My
Chapter 18
Lecture
Outline
1
Chapter 18: Electric Current and
Circuits
•Electric current
•EMF
•Current & Drift Velocity
•Resistance & Resistivity
•Kirchhoff’s Rules
•Series & Parallel Circuit Elements
•Applications of Kichhoff’s Rules
•Power & Energy
•Ammeters & Voltmeters
•RC Circuits
2
§18.1 Electric Current
e-
ee-
e-
e-
e-
e-
A metal wire.
e-
Assume electrons
flow to the right.
Current : I 
q
t
Current is a measure of the amount
of charge that passes though an area
perpendicular to the flow of charge.
Current units: 1C/sec = 1 amp
3
A current will flow until there is no potential difference.
The direction of current flow in a wire is opposite the flow of
the electrons. (In the previous drawing the current is to the
left.)
4
Example: If a current of 80.0 mA exists in a metal wire, how
many electrons flow past a given cross-section of the wire in
10.0 minutes?
I 
q
t

 q  I  t  80 . 0  10
3
# of electrons

A 600 sec   48 . 0 C
q

charge per electron

48 . 0 C
1.60  10
 3 . 00  10
 19
20
C/electron
electrons
5
§18.2 EMF and Circuits
An ideal battery maintains a constant potential difference.
This potential difference is called the battery’s EMF().
The work done by an ideal battery in pumping a charge q
is W = q.
6
At high potential
The circuit symbol for a
battery (EMF source) is

+
At low potential
Batteries do work by converting chemical energy into
electrical energy. A battery dies when it can no longer
sustain its chemical reactions and so can do no more work
to move charges.
7
§18.3 Microscopic View of Current
in a Metal
Electrons in a metal might have a speed of ~106 m/s, but
since the direction of travel is random, an electron has
vdrift = 0.
8
Only when the ends of a wire are at different potentials
(E  0) will there be a net flow of electrons along the
wire (vdrift  0). Typically, vdrift < 1 mm/sec.
9
Calculate the number of charges (Ne) that pass through the
shaded region in a time t:
N e  n ( Al )
l
The current in the wire is: I 
 nA ( v d  t )
q
t

eN e
t
 neAv
d
10
Example (text problem 18.19): A copper wire of crosssectional area 1.00 mm2 has a constant current of 2.0 A
flowing along its length. What is the drift speed of the
conduction electrons? Copper has 1.101029 electrons/m3.
I  neAv
vd 
I
d

neA
 1 . 1  10
1.10  10
4
2 .0 A
29
m
3
1 . 60  10
 19
C 1 . 00  10
6
m
2

m/sec  0.11 mm/sec
11
§18.4 Resistance and Resistivity
A material is considered ohmic if VI, where
 V  IR
The proportionality constant R is called resistance and is
measured in ohms (; and 1  = 1 V/A).
12
The resistance of a conductor is: R  
L
A
where  is the resistivity of the material, L is the length of
the conductor, and A is its cross sectional area.
With R a material is considered a conductor if  is “small”
and an insulator if  is “large”.
13
The resistivity of a material depends on its temperature:
   0 1   T  T0 
where 0 is the resistivity at the temperature T0, and  is
the temperature coefficient of resistivity.
A material is called a superconductor if  = 0.
14
Example (text problem 18.28): The resistance of a conductor
is 19.8  at 15.0 C and 25.0  at 85.0 C. What is the
temperature coefficient of resistivity?
Values of R are given at different temperatures, not
values of . But the two quantities are related.
R  
L
A
(1)
   0 1   T  T0 
(2)
Multiply both sides of equation (2) by L/A and use
equation (1) to get:
R  R 0 1   T  T0 
(3)
15
Example continued:
Solve equation (3) for  and evaluate using the given
quantities:
R
 
