1
Today’s agendum:
Electric Current.
You must know the definition of current, and be able to use it in solving problems.
Current Density.
You must understand the difference between current and current density, and be able to
use current density in solving problems.
Ohm’s “Law” and Resistance.
You must be able to use Ohm’s “Law” and electrical resistance in solving circuit problems.
Resistivity.
You must understand the relationship between resistance and resistivity, and be able to
calculate resistivity and associated quantities.
Temperature Dependence of Resistivity.
You must be able to use the temperature coefficient of resistivity to solve problems
involving changing temperatures.
2
Electric Current
Definition of Electric Current
The average current that passes any point in a conductor
during a time t is defined as
Q
Iav 
t
where Q is the amount of charge passing the point.
dQ
The instantaneous current is I =
.
dt
1C
.
One ampere of current is one coulomb per second: 1A =
1s 3
Currents in battery-operated devices are often in the milliamp
range: 1 mA = 10-3 A.
“m” for milli—another abbreviation to remember!
Here’s a really simple circuit:
+-
current
Don’t try that at home! (Why not?)
The current is in the direction of flow of positive charge…
…opposite to the flow of electrons, which are usually the charge
carriers.
4
+-
current
electrons
An electron flowing from – to + gives rise to the same
“conventional current” as a proton flowing from + to -.
“Conventional” refers to our convention, which is always to
consider the effect of + charges (for example, electric field
direction is defined relative to + charges).
5
“Hey, that figure you just showed me is confusing. Why don’t
electrons flow like this?”
+-
current
electrons
Good question.
6
+-
current
electrons
Electrons “want” to get away from - and go to +.
Chemical reactions (or whatever energy mechanism the battery
uses) “force” electrons to the negative terminal. The battery
won’t “let” electrons flow the wrong way inside it. So electrons
pick the easiest path—through the external wires towards the +
terminal.
Of course, real electrons don’t “want” anything.
7
Note!
Current is a scalar quantity, and it has a sign associated with it.
In diagrams, assume that a current indicated by a
symbol and an arrow is the conventional current.
I1
If your calculation produces a negative value for the current,
that means the conventional current actually flows opposite to
the direction indicated by the arrow.
8
Example: 3.8x1021 electrons pass through a point in a wire in 4
minutes. What was the average current?
Q Ne
I av 

t
t
Iav 
21
19
3.8

10
1.6

10



 4  60 
Iav  2.53A
“This is a piece of cake so far!”
Don’t worry, it gets “better” later.
9
Today’s agendum:
Electric Current.
You must know the definition of current, and be able to use it in solving problems.
Current Density.
You must understand the difference between current and current density, and be able to
use current density in solving problems.
Ohm’s “Law” and Resistance.
You must be able to use Ohm’s “Law” and electrical resistance in solving circuit problems.
Resistivity.
You must understand the relationship between resistance and resistivity, and be able to
use calculate resistivity and associated quantities.
Temperature Dependence of Resistivity.
You must be able to use the temperature coefficient of resistivity to solve problems
involving changing temperatures.
10
Current Density
When we study details of charge transport, we use the concept
of current density.
Current density is the amount of charge that flows across a unit
of area in a unit of time.
+
+
Current density: charge per area per time.
11
A current density J flowing through an infinitesimal area dA
produces an infinitesimal current dI.
dA
J
dI  J  dA
Current density is a vector. Its
direction is the direction of the
velocity of positive charge carriers.
The total current passing through A is just
I

