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Question
A capacitor of capacitance 100 μF is connected across a battery of emf 6 V through a resistance of 20 kΩ for 4 s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 4 s after the battery is disconnected?
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Solution
Given:-
Capacitance of capacitor ,C = 100 μF
Emf of battery ,E = 6V
Resistance ,R = 20 kΩ
Time for charging , t1 = 4 s
Time for discharging ,t2 = 4 s
During charging of the capacitor, the growth of charge across it,
\[Q = CE \left( 1 - e^{- \frac{t_1}{RC}} \right)\]
\[\frac{t_1}{RC} = \frac{4}{20 \times {10}^3 \times 100 \times {10}^{- 6}}\]
\[ = 2\]
\[ \Rightarrow Q = 6 \times {10}^{- 4} \left( 1 - e^{- 2} \right)\]
\[ = 5 . 187 \times {10}^{- 4} C\]
This is the amount of charge developed on the capacitor after 4s.
During discharging of the capacitor, the decay of charge across it,
\[Q' = Q\left( e^{- \frac{t}{RC}} \right)\]
\[ = 5 . 184 \times {10}^{- 4} \times e^{- 2} \]
\[ = 0 . 7 \times {10}^{- 4} C = 70 \mu C\]
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