25 . 0 
1
R0
T

1
3
1
19 . 8 
 3 . 75  10  C
85 . 0  C  15 . 0  C
16
§18.5 Kirchhoff’s Rules
Junction rule: The current that flows into a junction is the
same as the current that flows out. (Charge is conserved)
A junction is a place where two or more wires (or other
components) meet.
Loop rule: The sum of the voltage dropped around a closed
loop is zero. (Energy is conserved.)
17
For a resistor: If you cross a resistor in the direction of the
current flow, the voltage drops by an amount IR (write as
IR). There is a voltage rise if you cross the other way
(write as +IR).
A
I
B
If the current flows from A to B, then
the potential decreases from A to B.
The potential difference between A
and B is < 0 (V = IR) .
18
For batteries (or other sources of EMF): If you move from
the positive to the negative terminal the potential drops by 
(write as ). The potential rises if you cross in the other
direction (write as +).
At high potential


+
At low potential
19
A current will only flow around a closed loop.
A
Applying the loop rule:
V AB  IR  0
  Ir  IR  0
B
VAB is the
terminal voltage.
20
In a circuit, if the current always flows in the same direction
it is called a direct current (DC) circuit.
21
§18.6 Series and Parallel Circuits
Resistors:
The current through the two
resistors is the same. It is not
“used up” as it flows around
the circuit!
These resistors are in series.
Apply Kirchhoff’s loop rule:   IR 1  IR 2  0
  IR 1  IR 2  I ( R1  R 2 )  IR eq
22
The pair of resistors R1 and R2 can be replaced with a single
equivalent resistor provided that Req = R1 + R2.
In general, for resistors in series R eq  R1  R 2    R n
n

R.
i
i 1
23
Current only flows around closed loops. When the current
reaches point A it splits into two currents. R1 and R2 do not
have the same current through them, they are in parallel.
Apply Kirchhoff’s loop rule:
  I 1 R1  0
  I 2 R2  0
The potential drop across
each resistor is the same.
24
Applying the junction rule at A: I = I1+ I2.
From the loop rules:   I 1 R1  I 2 R 2
Substituting for I1 and I2 in the junction rule:
I 
I





R1
R2
1
1
R1

R2

1
R eq
25
The pair of resistors R1 and R2 can be replaced with a
single equivalent resistor provided that
1
1

R eq

R1
1
.
R2
In general, for resistors in parallel
1

R eq
1

R1
n


i 1
1
R2
1
 
1
Rn
.
Ri
26
Example (text problem 18.40): In the given circuit, what is
the total resistance between points A and B?
R1 = 15
A
R2 =
12 
R3 =
24 
B
R2 and R3 are in parallel. Replace
with an equivalent resistor R23.
1
R 23

1
R2

1
R3
R 23  8 
27
Example continued:
The circuit can now be redrawn:
R1 = 15
A
R23 = 8 
The resistors R23
and R1 are in series:
R123  R1  R 23
 23   R eq
B
A
R123 =
23 
B
Is the equivalent
circuit and the total
resistance is 23 .
28
Capacitors:
C1
C2
For capacitors in series the charge
on the plates is the same.

Apply Kirchhoff’s loop rule:
Q
 

C1

Q

1
C1
Q
0
C2

1
C2

1
C eq
29
The pair of capacitors C1 and C2 can be replaced with a
single equivalent capacitor provided that
1

C eq
1

C1
1
.
C2
In general, for capacitors in series
1

C eq
1
1

C1
n

C2
1
C
i 1
 
1
Cn
.
i
30
C2
C1
Apply Kirchhoff’s loop rule:
 
Q1
0
C1
 
Q2
0
C2

For capacitors in parallel the charge on the plates may be
different. Here
Q eq  Q 1  Q 2
 C eq   C 1   C 2
31
The pair of capacitors C1 and C2 can be replaced with a
single equivalent capacitor provided that Ceq= C1 + C2.
In general, for capacitors in parallel
n
C eq  C 1  C 2    C n 
C .
i
i 1
32
Example (text problem 18.49): Find the value of a single
capacitor that replaces the three in the circuit below if
C1 = C2 = C3 = 12 F.
C1
A
C2 and C3 are in parallel
C3
C2
C 23  C 2  C 3
 24  F
B
33
Example continued:
The circuit can be redrawn:
C1
The remaining two capacitors
are in series.
A
1
C23

