Consider the above circuit, which has an idealbattery(E = 3 V),a resistor( R = 1
ID: 1671631 • Letter: C
Question
Consider the above circuit, which has an idealbattery(E = 3 V),a resistor(R = 10000 ohms), and two capacitors(C1 =2 µF and C2 = 3µF).Both capacitors are initially uncharged. At timet= 0 s, the switch is closed.
(a) At t = 0+ what is the current in the circuit?
I(O+) = A
(b) After a long time, what is the current in the circuit?
I(inf) = A
(c) After a long time, what is the potentialdifferenceVab = Vb -Va acrossC2?
Vab(inf) = V
(d) What is the value of the time constant for this circuit?
t = s
Explanation / Answer
a. When the switch is justclosed, t = 0 +,capacitors behave as a short circuit, hence the resistance of thecircuit Rnet = R = 10000 Current I(0+) = /Rnet = 3/ 10000 = 3* 10-4 A b. At t = ,capacitors are fully charged and behave as an open circuit,hence Rnet = I() = /Rnet = 3/ = 0c. Net capacitance is givenby 1/C = 1/C1 + 1/C2 C = 2* 3 / ( 2 + 3) = 1.2 F C = 2* 3 / ( 2 + 3) = 1.2 F Netcharge Q = C * Q = 1.2F * 3.0 = 3.6 C Since capacitors are in series, charge oneach will be the same. Voltage acrossC2 Vab (inf) = Q/ C2 = 3.6C / 3.0 F = 1.2 V = 3.6C / 3.0 F = 1.2 V d. Timeconstant = R* C = 10000* 1.2 * 10-6 = 1.2* 10-2 s = 1.2* 10-2 s