surface
J  dA
12
Current density: charge per area per time.
J
A
If J is constant and parallel to dA (like in a wire), then
I
I   J  dA  J  dA  JA  J 
A
surface
surface
Now let’s take a “microscopic” view of current.
vt
q
v
A
If n is the number of charges
per volume, then the number of
charges that pass through a
surface A in a time t is
number
 volume   n  vt A 
volume
13
The total amount of charge passing through A is the number of
charges times the charge of each.
vt
q
v
Q  nqvt A
A
Divide by t to get the current…
Q
I
 nqv A
t
…and by A to get J:
J  nqv .
14
To account for the vector nature of the current density,
J  nqv
and if the charge carriers are electrons, q=-e so that
Je  n e v.
The – sign demonstrates that the velocity of the electrons is
antiparallel to the conventional current direction.
15
Currents in Materials
Metals are conductors because they have “free” electrons,
which are not bound to metal atoms.
In a cubic meter of a typical conductor there roughly 1028 free
electrons, moving with typical speeds of 1,000,000 m/s.
But the electrons move in random directions, and there is no
net flow of charge, until you apply an electric field...
16
E
electron “drift” velocity
just one
electron
shown, for
simplicity
inside a
conductor
The voltage accelerates the electron, but only until the
electron collides with a “scattering center.” Then the electron’s
velocity is randomized and the acceleration begins again.
Some predictions made by this theory are off by a factor or 10
or so, but it was the best we could do before quantum
mechanics.
We will see later in this course that the electron would follow curved trajectories,
17
but the idea here is still valid.
Even though the details of the model on the previous slide are
wrong, it points us in the right direction, and works when you
take quantum mechanics into account.
In particular, the velocity that should be used in
J  n q v.
is not the charge carrier’s velocity (electrons in this example).
Instead, we should the use net velocity of the collection of
electrons, the net velocity caused by the electric field.
This “net velocity” is like the terminal velocity of a parachutist;
we call it the “drift velocity.”
J  n q vd .
18
It’s the drift velocity that we should use in our equations for
current and current density in conductors:
J  n q vd
I  nqvd A
I
vd 
nqA
19
Example: the 12-gauge copper wire in a home has a crosssectional area of 3.31x10-6 m2 and carries a current of 10 A.
The conduction electron density in copper is 8.49x1028
electrons/m3. Calculate the drift speed of the electrons.
I
vd 
nqA
I
vd 
neA
10 C/s
vd 
(8.49 1028 m -3 )(1.60 1019 C)(3.31106 m 2 )
vd  2.22 104 m/s
20
Today’s agendum:
Electric Current.
You must know the definition of current, and be able to use it in solving problems.
Current Density.
You must understand the difference between current and current density, and be able to
use current density in solving problems.
Ohm’s “Law” and Resistance.
You must be able to use Ohm’s “Law” and electrical resistance in solving circuit problems.
Resistivity.
You must understand the relationship between resistance and resistivity, and be able to
use calculate resistivity and associated quantities.
Temperature Dependence of Resistivity.
You must be able to use the temperature coefficient of resistivity to solve problems
involving changing temperatures.
21
Resistance
The resistance of a material is a measure of how easily a
charge flows through it.
Resistance: how much “push” is needed to
get a given current to flow.
V
R
I
1V
.
The unit of resistance is the ohm: 1  
1A
Resistances of kilo-ohms and mega-ohms are common:
1 k  103 , 1 M=106.
22
Every circuit component has resistance.
This is the symbol we use for a “resistor:”
All wires have resistance. Obviously, for efficiency in carrying a
current, we want a wire having a low resistance. In idealized
problems, we will consider wire resistance to be zero.
Lamps, batteries, and other devices in circuits have resistance.
23
Resistors are often intentionally used in
circuits. The picture shows a strip of five
resistors (you tear off the paper and
solder the resistors into circuits).
The little bands of color on the resistors have meaning, namely
the amount of resistance.
24
Ohm’s “Law”
In some materials, the resistance is constant over a wide range
of voltages.
For such materials, we write V  IR, and call the equation
“Ohm’s Law.”
In fact, Ohm’s “Law” is not a “Law” in the same sense as
Newton’s “Laws”…
… and in advanced classes you will write something other than
V=IR when you write Ohm’s “Law.”
Newton’s Laws demand; Ohm’s “Law” suggests.
25
Today’s agendum:
Electric Current.
You must know the definition of current, and be able to use it in solving problems.
Current Density.
You must understand the difference between current and current density, and be able to
use current density in solving problems.
Ohm’s “Law” and Resistance.
You must be able to use Ohm’s “Law” and electrical resistance in solving circuit problems.
Resistivity.
You must understand the relationship between resistance and resistivity, and be able to
use calculate resistivity and associated quantities.
Temperature Dependence of Resistivity.
You must be able to use the temperature coefficient of resistivity to solve problems
involving changing temperatures.
26
Resistivity
It is also experimentally observed (and justified by quantum
mechanics) that the resistance of a metal wire is well-described
by
L
R
,
A
where  is a “constant” called the resistivity of the wire
material, L is the wire length, and A its cross-sectional area.
Notice that this relates R to a property of the material of which
the wire is made.
This makes sense: a longer wire or higher-resistivity wire should
have a greater resistance. A larger area means more “space”
for electrons to get through, hence lower resistance.
27
R = L / A,
units of 
are m

