RC Circuits 2
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
Quiz
Histogram
35
20
Frequency
15
10
5
Bin
80
95
60
65
40
50
20
35
0
80
0
20
Highest 97%
25
5
Mean 63%
Frequency
30
100%
RC circuits from yesterday
• When circuits contain capacitors, currents and voltages
vary with time
• Kirchoff’s Laws apply, they give differential equations!
• The differential equations have exponential solutions
– from the form of differential equation
• time constant t = RC
• series RC charging solution
• series RC discharging solution

Q  Ce 1  e
t
Q  Ce e
RC
t

RC
Question 1
• 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=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
(b) Q2 = 2Q1
a
I
b
I
R
e
C
R
(c) Q2 > 2Q1
Question 1
62%
35%
c
b
4%
a
1. a
2. b
3. c
Question 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
e
– 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
C
R
(c) Q2 > 2Q1
(b) Q2 = 2Q1
• Charge increases according to: Q  Ce (1  e
2 RC
)
2Q1
Q2
Q1
Q
• From the graph the charge increase 1
is not as fast as linear.
• The rate of increase is just
f( x ) 0.5
proportional to the current (dQ/dt)
Q
which decreases with time.
• Therefore, Q2 < 2Q1.
0
t
0
t1
2t2
3
4
Capacitor time constants
•The two circuits shown below contain identical fully charged
capacitors at t=0. Circuit 2 has twice as much resistance as circuit 1.
•How does the charge on
the two capacitors compare
a short time after t = 0?
•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 (twice as long).
•Therefore, at any given time, the charge on capacitor 2 is bigger
than that on capacitor 1.
Question 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
f ( x ) 0 .5q
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
4
t
00
0
t0
1
2
x
t/ RC
t
3
Question 2
1. a
2. b
3. c
54%
b
a
c
23%
23%
Question 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
f ( x ) 0 .5q
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
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
A very interesting RC circuit
I1
I2
I3
e
C
R2
R1
•First consider the short and long term behavior of this
circuit.
•Initially the capacitor acts like an ideal wire. Hence,
and
Question 3
•After a long time the current I1
is
a) I1 = 0
b) I1 = e/R1
c) I1 = e/(R1+ R2)
I1
I2
I3
e
C
R1
R2
Question 3
1. a
2. b
3. c
91%
c
2%
b
a
7%
Question 3
•After a long time the current I1 is
I1
I3
e
a) I1 = 0
b) I1 = e/R1
c) I1 = e/(R1+ R2)
I2
C
R1
•After a long time the capacitor becomes fully charged,
curent I3 becomes zero (the capacitor acts like an open
switch
•So, I1 = I2, and we have a one-loop circuit with two
e
resistors in series,
I1  I 2 
R1  R2
•C is in parallel with R2, therefore the voltage across each
is the same
Vc  I 2 R2 
e R2
R1  R2
e
R2
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
dQ
I1  I 3  I 2 
 I2
dt
Loop 1
I2
I3
C
R1
• Eliminate I1 in L1 and L2 using Node equation:
• Loop 1:
Q
 dQ

e   R1 
 I2   0
C
 dt

eliminate I2 from this
 dQ

e  I 2 R2  R1 
 I2   0
 dt

e dQ
Q


• Final differential eqn:
R1 dt  R1 R2 

C
 R1  R2 
• Loop 2:
R2
Very interesting RC circuit
Loop 2
• Final differential eqn:
dQ
Q
e


dt  R1 R2 
R1

C
 R1  R2 
dQ
Q
e


dt RC R1
• Try solution of the form:
continued
I1
e
I2
I3
Loop 1
C
R1

Q (t )  A 1  e
t
t

– and plug into differential equation to get parameters A and τ
A
t
e

t
t
t
A(1  e

CR
t
)

e
R1
t  CR  C
A
e

CR R1
R1 R2
R1  R2
A  Ce
R2
 R1  R2 
R2
Very interesting RC circuit
continued
Loop 2
• Final differential eqn:
dQ
Q
e


dt  R1 R2 
R1

C
 R1  R2 
I1
e
time constant: t parallel
combinationof R1 and R2
•Solution
t
Q (t )  A 1  e t


 R2 
A  Ce 

R

R
 1
2 
Loop 1
I2
I3
C
R2
R1

R R

1 2
t 
R R 
C
2 
 1
•C and R2 are in parallel
•After a long time the capacitor will be charged to the same
potential as the potential drop across R2. I3 will be 0 and the
current will just flow in the resistance loop
Very interesting RC circuit
continued
• What about discharging?
– Open the switch...
e
• Loop 1 does not exist any more
• I2 is only current
• only one loop
– start at x marks the spot...
•Solution
Q (t )  Ae
I2
 t
C
R2
R1
Q
dQ
Q
 I 2 R2   0  
R2 
C
dt
C
R2C
•Initial condition, Q(0)=A, is the same as the final condition before
Different time constant for
discharging
RC Circuits
• That is all about circuits for the time being
• After we have done magnetism we will come
back to consider circuits with inductance as
well as capacitance
Magnetism
The Magnetic Force


 
F  qE  qv  B
B
x x x x x x
x x x x x x
v
x x x x x x
q
F
B
B

v

 q
F
v
q
F=0
Magnetism
• Magnetic effects from natural magnets have been known
for a long time. Recorded observations from the Greeks
more than 2500 years ago.
• The word magnetism comes from the Greek word for a
certain type of stone (lodestone) containing iron oxide
found in Magnesia, a district in northern Greece.
• Properties of lodestones: could exert forces on similar
stones and could impart this property (magnetize) to a
piece of iron it touched.
• Small sliver of lodestone suspended with a string will
always align itself in a north-south direction—it detects
the earth’s magnetic field.
Bar Magnet
• Bar magnet ... two poles: N and S
Like poles repel; Unlike poles attract.
• Magnetic Field lines: (defined in same way as electric field
lines, direction and density)
S
N
• Looks like dipole in electrostatics?
Electric Field Lines
of an Electric Dipole
Magnetic Field Lines
of a bar magnet
S
N
Magnetic Monopoles
• Perhaps there exist magnetic charges, just like electric charges.
Such an entity would be called a magnetic monopole (having +
or - magnetic charge).
• How can you isolate this magnetic charge?
Try cutting a bar magnet in half:
S
N
S
N
S
N
Even an individual
electron has a
magnetic “dipole”!
• Many searches for magnetic monopoles—the existence
of which would explain (within framework of QM) the
quantization of electric charge (argument of Dirac)
• No monopoles have ever been found:
 B  dS  0
Source of Magnetic Fields?
• What is the source of magnetic fields, if not magnetic charge?
• Answer: electric charge in motion!
– e.g., current in wire surrounding cylinder (solenoid) produces
very similar field to that of bar magnet.
• Therefore, understanding source of field generated by bar
magnet lies in understanding currents at atomic level within bulk
matter.
Orbits of electrons about nuclei
Intrinsic “spin” of
electrons (more
important effect)
Summary
• Read Fishbane Chapter 28 on Magnetic Fields
• Do Fishbane problems Chapter 27, #57, 59, 78
• Have a good Spring Break