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SPH4UW
RC Circuits
Capacitors


Charge on Capacitors cannot change instantly.
Short term behavior of Capacitor:




If the capacitor starts with no charge, it has no potential
difference across it and acts as a wire.
If the capacitor starts with charge, it has a potential difference
across it and acts as a battery.
Current through a Capacitor eventually goes to zero.
Long term behavior of Capacitor:


If the capacitor is charging, when fully charged no current flows
and capacitor acts as an open circuit.
If capacitor is discharging, potential difference goes to zero and
no current flows.
Combine R+C Circuits

Gives time dependence
Current is not constant I(t)
 Charge is not constant q(t)


Used for timing
Pacemaker
 Intermittent windshield wipers


Models nervous system include R, C

Sports Trivia: How soon after starting gun
can you run w/o getting a False Start?
What is the time constant?

The time constant  = RC.

Given a capacitor starting with no charge, the
time constant is the amount of time an RC
circuit takes to charge a capacitor to about
63.2% of its final value.

The time constant is the amount of time an RC
circuit takes to discharge a capacitor by about
63.2% of its original value.
RC Circuits: Charging
The switches are originally open and the capacitor is uncharged.
Then switch S1 is closed.


KVL: + e - I(t)R - q(t) / C = 0
Just after S1 is closed: q =q0




R
+
Capacitor is uncharged (no time has e
passed so charge hasn’t changed yet)
- I
+
-
e - I0R = 0  I0 = e / R
Long time after: Ic= 0

Capacitor is fully charged

+ e - q/C =0  q = e C
q
1
Intermediate (more complex)
q(t) = q(1-e-t/RC)
I(t) = I0e-t/RC
C
S2
S1
Q

+
RC
2RC
q
f( x ) 0.5
00
0
1
t
2
3
4
Capacitor “Rules of Thumb”

Initially uncharged capacitor:
acts like a wire (short circuit) at t = 0
 acts like an open circuit (broken wire) as t



Initially charged capacitor:

acts like a battery at t = 0
Energy stored by a capacitor
is
1 Q2 1
1
U
 QV  CV 2
2 C 2
2
Time Constant Demo
Each circuit has a 1 F capacitor charged to 100 Volts.
1Q
When the switch is closed:
U
2. I 



2V
R
Which system will be brightest?
V
Which lights will stay on longest? 1. I  2 R
Which lights consumes more energy?
Same:
 2R
V
I
RT
 = 2RC

V
2R
2 C
1
 QV
2
1
 CV 2
2
1
2
 1F 100V 
2
 5000 J
1
1 1
 
RT R1 R2
RT  R1  R2
1
2
2
R
V
I
RT

2
 = RC/2

2V
R
Practice!
+
Calculate current immediately after switch is closed:
+ e - I0R - q0/C = 0
+ e - I0R - 0 = 0
I0 = e /R
20V
I0 
 2A
10W
I  I 0e

e
R
e
0.5
RC
+
-
0.5
20 100.5
20
 e 0.03  e100.03  0.38 Amps
10
10
C
E
S1
R=10W
C=30 mF
E =20 Volts
Calculate current after switch has been closed for a long time:
After a long time current through capacitor
is zero!
Calculate charge on capacitor after switch has been closed for a long time:
+ e - IR - q∞/C = 0
+ e - 0 - q∞ /C = 0
q∞ = eC
-
I
Calculate current after switch has been closed for 0.5 seconds:
t
RC
R
q   20V   30 103 F   0.6C
+
-
Understanding
After switch 1 has been closed for a long time, it is opened and
switch 2 is closed. What is the current through the right resistor just
after switch 2 is closed?
+ 2R -
1) IR = 0
3) IR = V/(2R)
IR
2) IR = V/(3R)
4) IR = V/R
e+
-
KVL: +q0/C-IR = 0
+
+
C
-
S1
R
-
S2
Recall q is charge on capacitor after charging:
q0=VC (since charged w/ switch 2 open!)
+V - IR = 0
 I = V/R
Understanding
Both switches are closed. What is the final charge on the
capacitor after the switches have been closed a long
time?
+ 2R -
1) Q = 0
3) Q = C ε /2
2) Q = C ε /3
4) Q = C ε
IR
e
+
-
+
+
C
-
R
-
KVL (right loop): +Q/C-IR = 0
KVL (outside loop): +ε - I(2R) - IR = 0
I = ε /(3R)
KVL: +Q/C - ε /(3R) R = 0
Q = C ε /3
S2
Charging: Intermediate Times
Calculate the charge on the capacitor 310-3 seconds
after switch 1 is closed.
R1 = 20 W
q(t) =
=
R2 = 40 W
q(1-e-t/R2C)
ε = 50 Volts
-3 /(4010010-6))
-310
q(1-e
)
+
= q (0.53)
Recall q = ε C
R2
-
Ib
e
+
C
-
= (50)(100x10-6) (0.53)
= 2.7 x10-3 Coulombs
C = 100mF
+
-
R1
S2
S1
RC Circuits: Discharging



KVL: - q(t) / C - I(t) R = 0
Just after…: q=q0
 Capacitor is still fully charged
 +q0 / C + I0 R = 0
 I0 = q0/(RC)
+
R
e
-
I
-
C
Long time after: Ic=0
 Capacitor is discharged
 q / C = 0  q = 0
Intermediate (more complex)
 q(t) = q0 e-t/RC
 Ic(t) = I0 e-t/RC
+
-
S1
RC
1

+
S2
2RC
1
q
f( x ) 0.5
0.0183156
0
0
1
t
2
x
3
4
4
RC Challenge
After being closed for a long time,
the switch is opened. What is the
charge Q on the capacitor 0.06
seconds after the switch is
opened?
1) 0.368 q0
2) 0.632 q0
3) 0.135 q0
4) 0.865 q0
q(t) = q0 e-t/RC
= q0
-3))
-0.06
/(4(1510
(e
)
= q0 (0.368)
E = 24 Volts
R=4W
C = 15 mF
2R
C
E
S1
R
RC Summary
Charging
q(t) = q(1-e-t/RC)
V(t) = V(1-e-t/RC)
I(t) = I0e-t/RC
Discharging
q(t) = q0e-t/RC
V(t) = V0e-t/RC
I(t) = I0e-t/RC
Time Constant  = RC
Large  means long time to charge/discharge
Short term: Charge doesn’t change (often zero or
max)
Long term: Current through capacitor is zero.