A
L
The longer a wire, the “harder” it is to push electrons through
it.
The greater the resistivity, the “harder” it is to push electrons
through it.
The greater the cross-sectional area, the “easier” it is to push
electrons through it.
Resistivity is a useful tool in physics because it depends on the
properties of the wire material, and not the geometry.
28
Resistivities range from roughly 10-8 ·m for copper wire to
1015 ·m for hard rubber. That’s an incredible range of 23
orders of magnitude, and doesn’t even include superconductors
(we might talk about them some time).
Example: Suppose you want to connect your stereo to remote
speakers.
(a) If each wire must be 20 m long, what diameter copper wire
should you use to make the resistance 0.10  per wire.
R = L / A
A = L / R
A =  (d/2)2
 (d/2)2 = L / R
geometry!
29
(d/2)2 = L / R
d/2= ( L / R )½
don’t skip steps!
d = 2 ( L / R )½
d = 2 [ (1.68x10-8) (20) /  (0.1) ]½
d = 0.0021 m = 2.1 mm
(b) If the current to each speaker is 4.0 A, what is the voltage
drop across each wire?
V=IR
V = (4.0) (0.10)
V = 0.4 V
30
Ohm’s “Law” Revisited
The equation for resistivity I introduced five slides back is a
semi-empirical one. Here’s how we define resistivity:
E
 .
J
Our equation relating R and  follows from the above equation.
We define conductivity  as the inverse of the resistivity:
1
1
  , or   .


31
With the above definitions,
E   J,
J  E.
The “official” Ohm’s “Law”, valid for non-ohmic materials.
Cautions!
In this context:
 is not volume density!
 is not surface density!
32
Example: the 12-gauge copper wire in a home has a crosssectional area of 3.31x10-6 m2 and carries a current of 10 A.
Calculate the magnitude of the electric field in the wire.
I
E  J  
A
(1.72 108   m) 10 C/s 
E
(3.31106 m 2 )
E  5.20 102 V/m
33
Today’s agendum:
Electric Current.
You must know the definition of current, and be able to use it in solving problems.
Current Density.
You must understand the difference between current and current density, and be able to
use current density in solving problems.
Ohm’s “Law” and Resistance.
You must be able to use Ohm’s “Law” and electrical resistance in solving circuit problems.
Resistivity.
You must understand the relationship between resistance and resistivity, and be able to
use calculate resistivity and associated quantities.
Temperature Dependence of Resistivity.
You must be able to use the temperature coefficient of resistivity to solve problems
involving changing temperatures.
34
Temperature Dependence of Resistivity
Many materials have resistivities that depend on temperature.
We can model* this temperature dependence by an equation
of the form
  0 1    T  T0   ,
where 0 is the resistivity at temperature T0, and  is the
temperature coefficient of resistivity.
*T0 is a reference temperature, often taken to be 0 °C or 20 °C. This approximation
can be used if the temperature range is “not too great;” i.e. 100 °C or so.
35
Example: a carbon resistance thermometer in the shape of a
cylinder 1 cm long and 4 mm in diameter is attached to a
sample. The thermometer has a resistance of 0.030 . What is
the temperature of the sample?
Look up resistivity of carbon, use it to calculate resistance.
0  3.519 105   m
This is the resistivity at 20 C.
T0  20C
L = 0.01 m
r = 0.002 m
0 L
R 0  2  0.028 
r
This is the resistance at 20 C.
36
  0.0005 C-1
RA
(R) 
L
1 
T(R)  T0    1
  0 
T(0.030)  122.6 C
The result is very sensitive to significant figures in resistivity
and .
37
Today’s agendum:
Emf, Terminal Voltage, and Internal Resistance.
You must be able to incorporate all of these quantities in your circuit calculations.
Electric Power.
You be able to calculate the electric power dissipated in circuit components, and
incorporate electric power in work-energy problems.
38
circuit components in series
Conservation of energy to implies that the voltage drop across
circuit components in series is the sum of the individual voltage
drops.
Vab
a
C1
V1
C2
V2
C3
b
-Q
+ V
Vab = V = V1 + V2 + V3
39
circuit components in series
Therefore, the voltage drop across resistors in series is the sum
of the individual voltage drops.
a
R1
R2
R3
b
V1
V2
V3
+ -
V
Vab = V = V1 + V2 + V3
This is a consequence of conservation of energy. Use this in
combination with Ohm’s “Law”, V=IR.
40
DC Currents
In Physics 104, whenever you work with currents in circuits,
you should assume (unless told otherwise) “direct current.”
Current in a dc circuit flows in one direction, from + to -.
We will not encounter ac circuits much in this course.
For any calculations involving household current, which is
ac, assuming dc will be “close enough” to give you “a feel”
for the physics.
If you need to learn about ac circuits, you’ll have courses
devoted to them.
The mathematical analysis is more complex. We have other
things to explore this semester.
41
emf, terminal voltage, and internal resistance
We have been making calculations with voltages from batteries
without asking detailed questions about the batteries. Now it’s
time to look inside the batteries.
We introduce a new term – emf – in this section.
Any device which transforms a form of energy into electric
energy is called a “source of emf.”
“emf” is an abbreviation for “electromotive force,” but emf does
not really refer to force!
The emf of a source is the voltage it produces when no current
is flowing.
42
The voltage you measure across the terminals of a battery (or
any source of emf) is less than the emf because of internal
resistance.
Here’s a battery with an emf. All batteries have an “internal
emf is the zero-current potential difference
resistance:”
a
+ -
b
The “battery” is everything
inside the green box.
Hook up a voltmeter to measure the emf:
emf
a
+ -
b
The “battery” is everything
inside the green box.
Getting ready to connect the
voltmeter (it’s not hooked up
yet).
43
Measuring the emf???
a
emf
+ -
I
b
The “battery” is everything
inside the green box.
As soon as you connect the
voltmeter, current flows.
You can’t measure voltage without some (however
small) current flowing, so you can’t measure emf
directly.
You can only measure Vab.
44
We model a battery as producing an emf, , and having an
internal resistance r:
a
+ -

