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Physics 212
Lecture 11
Today's Concept:
RC Circuits
(Circuits with resistors & capacitors & batteries)
Physics 212 Lecture 11, Slide 1
Main Point 1
First, just as in the last prelecture, to analyze any circuit, we only need apply two
rules: Kirchhoff’s Voltage and Current Rules. In this prelecture we applied
Kirchhoff’s voltage rule to solve for the current in RC circuits which have only one
node. The new wrinkle here is that the equation we obtained was a differential
equation which resulted in a time-dependent current.
Physics 212 Lecture 11, Slide 2
Main Point 2
Second, the solutions to the differential equations we obtained were exponential in
time, being characterized by a single time constant which is equal to the product of
an equivalent resistance and an equivalent capacitance. For example, for a circuit
that charges a single capacitor C through a single resistor R, we found that the
current decreases as e-t/t, while the charge on the capacitor increases as (1-e-t/t),
where the time constant t (tau) is just equal to R times C. Similar results are
Physics 212 Lecture 11, Slide 3
found for discharging the circuit.
Main Point 3
Third, in considering the energy flow in the RC charging circuit, we determined that
the rate at which energy is being supplied to the circuit by the battery is equal to
the rate at which energy is dissipated in the resistor plus the rate at which it is
stored in the capacitor. Since the current is decreasing exponentially over time, all
of these energy rates asymptotically approach zero as shown.
Physics 212 Lecture 11, Slide 4
RC Circuit (Charging)
• Switch is moved to position “a”
• Kirchoff’s Voltage Rule
a
E
• Short Term (Q=0)
C
b
R
• Long Term (Ic =0)
• Intermediate
Physics 212 Lecture 11, Slide 5
A circuit is wired up as shown below. The capacitor is initially uncharged and switches
S1 and S2 are initially open.
Checkpoint 1a
&
Checkpoint 1b
A) V1 = V
Close S1,
V1 = voltage across C immediately after
V2 = voltage across C a long time after
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V2 = V
B) V1 = 0
C) V1 = 0
V2 = V
V2 = 0
D) V1 = V
V2 = 0
Physics 212 Lecture 11, Slide 6
Physics 212 Lecture 11, Slide 7
A circuit is wired up as shown below. The capacitor is initially uncharged and switches
S1 and S2 are initially open.
Checkpoint 1c
IR
+
-
After being closed a long time, switch 1 is opened and switch 2 is closed. What is the
current through the right resistor immediately after switch 2 is closed?
A
A. IR = 0
B. IR = V/3R
C. IR = V/2R
D. IR = V/R
B
C
D
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Physics 212 Lecture 11, Slide 8
Physics 212 Lecture 11, Slide 9
A circuit is wired up as shown below. The capacitor is initially uncharged and switches
S1 and S2 are initially open.
Checkpoint 1d
Now suppose both switches are closed. What is the voltage across the capacitor after a
A
very long time?
B
A. VC = 0
B. VC = V
C. VC = 2V/3
C
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Physics 212 Lecture 11, Slide 10
Physics 212 Lecture 11, Slide 11
RC Circuit (discharging)
• Switch is moved to position “b”
• Kirchoff’s Voltage Rule
a
E
C
b
R
• Short Term (Q=Q0)
• Long Term (Ic=0)
• Intermediate
Physics 212 Lecture 11, Slide 12
Bulb DEMO
BB
V
Bulb 2
S
Bulb 1
R
R
C
What will happen after I close the switch?
A)
B)
C)
D)
Both bulbs come on and stay on.
Both bulbs come on but then bulb 2 fades out.
Both bulbs come on but then bulb 1 fades out.
Both bulbs come on and then both fade out.
Suppose the switch has been closed a long time.
Now what will happen after open the switch?
A)
B)
C)
D)
Both bulbs come on and stay on.
Both bulbs come on but then bulb 2 fades out.
Both bulbs come on but then bulb 1 fades out.
Both bulbs come on and then both fade out.
Physics 212 Lecture 11, Slide 13
Physics 212 Lecture 11, Slide 14
Calculation
S
R1
R2
V
C
In this circuit, assume V, C, and Ri are known.
C initially uncharged and then switch S is closed.
R3 After a long time S is opened:
What is the charge on the capacitor after
switch 1 has been closed a long time?
• Conceptual Analysis:
•Strategic Analysis
Physics 212 Lecture 11, Slide 15
Physics 212 Lecture 11, Slide 16
Physics 212 Lecture 11, Slide 17
Exponentials (e)
This is the value that makes slope equal to the function
1
x
 1
y     e x
e
0.9
0.8
e = 2.718281…
1/e = 0.367879…
0.7
0.6
0.5
dy
 y
dx
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
8
9
10
Physics 212 Lecture 11, Slide 18
Time Constant (t=RC)
Charge as a function of time for discharging two
different RC circuits.
Which circuit has the larger time constant?
A) A B) B C) Same
1
Q t 
0.9
Q0
0.7
Q  t   Q0e
0.8

t
RC
0.6
0.5
0.4
B
0.3
t
A
0.2
0.1
0
0
1
2
3
4
5
6
7
8
9
10
Physics 212 Lecture 11, Slide 19
The two circuits shown below contain identical capacitors that hold the same charge at t = 0. Circuit
2 has twice as much resistance as circuit 1.
Checkpoint 2a
Which circuit has the largest time constant?
A) Circuit 1
B)
Circuit 2
C)
Same
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Physics 212 Lecture 11, Slide 20
The two circuits shown below contain identical capacitors that hold the same charge at t = 0. Circuit
2 has twice as much resistance as circuit 1.
Checkpoint 2b
Which of the following statements best describes the charge remaining on each of the the two
capacitors for any time after t = 0?
A. Q1 < Q2
B. Q1 > Q2
C. Q1 = Q2
D. Q1 < Q2 at first, then Q1 > Q2 after long time
E. Q1 > Q2 at first, then Q1 < Q2 after long time
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Physics 212 Lecture 11, Slide 21
Physics 212 Lecture 11, Slide 22