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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.
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Solution
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,
\[q = Q . e^{- \frac{t}{RC}} \]
\[ \Rightarrow q = Q . e^{- \frac{1 \times {10}^{- 3}}{10 \times 10 \times {10}^{- 6}}} \]
\[ \Rightarrow q = 60 . e^{- 10} \]
\[ \Rightarrow q = 0 . 003 \mu C\]
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