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Question
By evaluating ∫i2Rdt, 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.
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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 t1 to t2,
\[U = \int_{t_1}^{t_2} i^2 Rdt\]
\[ = \frac{{Q_0}^2}{2C}\left( e^{- \frac{2 t_1}{RC}} - e^{- \frac{2 t_2}{RC}} \right)\]
\[ \because Q_0 = C V_{0,} \]
\[ U = \frac{1}{2}C {V_0}^2 \left( e^{- \frac{2 t_1}{RC}} - e^{- \frac{2 t_2}{RC}} \right)\]
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 t1 to t2,
\[E = \frac{1}{2}C {V_0}^2 \left( e^{- \frac{2 t_1}{RC}} \right) - \frac{1}{2}C {V_0}^2 \left( 1 - e^{- \frac{2 t_2}{RC}} \right)\]
\[ \Rightarrow E = \frac{1}{2}C {V_0}^2 \left( e^{- \frac{2 t_1}{RC}} - e^{- \frac{2 t_2}{RC}} \right)\]
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