=\frac{q^2}{2C}=\frac{{\epsilon }^2{C}^2}{2C}{\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)}^2 \\ \Rightarrow E=\frac{{\epsilon }^{2}C}{2}{\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)}^2 \end{gathered}$$

Let r be the rate of energy stored inside the capacitor. Then,
$$\begin{gathered} r=\frac{dE}{dt}=\frac{2{\epsilon }^{2}C}{2}\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left(-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left(-\frac{1}{RC}\right) \\ \Rightarrow r=\frac{{\epsilon }^2}{R}\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left({\mathrm{e}}^{-\frac{t}{RC}}\right) \end{gathered}$$
$\frac{dr}{dt}=\frac{{\epsilon }^2}{R}\left[\left(-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left(-\frac{1}{RC}\right)\left({\mathrm{e}}^{-\frac{t}{RC}}\right)+\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left({\mathrm{e}}^{-\frac{t}{RC}}\right)\left(-\frac{1}{RC}\right)\right]$
For r to be maximum, $\frac{dr}{dt}=0$
$$\begin{gathered} \Rightarrow \frac{{\epsilon }^2}{R}\left[\left(-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left(-\frac{1}{RC}\right)\left({\mathrm{e}}^{-\frac{t}{RC}}\right)+\left(1-{\mathrm{e}}^{-\frac{t}{RC}}\right)\left({\mathrm{e}}^{-\frac{t}{RC}}\right)\left(-\frac{1}{RC}\right)\right]=0 \\ \Rightarrow \left[\frac{{\mathrm{e}}^{-{\displaystyle \frac{2t}{RC}}}}{RC}+\frac{{\mathrm{e}}^{-{\displaystyle \frac{2t}{RC}}}}{RC}-\frac{{\mathrm{e}}^{{\displaystyle \frac{-t}{RC}}}}{RC}\right]=0 \\ \Rightarrow 2{\mathrm{e}}^{-\frac{2t}{RC}}={\mathrm{e}}^{-\frac{t}{RC}} \\ \Rightarrow {\mathrm{e}}^{-\frac{t}{RC}}=\frac{1}{2} \\ \Rightarrow -\frac{t}{RC}=-\ln 2 \\ \Rightarrow t=RC\ln 2 \end{gathered}$$