Names______________________ Group__________
RC Circuits – Group Worksheet
1. Find the time constant for the three circuits shown. In
each case R=10,000  and C=10  F.
a) C*2R = 0.2 s
b) C*(R/2) = 0.05 s
c) This circuit has different charging and discharging time constants. While charging, the branch
with only R in it would not play a role in the charging process of the capacitor, so the “charging”
time constant would simply be R*C = 0.1 s.
While discharging, we will basically have a loop consisting of one capacitor and two R resistors,
thus the “discharging” time constant would be C*2R = 0.2 s.
This type of RC circuit is actually very common in the industry, because most of the time we
need the capacitor in the circuit to charge and discharge in different rates, and with having
different R’s in a circuit like this, we will be able to make that happen.
2. Consider the circuit shown in the figure. The
capacitor is initially uncharged. At t=0, the switch
SW is set to position a. (a) How many time
constants must elapse for the capacitor to have
90% of its final, equilibrium charge? After the
capacitor is fully charged, the switch is now set to
position b. (b) How many time constants must
elapse for the capacitor to lose 90% of its charge?
(c) How much charge is left after twice this time? Thrice?
a) Vc= Vo(1-e-t/RC) = 0.9 Vo
1- e-t/RC = 0.9 ⇒ e-t/RC = 0.1 ⇒ t/RC = -ln(0.1) ≈ 2.3 ⇒ t ≈ 2.3 RC
b) Vc= Vo e-t/RC = 0.9 Vo
e-t/RC = 0.9 ⇒ t/RC = -ln(0.9) ≈ 0.1 ⇒ t ≈ 0.1 RC
c) t = 0.2 RC ⇒ Vc = Vo e-0.2 = 0.82 Vo
t = 0.3 RC ⇒ Vc = Vo e-0.3 = 0.74 Vo
3. Initially switch SW has been open for a long time.
(a) What is the charge on the capacitor when the
switch SW has been open for a long time? (b) Next
the switch SW is closed for a long time. What is the
charge on the capacitor after the switch SW has been closed for a long time?
a) Q = CV3 = 20 * 10-6 * 3 = 6 * 10-5 C
b) After the switch is closed for a long time the circuit basically comes down to one equivalent
battery of V3 – V2 = 1 V , and an equivalent resistor of 6 ohms. The charge on the capacitor
at this time is:
Q = CV = C (V3 – i4) = 20*10-6 F(0.3V)=6*10-6C where you solve for the current from the
Kirchoff loop rule: 2V+3V-6i=0
4.
Consider the RC-circuit shown to the right. Here,
Ɛ=120V, R1=4.2kΩ, R2=4kΩ, R3=1kΩ and
C=4.7mF. Initially the switch SW is open for a long
time and the capacitor C is uncharged. The switch
then is closed at t=0. (a) Find the values of i1 and i2
immediately after closing the switch, i.e., at t=0. (b)
Find the values of i1 and i2 a long time after closing
the switch.
a) Immediately after closing the switch, the capacitor will act like a wire; thus we just have a
circuit with three resistors. The equivalent resistance for this case will be:
1
1
Req = R1 + (𝑅 + 𝑅 ) -1 = 4.2 + 0.8 = 5 kΩ
2
3
i1 = Ɛ / Req = 120V / 5 kΩ = 24 mA
From Kirchhoff’s junction rule we have: i1 = i2 + i3 , and from the loop rule for the small
loop we have i3R3 – i2R2 = 0 . Thus:
i1 = i2 +
𝑅2
i2
𝑅3
⇒ i2 = 24/5 = 4.8 mA
b) After the switch has been closed for a long time, no current will flow through the branch
with the capacitor in it, so we will have a circuit with only resistors R1 and R2, thus i1 = i2.
Req = R1 + R2 = 8.2 kΩ
i1 = i2 = Ɛ / Req = 120V / 8.2 kΩ = 14.6 mA