Question

A capacitance C charged to a potential difference V is discharged by connecting its plates through a resistance R. Find the heat dissipated in one time constant after the connections are made. Do this by calculating ∫ i²R dt and also by finding the decrease in the energy stored in the capacitor.

Answer 1

Let Q₀ be the initial charge on the capacitor. Then,

Q₀ = CV

The charge on the capacitor at time t after the connections are made,

\[Q = Q_0 e^{- \frac{t}{RC}}\]

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

Heat dissipated during time t₁ to t₂,

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

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

Time constant = RC

Putting t₁ = 0 and t₂ = RC, we get:-

\[U = \frac{{Q_0}^2}{2C}\left( e^{- 0} - e^{- 2} \right)\]

\[ \because Q_0 = CV, \]

\[ U = \frac{CV^2}{2}\left( 1 - \frac{1}{e^2} \right)\]

Answer 2

Heat dissipated at any time = Energy stored at time 0 − Energy stored at time t

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

\[ \because t = RC, \]

\[ U = \frac{1}{2}C V^2 - \frac{1}{2}C V^2 e^{- 2} \]

\[ \Rightarrow U = \frac{1}{2}C V^2 \left( 1 - \frac{1}{e^2} \right)\]