By evaluating ∫i^{2}Rdt, show that when a capacitor is charged by connecting it to a battery through a resistor, the energy dissipated as heat equals the energy stored in the capacitor.

#### Solution

The growth of charge on the capacitor at time t ,

\[Q = Q_0 \left( 1 - e^{- \frac{t}{RC}} \right)\]

\[i = \frac{dQ}{dt} = \left( \frac{Q_0}{RC} \right) e^{- \frac{t}{RC}}\]

Heat dissipated during time t_{1} to t_{2},

\[U = \int_{t_1}^{t_2} i^2 Rdt\]

\[ = \frac{{Q_0}^2}{2C}\left( e^{- \frac{2 t_1}{RC}} - e^{- \frac{2 t_2}{RC}} \right)\]

\[ \because Q_0 = C V_{0,} \]

\[ U = \frac{1}{2}C {V_0}^2 \left( e^{- \frac{2 t_1}{RC}} - e^{- \frac{2 t_2}{RC}} \right)\]

The potential difference across a capacitor at any time t,

\[V = V_0 \left( 1 - e^{- \frac{t}{RC}} \right)\]

The energy stored in the capacitor at any time t,

\[E = \frac{1}{2}C V^2 = \frac{1}{2}C {V_0}^2 \left( 1 - e^{- \frac{2t}{RC}} \right)^2\]

∴ The energy stored in the capacitor from t_{1} to t_{2},

\[E = \frac{1}{2}C {V_0}^2 \left( e^{- \frac{2 t_1}{RC}} \right) - \frac{1}{2}C {V_0}^2 \left( 1 - e^{- \frac{2 t_2}{RC}} \right)\]

\[ \Rightarrow E = \frac{1}{2}C {V_0}^2 \left( e^{- \frac{2 t_1}{RC}} - e^{- \frac{2 t_2}{RC}} \right)\]