Question
If a parallel-plate capacitor in a RC circuit has a 10 C charge on each plate when it is disconnected from the battery, and in 10 sec drops to 10% of its initial charge, what is its capacitance? (Assuming R = 10 ohms)
Not sure which formula to use/
Not sure which formula to use/
Answers
V = Vi e^-(t/RC)
.1 = e^-(10/10C) = e^-(1/C)
ln .1 = -1/C
.1 = e^-(10/10C) = e^-(1/C)
ln .1 = -1/C
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