Question

The plates of a capacitor of capacitance 10 μF, charged to 60 μC, are joined together by a wire of resistance 10 Ω at t = 0. Find the charge on the capacitor in the circuit at (a) t = 0 (b) t = 30 μs (c) t = 120 μs and (d) t = 1.0 ms.

Answer 1

Given:
Capacitance of the capacitor, C = 10 μF
Initial charge on capacitor, Q = 60 μC
Resistance of the circuit, R = 10 Ω

(a)Decay of charge on the capacitor,
$q=Q{\mathrm{e}}^{-\frac{t}{RC}}$
At t = 0,
q = Q = 60 μC

(b) At t = 30 μs,
$$\begin{gathered} q=Q.{\mathrm{e}}^{-\frac{t}{RC}} \\ \Rightarrow q=Q.{\mathrm{e}}^{-\frac{30\times {10}^{-6}}{10\times 10\times {10}^{-6}}} \\ \Rightarrow \mathrm{q}=60.{\mathrm{e}}^{-0.3} \\ \Rightarrow \mathrm{q}=44\mathrm{\mu C} \end{gathered}$$

(c) At t = 120 μs,
$$\begin{gathered} q=Q.{\mathrm{e}}^{-\frac{t}{RC}} \\ \Rightarrow q=Q.{\mathrm{e}}^{-\frac{120\times {10}^{-6}}{10\times 10\times {10}^{-6}}} \\ \Rightarrow q=60.{\mathrm{e}}^{-1.2} \\ \Rightarrow q=18\mathrm{\mu C} \end{gathered}$$

(d) At t = 1 ms,
$$\begin{gathered} q=Q.{\mathrm{e}}^{-\frac{t}{RC}} \\ \Rightarrow q=Q.{\mathrm{e}}^{-\frac{1\times {10}^{-3}}{10\times 10\times {10}^{-6}}} \\ \Rightarrow q=60.{\mathrm{e}}^{-10} \\ \Rightarrow q=0.003\mathrm{\mu C} \end{gathered}$$