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,
$Q=Q_0{e}^{-\frac{t}{RC}}$
$i=\frac{dQ}{dt}=-\left(\frac{Q_0}{RC}\right){\mathrm{e}}^{-\frac{t}{RC}}$
Heat dissipated during time t₁ to t₂,
$$\begin{gathered} U={\int }_{t_1}^{t_2}{i}^{2}Rdt \\ =\frac{{Q_0}^2}{2C}\left({\mathrm{e}}^{-\frac{2t_1}{RC}}-{\mathrm{e}}^{-\frac{2t_2}{RC}}\right) \end{gathered}$$
Time constant = RC
Putting t₁ = 0 and t₂ = RC, we get:
$$\begin{gathered} U=\frac{{Q_0}^2}{2C}\left({\mathrm{e}}^{-0}-{\mathrm{e}}^{-2}\right) \\ \because Q_0=CV, \\ U=\frac{CV}{2}\left(1-\frac{1}{{\mathrm{e}}^2}\right) \end{gathered}$$

Alternative method:
Heat dissipated at any time = Energy stored at time 0 − Energy stored at time t

$$\begin{gathered} \Rightarrow U=\frac{1}{2}C{V}^2-\frac{1}{2}C{V}^2{\mathrm{e}}^{-\frac{2t}{RC}} \\ \because t=RC, \\ U=\frac{1}{2}C{V}^2-\frac{1}{2}C{V}^2{\mathrm{e}}^{-2} \\ \Rightarrow U=\frac{1}{2}C{V}^2\left(1-\frac{1}{{\mathrm{e}}^2}\right) \end{gathered}$$