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RC Circuits
I
a
I
a
I
I
R
C
e
RC
Ce1
R
b
b
+ +
C
e
2RC
RC
Ce
1 1
- -
2RC
Q
q  Cee  t / RC

f( x ) q
0.5
q  Ce 1  e
00
0
1
t
 t / RC

f( xq) 0.5
0.0183156 0
2
x
3
4
0
0
1
t
2
x
3
4
4
Today…
• Calculate Discharging of Capacitor
through a Resistor
• Calculate Charging of Capacitor
through a Resistor
Review: Behavior of Capacitors
• Capacitors resist change in charge and voltage
• Charging
– Initially, the capacitor behaves like a wire.
– After a long time, the capacitor behaves like an open switch.
• Discharging
– Initially, the capacitor behaves like a battery.
– After a long time, the capacitor behaves like a wire.
Preflight 11:
The capacitor is initially
uncharged, and the two
switches are open.
E
3) What is the voltage across the capacitor immediately after switch
S1 is closed?
a) Vc = 0
b) Vc = E
c) Vc = 1/2 E
Initially: Q = 0
VC = 0
I = E/(2R)
4) Find the voltage across the capacitor after the switch has been
closed for a very long time.
a) Vc = 0
c) Vc = 1/2 E
b) Vc = E
Q=EC
I=0
Preflight 11:
E
6) 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
the switch 2 is closed?
a) IR= 0
b) IR=E/(3R)
c) IR=E/(2R)
d) IR=E/R
Now, the battery and the
resistor 2R are disconnected
from the circuit, so we have a
different circuit.
Since C is fully charged, VC = E.
Initially, C acts like a battery,
and I = VC/R.
RC Circuits
(Time-varying currents -- discharging)
• Discharge capacitor:
C initially charged with
Q = Q0 = Ce
b
Connect switch to b at t = 0.
Calculate current and
charge as function of time.
•
Loop theorem 
I
a
e
I
R
+ +
C
Q
IR   0
C
• Convert to differential equation for Q:
Note: Although we know
the current is flowing off
the cap., we define it as
shown so that …
dQ 
I 
dt
dQ Q
R
 0
dt C
- -
Discharging Capacitor
I
a
dQ Q
R
 0
dt C
b
I
R
C
e
• Guess solution:
+ +
Q  Q0 et /  Ce et / RC
• Check that it is a solution:
Note that this “guess”
incorporates the
boundary conditions:
dQ
1 

 Ce e  t / RC  

dt
 RC 

dQ Q


R dt    e e t / RC  e e t / RC  0
C
!
t  0  Q  Ce
t Q0
- -
Discharging Capacitor
• Discharge capacitor:
Q  Q0 et /  Ce et / RC
• Current is found from
differentiation:
dQ
e t / RC
I
 e
dt
R

Minus sign:
Current is opposite to
original definition,
i.e., charges flow
away from capacitor.
I
a
b
I
R
+ +
C
e
- -
Conclusion:
• Capacitor discharges
exponentially with time constant
 = RC
• Current decays from initial max
value (= -e/R) with same time
constant
Discharging Capacitor
Ce1
Charge on C
RC
2RC
1
2
1
Q  Ce et / RC
Max = Ce
f( x ) 0.5
Q
37% Max at t = RC
0.0183156
0
zero
0
01 0
t
3
x
4
4
dQ
e
I
  e t / RC
dt
R
Q
Current
I
f( x ) 0.5
“Max” = -e/R
37% Max at t = RC
-e /R0
0
1
2
x
t/RC
t
3
4
1
Preflight 11:
The two circuits shown below contain identical fully charged
capacitors at t=0. Circuit 2 has twice as much resistance as circuit 1.
8) Compare the charge on the two capacitors a short time after t = 0
a) Q1 > Q2
b) Q1 = Q2
c) Q1 < Q2
Initially, the charges on the two capacitors are the same.
But the two circuits have different time constants:
1 = RC and 2 = 2RC. Since 2 > 1 it takes circuit 2
longer to discharge its capacitor. Therefore, at any given
time, the charge on capacitor 2 is bigger than that on
capacitor 1.
Lecture 11, ACT 1
a
• The capacitor in the circuit shown is
initially charged to Q = Q0. At t = 0 the
switch is connected to position a.
• At t = t0 the switch is immediately
flipped from position a to position b.
3R
C
Q0
Q0
Q
Q
time
(c)
(b)
Q
(a)
t0
1B
R
Which of the following graphs best represents
the time dependence of the charge on C?
Q0
1A
b
t0
time
t0
time
Which of the following correctly relates the value of t0 to the
time constant  while the switch is at a?
(a) t0 < 
(b) t0 = 
(c) t0 > 
Lecture 11, ACT 1
a
• The capacitor in the circuit shown is
initially charged to Q = Q0. At t = 0 the
switch is connected to position a.
• At t = t0 the switch is immediately
flipped from position a to position b.
R
3R
C
Q0
Q
Q
t0
time
(c)
(b)
Q
(a)
Q0
Which of the following graphs best represents
the time dependence of the charge on C?
Q0
1A
b
t0
time
t0
time
• For 0 < t < t0, the capacitor is discharging with time constant  = RC
• For t > t0, the capacitor is discharging with time constant  = 3RC, i.e.,
much more slowly. Therefore, the answer is (a)
• (b) has equal discharging rates
• (c) has faster discharging after t0 than before.
Lecture 11, ACT 1
a
R
3R
C
Q
Q0
• The capacitor in the circuit shown is
initially charged to Q = Q0. At t = 0 the
switch is connected to position a.
• At t = t0 the switch is immediately
flipped from position a to position b.
b
t0
1B
time
Which of the following correctly relates the value of t0 to the
time constant  while the switch is at a?
(a) t0 < 
(b) t0 = 
(c) t0 > 
We know that for t = , the value of the charge is e-1 = 0.37 of the value
at t = 0. Since the curve shows Q(t0) ~ 0.6 Q0, t0 must be less than .
RC Circuits
(Time-varying currents, charging)
I
a
• Charge capacitor:
I
R
C initially uncharged;
connect switch to a at t=0
Calculate current and
b
C
e
charge as function of time.
Q
e  IR   0
C
•Convert to differential equation for Q:
• Loop theorem 
dQ
I
dt