r
b
The “battery” is everything
inside the green box.

Vab
The terminal voltage, Vab, is the voltage you measure with
current flowing. When a current I flows through the battery,
Vab is related to the emf, , by
Vab = ε ± I r .
45
Why the  sign? If the battery is delivering current, the V it
delivers is less than the emf, so the – sign is necessary.
If the battery is being charged, you have to “force” the current
through the battery, and the V to “force” the current through is
greater than the emf, so the + sign is necessary.
Your text writes V = ε - I r and expects you to put the
correct sign (+ or -) on the I. I’ll go along with your text, so our
equation is
Vab = ε - I r .
46
Useful facts:
When current passes through a battery in the direction from
the - terminal toward the + terminal, the terminal voltage
Vab of the battery is Vab = ε - I r .
The sum of the potential changes around a circuit loop is
zero. Potential decreases by IR when going through a
resistor in the direction of the current and increases by 
when passing through an emf in the direction from the - to
+ terminal.
+
I
V is +
+
I
V is -
47
To model a battery, simply include an extra resistor to represent
the internal resistance, and label the voltage source* as an emf
instead of V (units are still volts):
+ -
r

*Remember, all sources of emf—not just batteries—have an internal resistance.
48
Example: a battery is known to have an emf of 9 volts. If a 1
ohm resistor is connected to the battery, the terminal voltage is
measured to be 3 volts. What is the internal resistance of the
battery?
Because the voltmeter draws
“no” current, r and R are in
series with a current I flowing
through both.
ε = Ir + IR
IR, the potential drop across
the resistor, is just the
potential difference Vab.
Vab = IR
R=1 
I
emf
+ -
a
internal resistance r
b
terminal voltage Vab
the voltmeter’s resistance is so
large that approximately zero
current flows through the voltmeter49
ε = Ir + IR
Ir = ε - IR
ε - IR
r=
I
ε
r= -R
I
r=
Vab = IR
Vab
I=
R
εR
-R
Vab
I
R=1 
emf
a
+ -
b
 ε

r = R
- 1
 Vab 
9 
r = 1  - 1  =  3 - 1  = 2
3 
A rather unrealistically large value for
the internal resistance of a 9V battery.
50
By the way, the experiment described in the previous example
is not a very good idea.
I=
I=
Vab
R
3
= 3A
1
51
Today’s agendum:
Emf, Terminal Voltage, and Internal Resistance.
You must be able to incorporate all of these quantities in your circuit calculations.
Electric Power.
You be able to calculate the electric power dissipated in circuit components, and
incorporate electric power in work-energy problems.
52
Electric Power
Last semester you defined power in terms of the work done by
a force.
dWF
PF 
dt
We’d better use the same definition this semester! So we will.
We focus here on the interpretation that power is energy
transformed per time, instead of work by a force per time.
energy transformed
P
time
53
However, we begin with the work aspect. We know the work
done by the electric force in moving a charge q through a
potential difference:
Wif  Uif  qVif .
The work done by the electric force in moving an infinitesimal
charge dq through a potential difference is:
dWif  dq Vif .
The instantaneous power, which is the work per time done by
the electric force, is
dWif
dq Vi f
P

