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 S is now closed for more than 10 time constants.

a) 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)

b) what is the time constant of the discharging circuit?

c) what is the voltage at time T= 0.35 seconds?

d) at what time will the voltage on the capacitor be 4 volts?

Explanation / Answer

solution:-

V = Vo(1 - e-t/RC )

THIS IS FOR A CHARGING CIRCUIT

SO, after 10 time constants

V = Vo ( 1 - e-10 ) which is almost equal to Vo

so Vo across capacitor should be 5/10 + 10 + 5 *30

= 6 V( Maximum across capacitor)

now to find time constant of discharging circuit replace battery by a straight wire and see how many resistors are in series with capacitor

   so time constant = (10+10) * 10uF

= 0.2 seconds

c. V = Vo e - t/time const. ( t = 0.35 sec)

= 1.05 V

d. 4 = 6 e- t/0.2

so t = 0.08 sec

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