For the capacitor circuit drawn to the right, the capacitor is initially uncharg

Question

For the capacitor circuit drawn to the right, the capacitor is initially uncharged. The switch labeled S is now closed for more than ten time constants.

For the capacitor circuit drawn to the right, the capacitor is initially uncharged. The switch S is now closed for more than 10 time constants. What is the final voltage that the capacitor charges to? At time t = 0 switch S is now opened, discharging the capacitor. (if you did not get an answer for Part A) use a maximum voltage of 5 volts for the problem below ) What is the time constant of the discharging circuit? What is the voltage at time T = 0.35 seconds? at what time will the voltage on the capacitor be 4 volts?

Explanation / Answer

A) Final volatge is V = 30/3 = 10 V

B)
For discharging Time constant is T = R*C = (10+10)*10^3*10^10^-6 = 0.2 S

C) v = 10*e^(-t/T)

v = 10*e^(-0.35/0.2) = 1.73 V

d) v = 10*e^(-t/T)

4 = 10*e^(-t/T)

-t/T = ln(4/10) = -0.916

t = 0.916*T = 0.916*0.2 = 0.183 S

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