.
dt
dt
54
Let’s get lazy and drop the  in front of the V, but keep in the
back of our heads the understanding that we are talking about
potential difference. Then
dW
dq
P
  V.
dt
dt
But wait! We defined I = dQ/dt. So
P  IV.
And one more thing… the negative sign means energy is being
“lost.” So everybody writes
P  IV
and understands that P<0 means energy out, and P>0 means
energy in.
55
Also, using Ohm’s “law” V=IR, we can write P = I2R = V2/R.
Truth in Advertising I. The V in P=IV is a potential
difference, or voltage drop. It is really a V.
Truth in Advertising II. Your power
company doesn’t sell you power. It sells
energy. Energy is power times time, so a
kilowatt-hour (what you buy from your
energy company) is an amount of energy.
56
Example: an electric heater draws 15.0 A on a 120 V line. How
much power does it use and how much does it cost per 30 day
month if it operates 3.0 h per day and the electric company
charges 10.5 cents per kWh. For simplicity assume the current
flows steadily in one direction.
What’s the meaning of this assumption about
the current direction?
The current in your household wiring doesn’t flow in one
direction, but because we haven’t talked about current other
than a steady flow of charge, we’ll make the assumption. Our
calculation will be a reasonable approximation to reality.
57
An electric heater draws 15.0 A on a 120 V line. How much
power does it use.
P  IV
P  15 A 120 V   1800 W = 1.8 kW
How much does it cost per 30 day month if it operates 3.0 h
per day and the electric company charges 10.5 cents per kWh.
 3 h  $0.105 

cost  1.8kW 30 days 

 day  kWh 
cost  $17.00
58
How much energy is a kilowatt hour (kWh)?
1 kW 1 h   1000 W 3600 s 
J

 1000   3600 s 
s

= 3.6 106 J
So a kWh is a “funny” unit of energy. K (kilo) and h (hours) are
lowercase, and W (James Watt) is uppercase.
59
How much energy did the electric heater use?
Paverage 
Wdone by force
time
Energy Transformed

time
Energy Transformed   Paverage   time 
 3 h used   3600 s 
J

Energy Transformed  1800   30 days  


s

 day   h 
Energy Transformed  583, 200, 000 Joules used
60
Energy Transformed  583, 200, 000 Joules used
That’s a ton of joules! Good bargain for $17. That’s about
34,000,000 joules per dollar (or 0.0000029¢/joule).
OK, “used” is not an SI unit, but I stuck it in there to help me
understand. And joules don’t come by the ton.
One last quibble. You know from energy conservation that you
don’t “use up” energy. You just transform it from one form to
another.
61
Example: a typical lightning bolt can
transfer 109 J of energy across a
potential difference of perhaps 5x107 V
during a time interval of 0.2 s. Estimate
the total amount of charge transferred,
the current and the average power over
the 0.2 s.
What kind of a problem is this?
learn about
lightning at
howstuffworks
You are given energy, potential difference, and time.
You need to calculate charge transferred, current, and average
power. Equations for current and power are “obvious:”
Iavg
Q

