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 ,
$Q = Q_{0} (1 - e^{- \frac{t}{R C}})$
$i = \frac{d Q}{d t} = (\frac{Q_{0}}{R C}) e^{- \frac{t}{R C}}$
Heat dissipated during time t₁ to t₂,
$U = \int_{t_{1}}^{t_{2}} i^{2} R d t = \frac{{Q_{0}}^{2}}{2 C} (e^{- \frac{2 t_{1}}{R C}} - e^{- \frac{2 t_{2}}{R C}}) ∵ Q_{0} = C V_{0,} U = \frac{1}{2} C {V_{0}}^{2} (e^{- \frac{2 t_{1}}{R C}} - e^{- \frac{2 t_{2}}{R C}})$

The potential difference across a capacitor at any time t,
$V = V_{0} (1 - e^{- \frac{t}{R C}})$
The energy stored in the capacitor at any time t,
$E = \frac{1}{2} C V^{2} = \frac{1}{2} C {V_{0}}^{2} {(1 - e^{- \frac{2 t}{R C}})}^{2}$
∴ The energy stored in the capacitor from t₁ to t₂,
$E = \frac{1}{2} C {V_{0}}^{2} (e^{- \frac{2 t_{1}}{R C}}) - \frac{1}{2} C {V_{0}}^{2} (1 - e^{- \frac{2 t_{2}}{R C}}) \Rightarrow E = \frac{1}{2} C {V_{0}}^{2} (e^{- \frac{2 t_{1}}{R C}} - e^{- \frac{2 t_{2}}{R C}})$

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