You have the following RC circuit: V = 52V C = 0.1 times 10-6F and R = 3.1M Ohm

Question

You have the following RC circuit: V = 52V C = 0.1 times 10-6F and R = 3.1M Ohm The circuit is initially uncharged. It is then allowed to charge. It is then left hooked up to the battery for a long time. Then its allowed to discharge completely. Answer the following: What is the value of the time constant? What is the maximum charge on the capacitor? At what time, after it began charging, is voltage across the capacitor 35V? At what time after it started discharging is its charge 40 percent of its maximum charge?

Explanation / Answer

time constant, = R*C = 0.31 s
maximum charge = C*V = .1*10^-6*52 = 5.2*10^-6 C

using the equation,
V = V0(e^(-t/))     where V0 = 52

so, 35 = 52*(e^(-t/0.31))

so, t = 0.122 s <------answer(c)

using the same equation,

.4 = 1(e^-t/.31)

so, t = 0.284 s    <---------------answer(d)

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