t
Pavg
W

t
62
We could calculate power right now, but let’s do this in the
order requested. Besides, we can’t get current without Q,
charge transferred.
U  qV
We need to think in terms of energy transformations rather
than work done by forces. The equation above tells us that
potential energy stored in clouds can be transferred to the
ground (at a different potential) by moving charge from cloud
to ground. We are given energy transferred and potential
difference, so we can calculate q.
“Could I think of the cloud-earth system as a giant capacitor which stores energy?”
You could, except our capacitor equation U=QV/2 assumes the same charge on both
63
plates; that’s untrue here.
Continuing with our energy transformation idea:
Etransferred= Qtransferred Vif
Qtransferred = Etransferred / Vif
Qtransferred = 109 J / 5x107 V
Qtransferred = 20 C
That’s a lot of charge (remember, typical charges are 10-6 C).
Once we have the charge transferred, the current is easy.
ΔQ
I=
Δt
20 C
I=
= 100 A
0.2 s
64
Average power is just the total energy transferred divided by
the total time.
WF
PF =
t
E transferred
P=
t
109 J
P=
0.2 s
P = 5×109 W
P = 5 GW
The numbers in this calculation differ substantially
than the numbers in a homework problem(not
necessarily assigned this semester). “This”
lightning bolt carries relatively low current for a
long time through a high potential difference, and
transports a lot of energy.
In reality, there is no such thing as a universallytypical lightning bolt, so expect different results for
different bolts.
Holy ****, Batman. That’s the power output of five enormous
power plants!
65
Example: A 12 V battery with 2  internal resistance is
connected to a 4  resistor. Calculate (a) the rate at which
chemical energy is converted to electrical energy in the battery,
(b) the power dissipated internally in the battery, and (c) the
power output of the battery.
(a) Rate of energy conversion.
The total resistance in the
circuit is 6 , so
R=4
+-
V = I Rtotal
I
r=2
 = 12 V
I =  / Rtotal = 12 V / 6  = 2 A
Energy is converted at the rate Pconverted=I=(2 A)(12 V)=24W.
66
(b) Power dissipated internally in the battery.
R=4
+-
I=2A
r=2
 = 12 V
Pdissipated = I2r = (2 A)2 (2 ) = 8 W.
(c) Power output of the battery.
Poutput = Pconverted - Pdissipated = 24 W - 8 W = 16W.
67
(c) Power output of the battery (double-check).
I=2A
R=4
+-
I=2A
r=2
 = 12 V
The output power is delivered to (and dissipated by) the
resistor:
Poutput = Presistor = I2 R = (2 A)2 (4 ) = 16W.
68
Example: the electric utility company
supplies your house with electricity
from the main power lines at 120 V.
The wire from the pole to your house
has a resistance of 0.03 . Suppose
your house is drawing 110 A of
current…
I
VT
VH
R
(a) Find the voltage at the point where the power wire enters
your house.
VHT = IR
VT-VH = IR
VH = VT-IR
VH = (120 V) – (110 A) (0.03) = 116.7 V
69
(b) How much power is being dissipated in the wire from the
pole to your house?
I
VT
VH
Three different ways
to solve; all will give
the correct answer.
R
P = IV = I2R = (V)2/R
P = I(VT-VH) = I2R = (VT-VH)2/R
P = (110 A) (120 V -116.7 V) = 363 W
or P = (110 A)2 (0.03) = 363 W
or P = (120 V – 116.7 V)2 / (0.03) = 363 W
70
(c) How much power are you using inside your house?
I
VT
VH
R
You need to understand that your household voltage represents
the potential difference between the “incoming” and “outgoing”
power lines, and the “outgoing” is at ground (0 V in this
case)…except…
…because the “outgoing” power line is at 0 V, you can
“accidentally” get this correct if you simply multiply the current
by the voltage at the point where the power wire enters your
house.
71
(c) How much power are you using inside your house?
I
VT
VH
R
P = IV
P = (110 A) (116.7 V – 0 V)
P = 12840 W
You don’t want to use the P=I2R=V2/R equations because you
don’t know the effective resistance of your house (although you
could calculate it).
P = (110 A) (120 V) – (110 A)(3.3 V) is
also a reasonable way to work this part.
72
Today’s agendum:
Resistors in Series and Parallel.
You must be able to calculate currents and voltages in circuit components in series and in
parallel.
Kirchoff’s Rules.
You must be able to use Kirchoff’s Rules to calculate currents and voltages in circuit
components that are not simply in series or in parallel.
73
Resistances in Circuits
There are “two” ways to connect circuit elements.
Series:
A
B
Put your finger on the wire at A. If you can move along the
wires to B without ever having a choice of which wire to
follow, the circuit components are connected in series.
Truth in advertising: it is possible to have circuit elements that are connected
neither in series nor in parallel. See problem 24.73 for an example with capacitors.
74
Parallel:
A
B
Put your finger on the wire at A. If in moving along the
wires to B you ever have a choice of which wire to follow,
the circuit components are connected in parallel.*
*Truth in advertising: actually, the circuit components are
not connected in series, and may be connected in parallel.
75
Are these resistors in series or parallel?
76
Here’s a circuit with three resistors and a battery:
I
I
I
R1
R2
R3
V1
V2
V3
+ -
I
V
Current flows…
…in the steady state, the same current flows through all
resistors…
…there is a potential difference (voltage drop) across each
resistor.
77
Applying conservation of energy allows us to calculate the
equivalent resistance of the series resistors.
I am including the derivation in these notes, for the benefit of
students who want to look at it.
78
I
I
I
R1
R2
R3
V1
V2
V3
+ -
I
V
An electric charge q is given a potential energy qV by the
battery.
As it moves through the circuit, the charge loses potential
energy qV1 as it passes through R1, etc.
The charge ends up where it started, so the total energy lost
must equal the initial potential energy input:
qV = qV1 + qV2 + qV3 .
79
I
I
I
R1
R2
R3
V1
V2
V3
+ -
I
V
qV = qV1 + qV2 + qV3
V = V1 + V2 + V3
V = IR1 + IR2 + IR3
Now imagine replacing the three resistors by a single resistor,
having a resistance R such that it draws the same current as
the three resistors in series.
80
I
Req
V
+ -
I
As above:
From before:
Combining:
V
V = IReq
V = IR1 + IR2 + IR3
IReq = IR1 + IR2 + IR3
Req = R1 + R2 + R3
For resistors in series, the total resistance is the sum of the
separate resistances.
81
We can generalize this to any number of resistors:
R eq   R i
(resistors in series)
a consequence of
conservation of energy
i
R1
R2
R3
+ -
V
82
Here’s another circuit with
three resistors and a battery.
I1
R1
V
I2
Current flows…
R2
V
…different currents flows through
different resistors…
R3
I3
V
…but the voltage drop across
each resistor is the same.
+ -
I
V
83
Applying conservation of charge allows us to calculate the
equivalent resistance of the parallel resistors.
I am including the derivation in these notes, for the benefit of
students who want to look at it.
84
I1
In the steady state, the
current I “splits” into I1, I2,
and I3 at point A.
V
I2
A
I1, I2, and I3 “recombine” to
make a current I at point B.
B
R3
I3
I2 =
V
+ V
I
Because the voltage drop across
each resistor is V:
V
R1
R2
V
Therefore, the net current
flowing out of A and into B is I
= I1 + I2 + I3 .
I1 =
R1
V
R2
I3 =
V
R3
I
85
I
Now imagine replacing the
three resistors by a single
resistor, having a resistance R
such that it draws the same
current as the three resistors in
parallel.
Req
A
B
V
+ -
From above, I = I1 + I2 + I3, and
V
I1 =
R1
So that
V
I2 =
R2
I
V
I
V
I3 =
.
R3
V
V
V
V
= + + .
R eq R1 R 2 R 3
86
Dividing both sides by V gives
1
1
1
1
= + + .
R eq R1 R 2 R 3
We can generalize this to any number of resistors:
1
1

