Assume that the initial voltage of the capacitor is V(t = 0) = V_0 = 10.0 V. The

Question

Assume that the initial voltage of the capacitor is V(t = 0) = V_0 = 10.0 V. The capacitor has a value C = 10 nF and the resistor has a value R = 1 k Ohm. The switch is closed at t = 0. R What is the time constant, t, in seconds? Determine the maximum current that flows through the resistor. The capacitor is fully charged and then the switch is opened to discharge it. If the voltage across the capacitor, V_c, needs to be less than 0.01% of the fully charged voltage before you can consider it settled, how many time constants will you have to wait to be sure? N is the smallest integer.

Explanation / Answer

a) time constant is T = R*C = 1000*10*10^-9 = 10^-5 sec

b) amximum current is Imax = V/R = 10/1000 = 0.01 A

C) while discharging

V = Vo*e^(-t/T)

(0.01/100)*Vo = Vo*e^(-t/T)

0.01/100 = e^(-t/10^-5)

solving we get t = 9.2*10^-5 sec

So required time constants are N = 9.2 = 9 time constants

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