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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 Charging of Capacitor
through a Resistor
• Calculate Discharging of Capacitor
through a Resistor
Text Reference: Chapter 26.6
Examples: 26.17,18 and 19
Last time--Behavior of Capacitors
(from Lect. 10)
• 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
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
Initially: Q = 0
VC = 0
I = E/(2R)
After a long time:
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
After C is fully charged, S1 is opened and S2 is closed.
Now, the battery and the resistor 2R are disconnected
from the circuit. So we now 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)
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
RC Circuits
(Time-varying currents)
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
RC Circuits
(Time-varying currents)
• 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 t = 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
1
x
t/RC
e
f( x ) 0.5
I
37% Max at t=RC
0.0183156 0
0
0
1
2
x
t
3
4
4
Lecture 11, ACT 1
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=t, 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=2t ?
(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
t1
2t2
3
4
RC Circuits
(Time-varying currents)
• Discharge capacitor:
C initially charged with
Q=Ce
b
Connect switch to b at t=0.
e
Calculate current and
charge as function of
time.
•
Loop theorem 
IR 
I
a
Q
0
C
• Convert to differential equation for Q:
dQ 
I
dt
dQ Q
R
 0
dt C
I
R
+ +
C
- -
RC Circuits
(Time-varying currents)
• Discharge capacitor:
I
a
dQ Q
R
 0
dt C
b
I
R
C
e
• Guess solution:
+ +
Q = 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
- -
RC Circuits
(Time-varying currents)
• Discharge capacitor:
-t/RC
ee
Q=C
b
• Current is found from
differentiation:
dQ
e t / RC
I
 e
dt
R
I
a

Minus sign:
original definition
of current “I” direction
I
R
+ +
C
e
- -
Conclusion:
• Capacitor discharges
exponentially with time constant
t = 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 = C e 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
2
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:
t1 = RC and t2 = 2RC. Since t2 > t1 it takes circuit 2 longer to discharge its
capacitor. Therefore, at any given time, the charge on capacitor is bigger than that
on capacitor 1.
Lecture 11, ACT 2
a
Ce1
0
t01
2
x
t/RC
C
Ce
1
(c)
(b)
(a)
00
2R
e
Ce1
3
4
f( x )q
0.5
t
00
Q
f( x )q
0.5
b
R
– At t = t0, the switch is thrown from
position a to position b.
– Which of the following graphs best
represents the time dependence of the
charge on C?
Q
Q
• At t=0 the switch is connected to
position a in the circuit shown: The
capacitor is initially uncharged.
0
t01
2
x
t/RC
3
f ( x ) 0 .5q
4
t
00
0
t0
1
2
x
t/ RC
• For 0 < t < t0, the capacitor is charging with time constant t = RC
• For t > t0, the capacitor is discharging with time constant t = 2RC
• (a) has equal charging and discharging time constants
• (b) has a larger discharging t than a charging t
• (c) has a smaller discharging t than a charging t
t
3
Charging
RC
Ce1
2RC
Q  Ce 1  et / RC 
2RC
Q = C e 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!!
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)
What is voltage across C after a long time? C is in parallel with R2 !!
VC = I1R2 = E R2/(R1+R2) < E
Very interesting RC circuit continued
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: t
parallel combination
of R1 and R2
• Try solution of the form:
continued
I2
I3
C
R2
R1

Q(t )  A 1  e t / 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 
 t  
 R1  R2 
 R1  R2

C

Very interesting RC circuit
continued
Loop 2
I1
e
• What about discharging?
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 t = RC
– what R, what C?? You must analyze the problem!
• series RC charging solution
Q  Ce 1  e t / RC 
• series RC discharging solution
Q = C e e -t/RC
Next time: Start Magnetism
Reading assignment: Ch. 28.1-2, 28.4
Examples: 28.1,4,5 and 6