The switch S is initially in the open position shown, and the capacitors uncharg

ID: 2208456 • Letter: T

Question

The switch S is initially in the

open position shown, and the capacitors uncharged. A voltmeter (not

shown) is used to measure the potential difference across resistor R1.

a. At time t = 0, the switch is moved to position A. Determine the

voltmeter reading for the time immediately after t = 0.

b. How much time has passed when the voltmeter across resistor R1 reads 4.0 Volts? What is this time in terms of the

time constant, ?, of this circuit?

c. After a time much longer than 5?, a measurement of potential difference across R1 is again taken. Determine for

this later time each of the following.

i. The voltmeter reading

ii. The charge on the capacitor

Please ask if you have any doubt.I will help you.

Switch is moved to A :

The time constant of the circuit is given by

= Req C = (10 + 20)(15 F) = 450 s or 0.45 ms.

At time t = 0,capacitor acts as short circuit ,

therefore :

I = 20V/30 ,

a) V(R1) = (10)(20V/30) = 20/3 V.

b) First lets find voltage across the capacitor .

Vc(t) = Vc() + (Vc(0) - Vc())e-t/ .

Now Vc(0) = 0 and at t = ,capacitor becomes open circuit.

Thus Vc() = 20V.

Vc(t) = 20(1 - e-t/) ,where = 0.45 ms.

The voltage across R1 is given by :

V(R1) = (1/3) (20 V - Vc(t)) = (20/3) e-t/ .

4 = (20/3) e-t/ gives

t = 0.511 = 0.23 ms .

c) at t > 5 or t = the capacutor would have reached its saturation .

Vc() = 20V

i . V(R1) (t = ) = 0.

ii . Q = CVc = (15 F)(20V) = 300 C .

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