R eq
i Ri
(resistors in parallel)
a consequence of
conservation of charge
87
Summary:
Series
B
A
i
same I, V’s add
Parallel
A
R eq   R i
B
1
1

R eq
i Ri
same V, I’s add
88
Example: calculate the equivalent resistance of the resistor
“ladder” shown. All resistors have the same resistance R.
I’ll work this “conceptually.
A
B
Here’s the key to solving Physics problems: don’t bite off more
than you can chew. Bite off little bite-sized chunks.
89
A hot dog. Where do you take the first bite?
A
B
90
Not a “law” of physics, but sometimes helps with circuits: look
for “bite-sized” chunks sticking out at one end.
Series
A
B
91
The new color indicates the value of the resistance is not R. In
a real problem, you would calculate the “new color” resistor’s
resistance.
Parallel
A
B
Any more bite-sized chunks?
92
Series
A
B
93
Parallel
A
B
94
Series
A
B
95
All done!
A
B
96
Example: For the circuit below, calculate the current drawn
from the battery and the current in the 6  resistor.
10 
8
6
8
1
3
9V
The lecture notes are a “conceptual” solution. I may work the
97
problem on the blackboard in lecture.
In a few minutes, we will learn a general technique for solving
circuit problems. For now, we break the circuit into
manageable bits. “Bite-sized chunks.”
10 
8
6
8
3
1
9V
Replace the parallel combination (green) by its equivalent.
Do you see any bite-sized chunks that are simple series or
parallel?
98
Any more “bite-sized chunks?” Remember that everything
inside the green box is equivalent to a single resistor.
10 
8
6
8
1
3
9V
Replace the series combination (blue box) by its equivalent.
99
We are left with an equivalent circuit of 3 resistors in series,
which is easy handle.
10 
8
6
8
1
3
9V
Next bite-sized chunk. Inside the blue box is “a” resistor.
Replace the parallel combination (orange) by its equivalent. 100
Example: two 100  light bulbs are connected (a) in series
and (b) in parallel to a 24 V battery. For which circuit will
the bulbs be brighter?
Please make your selection...
1.
2.
parallel (left)
series (right)
101
Example: two 100  light bulbs are connected (a) in series and
(b) in parallel to a 24 V battery. What is the current through
each bulb and what is the equivalent resistance of each circuit?
(a) Series combination.
Req = Ri
R1
Req = R1 + R2
V = I Req
R2
+ -
I
V = 24 V
V = I (R1 + R2)
I = V / (R1 + R2) = 24 V / (100  + 100 ) = 0.12 A
102
The same current of 0.12 A flows through each bulb.
The equivalent resistance is
Req = R1 + R2
Req = 100  + 100  = 200  .
(b) Parallel combination.
1
=
R eq

i
1
Ri
I1
R1
V
I2
R2
V
1
1
1
=
+
R eq
R1
R2
+ -
I
V = 24 V
I
103
V = I Req
V=
I
1
1
+
R1
R2
1
1 
I= V  +

R
R
2 
 1
1 
 1
I = 24V 
+

100
Ω
100
Ω


 200 
I = 24 
 = 0.48 A
 10000 
104
The equivalent resistance is
1
1   200 Ω 
 1
= 
+
=
2 
R eq
100
Ω
100
Ω
10000
Ω

 

