A capacitor of capacitance 200 µF is connected across a battery of emf 10.0 V through a resistance of 40 k Ω for 16.0 s. The battery is then replaced by a thick wire. What will be the charge on the capacitor 16.0 s after the battery is disconnected? (Given: e –2 = 0.135)
Text Solution
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(q = 20 × 10 – 4 (1 – e – 2 )e – 2 = 233.55 µC)
Sol. For charging
q 1 = CV(1 – e – t/RC ) = 20 × 10 – 4 (1 – e – 16/200 × 10 – 6 × 40 × 103 )
= 20 × 10 – 4 (1 – e – 2 )
For discharging
q 2 = q 1 e – t/RC = 20 × 10
– 4 (1 – e – 2 )
= 20 × 10 – 4 (1 – e – 2 ) e – 2 = 20 × 10 – 4 
= 233.55 µC
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