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,

\[q = \epsilon C \left(1 − e^\frac{- t}{RC}\right)\]

Let E be the energy stored inside the capacitor. Then,

\[E = \frac{q^2}{2C} = \frac{\epsilon^2 C^2}{2C} \left( 1 - e^{- \frac{t}{RC}} \right)^2 \]

\[ \Rightarrow E = \frac{\epsilon^2 C}{2} \left( 1 - e^{- \frac{t}{RC}} \right)^2\]

Let r be the rate of energy stored inside the capacitor. Then,

\[r = \frac{dE}{dt} = \frac{2 \epsilon^2 C}{2}\left( 1 - e^{- \frac{t}{RC}} \right)\left( - e^{- \frac{t}{RC}} \right)\left( - \frac{1}{RC} \right)\]

\[ \Rightarrow r = \frac{\epsilon^2}{R}\left( 1 - e^{- \frac{t}{RC}} \right)\left( e^{- \frac{t}{RC}} \right)\]

\[\frac{dr}{dt} = \frac{\epsilon^2}{R}\left[ \left( - e^{- \frac{t}{RC}} \right)\left( - \frac{1}{RC} \right)\left( e^{- \frac{t}{RC}} \right) + \left( 1 - e^{- \frac{t}{RC}} \right)\left( e^{- \frac{t}{RC}} \right)\left( - \frac{1}{RC} \right) \right]\]

For r to be maximum,

\[\frac{dr}{dt} = 0\]

\[\Rightarrow \frac{\epsilon^2}{R}\left[ \left( - e^{- \frac{t}{RC}} \right)\left( - \frac{1}{RC} \right)\left( e^{- \frac{t}{RC}} \right) + \left( 1 - e^{- \frac{t}{RC}} \right)\left( e^{- \frac{t}{RC}} \right)\left( - \frac{1}{RC} \right) \right] = 0\]

\[ \Rightarrow \left[ \frac{e^{- \frac{2t}{RC}}}{RC} + \frac{e^{- \frac{2t}{RC}}}{RC} - \frac{e^\frac{- t}{RC}}{RC} \right] = 0\]

\[ \Rightarrow 2 e^{- \frac{2t}{RC}} = e^{- \frac{t}{RC}} \]

\[ \Rightarrow e^{- \frac{t}{RC}} = \frac{1}{2}\]

\[ \Rightarrow - \frac{t}{RC} = - \ln2\]

\[ \Rightarrow t = RC\ln2\]