Would it matter where R
is placed in the loop??
dQ Q
e R 
dt C
Charging Capacitor
I
a
• Charge capacitor:
I
R
dQ Q
e R

dt C
b
e
C
• Guess solution:
Q  Ce (1  e
t
RC
)
•Check that it is a solution:
dQ
1 
 t / RC 
 Ce e


dt
RC



t
dQ Q
t / RC
R
  ee
 e (1  e RC )  e !
dt C
Note that this “guess”
incorporates the
boundary conditions:
t 0Q0
t    Q  Ce
Charging Capacitor
• Charge capacitor:
I
a
R
Q  Ce 1  et / RC 
b
C
e
• Current is found from
differentiation:
dQ e t / RC
I
 e
dt R
I

Conclusion:
• Capacitor reaches its final
charge(Q=Ce ) exponentially
with time constant  = RC.
• Current decays from max
(=e /R) with same time
constant.
Charging Capacitor
RC
Charge on C
Q
Q  Ce 1  et / RC 
Max = Ce
Ce
2RC
1
f( x ) 0.5
Q
63% Max at t = RC
00
0
I
Current
dQ e  t / RC
dt

Max = e /R
R
1
e 1/R1
2
t
3
4
x
t/RC
e
f( x ) 0.5
I
37% Max at t = RC
2
0.0183156 0
0
0
1
2
x
t
3
4
4
Lecture 11, ACT 2
• At t=0 the switch is thrown from position b to
position a in the circuit shown: The capacitor
is initially uncharged.
– At time t = t1 =, the charge Q1 on the capacitor
is (1-1/e) of its asymptotic charge Qf = Ce.
– What is the relation between Q1 and Q2 , the
charge on the capacitor at time t = t2 = 2 ?
(a) Q2 < 2Q1
(b) Q2 = 2Q1
a
I
b
I
R
e
C
R
(c) Q2 > 2Q1
Lecture 11, ACT 2
a
• At t=0 the switch is thrown from position b to
position a in the circuit shown: The capacitor
is initially uncharged.
I
R
b
– At time t = t1 =, the charge Q1 on the capacitor
is (1-1/e) of its asymptotic charge Qf = Ce.
– What is the relation between Q1 and Q2 , the
charge on the capacitor at time t = t2 = 2 ?
(a) Q2 < 2Q1
I
e
C
R
(c) Q2 > 2Q1
(b) Q2 = 2Q1
• The point of this ACT is to test your understanding of the exact time
dependence of the charging of the capacitor.
• Charge increases according to:
Q  Ce (1  e
• So the question is: how does this charge
increase differ from a linear increase?
t
2 RC
)
2Q1
1
Q2
• From the graph at the right, it is clear that the
charge increase is not as fast as linear.
0.5
• In fact the rate of increase is just proportional to f( x )Q
the current (dQ/dt) which decreases with time.
• Therefore, Q2 < 2Q1.
Q
Q1
0
0
1
22
3
4
Charging
RC
Ce1
2RC
Q  Ce 1  et / RC 
2RC
Q  Ce et / RC
f( x ) Q
0.5
0.0183156 0
00
0
1
1e /R1
2
t
3
0
1
dQ e t / RC
 e
dt R
2
1
01 0
Q
( x ) 0.5
I
0
4
x
t/RC
I
0
RC
Ce 1
1
f( x ) Q
0.5
156
Discharging
t
2
4
4
dQ
e t / RC
I
 e
dt
R
I
f( x ) 0.5
4
3
x
-e /R0
3
t
0
1
2
t
3
4
A very interesting RC circuit
I1
I2
I3
e
C
R2
R1
First consider the short and long term behavior of this
circuit.
• Short term behavior:
Initially the capacitor acts like an ideal wire. Hence,
and
•Long term behavior:
Exercise for the student!!
3
Preflight 11:
The circuit below contains a
battery, a switch, a capacitor
and two resistors
10) Find the current through R1 after the switch has been closed
for a long time.
a) I1 = 0
b) I1 = E/R1
c) I1 = E/(R1+ R2)
After the switch is closed for a long time …..
The capacitor will be fully charged, and I3 = 0.
(The capacitor acts like an open switch).
So, I1 = I2, and we have a one-loop circuit with two resistors in series,
hence I1 = E/(R1+R2)
Lecture 11, ACT 3
• At t = 0 the switch is closed in
the circuit shown. The initially
uncharged capacitor then
begins to charge.
I1
I2
I3
e
C
R1
3A
