close switch Problem with capacitors discharging and charging When the switch is

Question

close switch Problem with capacitors discharging and charging When the switch is closed at t-0s, the capacitors in the fig ure are charged. 40 25 a) What is the time constant of this circuit? 15 10 1 5 b) At what time has the current in the 15 resistor decayed to half the value it had immediately after the switch was closed? 9V A 9V battery is added to the circuit as shown in the picture If the capacitors are not charged, right after the battery is connected... 40 25 c) What is the initial current through the 15 resistor?? 15 40 F 10 1 5 d) How long does it take for the current to drop to 1/e times its initial value?

Explanation / Answer

c) initial current through 15 ohm resistor is i =V/R = 9/55.4 = 0.163 A

d) i = Imax*(1-e^(-t/T))

Time constant is T = R*C = 55.4*20*10^-6 = 1.108*10^-3 sec

then

i = Imax/e

then

Imax/e = Imax*(1-e^(-t/T))

1/e = (1-e^(-t/(1.108*10^-3))

e = 2.718

then

1/2.178 = (1-2.718^(-t/(1.108*10^-3))

t = 6.81*10^-4 sec

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