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Physics 1161 Lecture 11
RC Circuits
Time Constant Demo
Each circuit has a 0.5 F capacitor charged to 9 Volts.
When the switch is closed:
• Which system will be brightest?
• Which lights will stay on longest?
• Which lights consume more energy?
1
t = 2RC
2
t = RC/2
Time Constant Demo
Each circuit has a 0.5 F capacitor charged to 9 Volts.
When the switch is closed:
• Which system will be brightest?
• Which lights will stay on longest?
• Which lights consume more energy?
1
t = 2RC
2
t = RC/2
2 I=2V/R
1
Same U=1/2 CV2
Preflight 11.1, 11.3
Both switches are initially open, and the capacitor is uncharged.
What is the current through the battery just after switch S1 is
closed?
1) Ib = 0
+
15%
2) Ib = e /(3R) 4%
Ib
3) Ib = e /(2R) 62%
4) Ib = e /R
e+
-
19%
Both switches are initially open, and the
capacitor is uncharged.
What is the current through the battery
after switch 1 has been closed a long time?
50%
1) Ib = 0
2R -
0%
2) Ib = V/(3R)
35%
3) Ib = V/(2R)
C
+
-
R
S2
S1
15%
4) Ib = V/R
Practice!
R
Calculate current immediately after switch is closed:
C
E
Calculate current after switch has been closed for 0.5 seconds:
S1
R=10W
C=30 mF
E =20 Volts
Calculate current after switch has been closed for a long time:
Calculate charge on capacitor after switch has been closed for a long time:
-
Practice
+
Calculate current immediately after switch is closed:
+
-
e - I0 R - 0 = 0
I0 = e /R
Calculate current after switch has been closed for 0.5 seconds:
I  I 0e

e
R
e
0.5
RC
0.5
20 100.5
20

e 0.03 
e100.03
10
10
 0.38 A
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
e - I0R - q0/C = 0
t
RC
R
+
-
Both switches are closed. What is the final
charge on the capacitor after the switches have
been closed a long time?
1. Q = 0
2. Q = C e /3
3. Q = C e /2
4. Q = C e
+ 2R IR
+
+
C
-
R
-
25%
S1
25%
25%
2
3
25%
S2
1
4
Charging: Intermediate Times
Calculate the charge on the capacitor 310-3 seconds after switch 1 is closed.
R1 = 20 W
q(t) = q(1-e-t/RC)
R2 = 40 W
ε = 50 Volts
+
R2
-
C = 100mF
Ib
e+
-
C
+
-
R1
S2
S1
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
•
•
•
KLR: ____________
Just after…: ________
– Capacitor is still fully charged
R
e +
-
C
+
-
S1
Long time after: ____________
1
•
I
RC
1
S2
2RC
Intermediate (more complex)
– q(t) = q0 e-t/RC
– Ic(t) = I0 e-t/RC
q
f( x ) 0.5
0.0183156
0
0
1
t
2
x
3
4
4
RC Circuits: Discharging
•
•
•
•
KLR: q(t) / C - I(t) R = 0
Just after…: q=q0
– Capacitor is still fully charged
– q0 / C - I 0 R = 0
 I0 = q0/(RC)
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
R
e +
-
I
C
+
-
S1
RC
1
S2
2RC
1
q
f( x ) 0.5
0.0183156
0
0
1
t
2
x
3
4
4
Preflight 11.5
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?
1) IR = 0
+
2R -
27%
IR
2) IR = e /(3R) 27%
3) IR = e /(2R) 4%
4)
IR = e /R
42%
e+
-
KLR: -q0/C+IR = 0
Recall q is charge on capacitor after charging:
q0= e C (since charged w/ switch 2 open!)
- e + IR = 0
 I = e /R
+
+
C
-
R
-
S1
S2
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
E = 24 Volts
R=4W
2. 0.632 q0
C = 15 mF
2R
3. 0.135 q0
4. 0.865 q0
C
E
S1
R
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
E = 24 Volts
R=4W
2. 0.632 q0
C = 15 mF
2R
3. 0.135 q0
4. 0.865 q0
25%
1
25%
25%
2
3
25%
C
4
E
q(t) = q0 e-t/RC
= q0
-3))
-0.06
/(4(1510
(e
)
= q0 (0.368)
S1
R