C 123
1

C1

B
1
C 23
1
12  F

1
24  F
C 123  8  F
A
C123
B
Is the final, equivalent circuit.
34
§18.8 Power and Energy in Circuits
The energy dissipation rate is:
For an EMF source:
P 
U
t

q
t
 V  I V
P  I
For a resistor: P  I  V  I 2 R 
V
2
R
35
Example: Use the results of the example starting on slide 35
to determine the power dissipated by the three resistors in
that circuit.
P I R
2
Resistance () Current (A) Power (W)
122
0.199
4.83
5.6
75
0.123
0.0760
0.0847
0.433
36
§18.9 Measuring Currents and
Voltages
Current is measured with an ammeter. An ammeter is placed
in series with a circuit component.
A1
An ammeter
has a low
internal
resistance.
R1
A1 measures the
current through R1.
R2
A2 measures the
current through R2.
A2
A3

A3 measures the
current drawn from
the EMF.
37
A voltmeter is used to measure the potential drop across a
circuit element. It is placed in parallel with the component.
A voltmeter has a large internal resistance.
V
The voltmeter measures the
voltage drop across R1.
R1
R2

38
§18.10 RC Circuits
Switch
R

+
+
C
Close the switch at t = 0 to
start the flow of current. The
capacitor is being charged.


Apply Kirchhoff’s loop rule:
  IR 
Q
0
C
Note : I 
Q
t
39
The current I(t) that satisfies Kirchhoff’s loop rule is:
I t   I 0 e
where
I0 

 t

and   RC .
R
 is the RC time constant and is a measure of
the charge (and discharge) rate of a capacitor.
40
The voltage drop
across the capacitor is:
t 

V C t     1  e  


The voltage drop
across the resistor is:
V R t   I t  R
The charge on the
capacitor is:
Q C t   CV C t 
Note: Kirchhoff’s loop rule must be satisfied for all times.
41
Plots of the voltage drop across the (charging) capacitor
and current in the circuit.
42
While the capacitor is charging S2 is open. After the capacitor
is fully charged S1 is opened at the same time S2 is closed:
this removes the battery from the circuit. Current will now
flow in the right hand loop only, discharging the capacitor.
I
S1
Apply Kirchhoff’s loop rule:
R
S2
+

C
 IR 
Q
0
C

The current in the circuit is
I t   I 0 e
 t

.
But the voltage drop across the capacitor is now V C t    e
 t
43

.
The voltage drop across the discharging capacitor:
44
Example (text problem 18.85): A capacitor is charged to an
initial voltage of V0 = 9.0 volts. The capacitor is then
discharged through a resistor. The current is measured and
is shown in the figure.
45
Example continued:
(a) Find C, R, and the total energy dissipated in the resistor.
Use the graph to determine . I0 = 100 mA; the current is
I0/e = 36.8 mA at t = 13 msec.
I 0  100 mA 

R
  13 msec  RC
Since  = V0 = 9.0 volts, R = 90  and C = 144 F.
All of the energy stored in the capacitor is eventually
dissipated by the resistor.
U 
1
2
2
CV 0  5 . 8  10
3
J
46
Example continued:
(b) At what time is the energy in the capacitor half of the
initial value?
U (t ) 
1
CV ( t ) and U ( t  0 ) 
2
2
U (t ) 
Want:
2
1
U (t  0 ) 
2
1
CV ( t ) 
2
2
1
4
V (t ) 
1
1
4
2
CV 0
2
CV 0
2
CV 0
1
2
V0
47
Example continued:
Solve for t:
V (t )  V 0 e
t   ln
 t


1
2
V0
2  13 msec  ln
2  4 . 5 msec
48
Summary
•Current & Drift Velocity
•Resistance & Resistivity
•Ohm’s Law
•Kirchhoff’s Rules
•Series/Parallel Resistors/Capacitors
•Power
•Voltmeters & Ammeters
• RC Circuits
49