Download RC Circuits

Physics 102: Lecture 7
RC Circuits
• Textbook Section 18.10-11
Physics 102: Lecture 7, Slide 1
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?
Physics 102: Lecture 7, Slide 2
RC Circuits: Charging
The switches are originally open and the capacitor is uncharged. Then
switch S1 is closed.
• KVL:
• Just after S1 is closed: ______
+
R
e
– Capacitor is uncharged (no time has
– passed so charge hasn’t changed yet)
+
- I
-
+
-
– Capacitor is fully charged
q(t) = q(1-e-t/RC)
I(t) = I0e-t/RC
Physics 102: Lecture 7, Slide 3
S2
S1
• Long time after: __________
q
1
Q
• Intermediate (more complex)
C
RC
2RC
q
f( x ) 0.5
00
0
1
t
2
x
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
Physics 102: Lecture 7, Slide 4
Time Constant Demo
Each circuit has a 1 F capacitor charged to 100 Volts.
When the switch is closed:
• Which system will be brightest?
• Which lights will stay on longest?
• Which lights consumes more energy?
1
t = 2RC
Physics 102: Lecture 7, Slide 5
2
t = RC/2
Time Constant Demo
Each circuit has a 1 F capacitor charged to 100 Volts.
When the switch is closed:
• Which system will be brightest?
• Which lights will stay on longest?
• Which lights consumes more energy?
1
t = 2RC
Physics 102: Lecture 7, Slide 6
2 I=2V/R
1
Same
U=1/2 CV2
2
t = RC/2
Preflight 7.1, 7.3
Both switches are initially open, and the capacitor is uncharged.
What is the current through the battery just after switch S1 is
closed?
+ 2R -
1) Ib = 0
2) Ib = e /(3R)
3) Ib = e /(2R)
4) Ib = e /R
Ib
e
+
C
-
+
-
R
S2
S1
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? 1) I = 0
2) I = V/(3R)
b
Physics 102: Lecture 7, Slide 7
3) Ib = V/(2R)
b
4) Ib = V/R
ACT: RC Circuits
• Both switches are closed. What is the final
charge on the capacitor after the switches
have been closed a long time? + 2R IR
1) Q = 0
3) Q = C e /2
2) Q = C e /3
e
+
-
R
-
4) Q = C e
S1
Physics 102: Lecture 7, Slide 8
+
+
C
-
S2
ACT: RC Circuits
• Both switches are closed. What is the final
charge on the capacitor after the switches
+ 2R have been closed a long time?
1) Q = 0
2) Q = C e / 3
e
3) Q = C e /2
4) Q = C e
IR
+
+
C
-
+
-
R
-
KVL (right loop): -Q/C+IR = 0
S1
KVL (outside loop), - e + I(2R) + IR = 0
I = e /(3R)
combine: -Q/C + e /(3R) R = 0
Q=Cε/ 3
Physics 102: Lecture 7, Slide 9
S2
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:
Physics 102: Lecture 7, Slide 10
-
Practice!
+
Calculate current immediately after switch is closed:
- e + I0R + q0/C = 0
- e + I0R + 0 = 0
I0 = e /R
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
+
-
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
Physics 102: Lecture 7, Slide 12
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
Physics 102: Lecture 7, Slide 13
C = 100mF
+
-
R1
S2
S1
RC Circuits: Discharging
+
• KVL: ____________
R
• Just after…: ________
– Capacitor is still fully charged
e
+
-
I
-
C
S1
• Long time after: ____________
1
• Intermediate (more complex)
– q(t) = q0
– Ic(t) = I0 e-t/RC
e-t/RC
Physics 102: Lecture 7, Slide 14
+
-
RC
1
S2
2RC
q
f( x ) 0.5
0.0183156
0
0
1
t
2
x
3
4
4
RC Circuits: Discharging
+
• KVL: - q(t) / C - I(t) R = 0
R
• Just after…: q=q0
– Capacitor is still fully charged
– -q0 / C - I0 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
e
+
-
I
-
C
S1
1
RC
1
S2
2RC
q
f( x ) 0.5
0.0183156
0
Physics 102: Lecture 7, Slide 15
+
-
0
1
t
2
x
3
4
4
Preflight 7.5
• 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
3) IR = e /(2R)
Physics 102: Lecture 7, Slide 16
2) IR = e /(3R)
4) IR = e /R
+ 2R IR
+
+
C
-
e+
-
S1
R
-
S2
Preflight 7.5
• 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
2) IR = e /(3R)
3) IR = e /(2R)
KVL: -q0/C+IR = 0
4) IR = e /R
+ 2R IR
+
+
C
-
e+
-
S1
R
-
S2
Recall q is charge on capacitor after charging:
q0= e C (since charged w/ switch 2 open!)
- e + IR = 0
 I = e /R
Physics 102: Lecture 7, Slide 17
ACT: RC Challenge
E = 24 Volts
R=4W
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
Physics 102: Lecture 7, Slide 18
C = 15 mF
2R
C
E
S1
R
ACT: RC Challenge
E = 24 Volts
R=4W
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)
Physics 102: Lecture 7, Slide 19
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 t = RC
Large t means long time to charge/discharge
Short term: Charge doesn’t change (often zero or
max)
Long term: Current through capacitor is zero.
Physics 102: Lecture 7, Slide 20
See you next lecture!
• Read Textbook Sections 19.1-4
Physics 102: Lecture 7, Slide 21