• What will be the voltage across the capacitor a long time
after the switch is closed?
(a) VC = 0
3B
(b) VC = e R2/(R1+ R2)
(c) VC = e
• What is the charging time constant  ?
(a)   R1C
(b)   ( R1  R2 )C
 R1R2 
(c)   
C
 R1  R2 
R2
Lecture 11, ACT 3
• At t = 0 the switch is closed in
the circuit shown. The initially
uncharged capacitor then
begins to charge.
I1
I2
I3
e
C
R1
3A
• What will be the voltage across the capacitor a long time
after the switch is closed?
(a) VC = 0
(b) VC = e R2/(R1+ R2)
(c) VC = e
After a long time the capacitor is completely charged, so no
current flows through it. The circuit is then equivalent to a
battery with two resistors in series. The voltage across the
capacitor equals the voltage across R2 (since C and R2 are in
parallel). Either from direct calculation, or remembering the
“Voltage Divider Circuit”, VC = VR2 = e R2/(R1+ R2).
R2
Lecture 11, ACT 3
• At t = 0 the switch is closed in the
circuit shown. The initially uncharged
e
capacitor then begins to charge.
3B
– What is the charging time
constant  ?
(a)   R1C
(b)   ( R1  R2 )C
I1
I2
I3
C
R2
R1
 RR 
(c)    1 2  C
 R1  R2 
• An ideal voltage source contributes no resistance or capacitance
 time constant is entirely determined by C, R1, and R2.
• I personally find it easier to think about the circuit as if C was discharging
than charging; I imagine that the capacitor is charged, and that the battery is
replaced by a wire (which also has no resistance or capacitance). Since the
battery supplies a constant voltage, it doesn’t affect the time constant.
• We simply need to find the effective resistance Reff through which the
capacitor (dis)charges. Looking at the new circuit, it is clear that the
capacitor would be (dis)charging through both R1 and R2, which are in
parallel  their effective resistance is Reff = R1R2/(R1 + R2) and  = ReffC.
Very interesting RC circuit , detailed
Loop 2
• Loop 1:
• Loop 2:
• Node:
Q
e   I1 R1  0
C
e  I 2 R2  I1R1  0
I1
e
I1  I 2  I 3
Loop 1
I2
I3
C
R1
• Eliminate I1 in L1 and L2 using Node equation:
• Loop 1:
• Loop 2:
Q
 dQ

e
 R1 
 I2   0
C
 dt

eliminate I2 from this
 dQ

e  I 2 R2  R1 
 I2   0
 dt

• Final differential eqn:
e dQ
Q


R1 dt  R1 R2 

C
 R1  R2 
R2
Very interesting RC circuit
Loop 2
• Final differential eqn:
dQ
Q
e


dt  R1 R2 
R1

C
 R1  R2 
I1
e
Loop 1
time constant: 
parallel combination
of R1 and R2
• Try solution of the form:
continued
I2
I3
C
R2
R1

Q(t )  A 1  e t / 

– and plug into ODE to get parameters A and τ
• Obtain results that agree with initial and final conditions:
 R1 R2
 R2 
A  Ce 
   
 R1  R2 
 R1  R2

C

Very interesting RC circuit
continued
Loop 2
I1
• What happens when we
discharge the capacitor?
e
Loop 1
I2
I3
C
R2
– Open the switch...
R1
• Loop 1 and Loop 2 do not exist!
• I2 is only current
• only one loop
e
– start at x marks the spot...
 I 2 R2 
Q
0
C
but
I2
I2  
dQ
dt
C
R1
Different time constant for discharging
R2
Summary
• Kirchoff’s Laws apply to time dependent circuits
they give differential equations!
• Exponential solutions
– from form of differential equation
• time constant  = RC
– what R, what C?? You must analyze the problem!

• series RC charging solution Q  Q(t  ) 1  e t / RC
• series RC discharging solution
Next time: Start Magnetism
Q  Q(t  0) et / RC
