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,

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

At t = 0,

q = Q = 60 μC

(b) At t = 30 μs,

\[q = Q .  e^{- \frac{t}{RC}} \]

\[ \Rightarrow q = Q .  e^{- \frac{30 \times {10}^{- 6}}{10 \times 10 \times {10}^{- 6}}} \]

\[ \Rightarrow q = 60 .  e^{- 0 . 3} \]

\[ \Rightarrow q = 44  \mu C\]

(c) At t = 120 μs,

\[q = Q .  e^{- \frac{t}{RC}} \]

\[ \Rightarrow q = Q .  e^{- \frac{120 \times {10}^{- 6}}{10 \times 10 \times {10}^{- 6}}} \]

\[ \Rightarrow q = 60 .  e^{- 1 . 2} \]

\[ \Rightarrow q = 18  \mu C\]

(d) At t = 1 ms,

\[q = Q .  e^{- \frac{t}{RC}} \]

\[ \Rightarrow q = Q .  e^{- \frac{1 \times {10}^{- 3}}{10 \times 10 \times {10}^{- 6}}} \]

\[ \Rightarrow q = 60 .  e^{- 10} \]

\[ \Rightarrow q = 0 . 003  \mu C\]