R eq = 50 Ω
For which of the two circuits above would the bulb(s) be
brighter…
105
To answer the question, we must calculate the power dissipated
in the bulbs for each circuit. The more power “consumed,” the
brighter the bulb.
In both circuits, the bulbs are identical and have identical
currents passing through them. We pick either bulb for the
calculation.
Series circuit: we know the resistance and current through
each bulb, so we use:
P = I2 R
P = (0.12 A)2 (100 )
P = 1.44 W
106
Parallel circuit: we know the resistance and voltage drop across
each bulb, so we use:
P = V2 / R
P = (24 V)2 / ( 100 )
P = 5.76 W
Compare:
Pseries = 1.44 W
Pparallel = 5.76 W
The bulbs in parallel are brighter.
107
This is what you see if you connect 40 W bulbs directly to a 120
V outlet. (DO NOT TRY AT HOME.)
Off
On
108
Today’s agendum:
Resistors in Series and Parallel.
You must be able to calculate currents and voltages in circuit components in series and in
parallel.
Kirchhoff’s Rules.
You must be able to use Kirchoff’s Rules to calculate currents and voltages in circuit
components that are not simply in series or in parallel.
109
Kirchhoff’s Rules
No, it is pronounced “KEERK-HOFF’s” rules. The ch sounds like
“k,” not like “ch.” [Means church yard, cemetery.]
Analyze this circuit for me, please. Find the currents I1, I2, and
I3.
h
30 
I1
I3
40 
a
1
2 = 45 V
c
b
d
20 
I2
1 = 85 V 1 
g
f
e
110
I see two sets of resistors in series.
This.
And this.
You know how to analyze those.
Further analysis is difficult. For example, series1 seems to be in
parallel with the 30  resistor, but what about 2? You don’t
know how to analyze that combination.
h
30 
I1
I3
40 
a
1
2 = 45 V
c
b
series1
I2
1 = 85 V 1 
g
f
series2
e
d
20 
111
A new technique is needed to analyze this, and far more
complex circuits.
Kirchhoff’s Rules
Kirchhoff’s Junction Rule: at any junction point, the sum of all
currents entering the junction must equal the sum of all
currents leaving the junction. Also called Kirchhoff’s First Rule.*
Kirchhoff’s Loop Rule: the sum of the changes of potential
around any closed path of a circuit must be zero. Also called
Kirchhoff’s Second Rule.**
*This is just conservation of charge: charge in = charge out.
**This is just conservation of energy: a charge ending up
where it started out neither gains nor loses energy (Ei = Ef ).112
Kirchhoff’s Rules
Starting Equations
I = 0
V =0
at any junction
around any closed loop
simple… but there are details to worry about…
113
Solving Problems with Kirchhoff’s Rules
If this were Physics 103, you would have a procedure for circuit
problems.
Procedure for Circuit Problems
1. Draw the circuit.
2. Label + and – for each battery (the short side is -).
3. Label the current in each branch of the circuit with a
symbol and an arrow. You may choose whichever
direction you wish for the arrow.
4. Apply Kirchoff’s Junction Rule at each junction. The
direction of the current arrows tell you whether current is
flowing in (+) or out (-).
Step 4 will probably give you fewer equations than variables.
114
Proceed to step 5 go get additional equations.
5. Apply Kirchhoff’s Loop Rule for as many loops as
necessary to get enough equations to solve for your
unknowns. Follow each loop in one direction only—your
choice.
5a. For a resistor, the sign of the potential difference
is negative if your chosen loop direction is the same
as the chosen current direction through that resistor;
positive if opposite. 5a. Resistor:
I
V is 5b. For a battery, the sign of the potential difference
is positive if your chosen loop direction moves from
the negative terminal towards the positive; negative
+
if opposite.
5b. Battery:
- V is +
loop
115
6. Collect equations, solve, and check results.
h
30 
I1
I3
40 
a
1
2 = 45 V
c
b
dd
20 
I2
1 = 85 V
g
1
f
e
Back to our circuit: we have 3 unknowns (I1, I2, and I3), so we
will need 3 equations. We begin with the junctions.
Junction a:
I3 – I1 – I 2 = 0
Junction d:
-I3 + I1 + I2 = 0
--eq. 1
Junction d gave no new information, so we still need two more equations.116
h
30 
I1
I3
40 
a
1
2 = 45 V
c
b
d
20 
I2
1 = 85 V
g
There are three loops.
1
e
f
Loop 1.
Loop 2.
Loop 3.
Any two loops will produce independent equations. Using the
third loop will provide no new information.
117
Reminders:
I
V is -
loop
+-
V is +
loop
5
The “green” loop (a-h-d-c-b-a):
(- 30 I1) + (+45) + (-1 I3) + (- 40 I3) = 0
- 30 I1 + 45 - 41 I3 = 0
--eq. 2
The “blue” loop (a-b-c-d-e-f-g):
(+ 40 I3) + ( +1 I3) + (-45) + (+20 I2) + (+1 I2) + (-85) = 0
41 I3 -130 + 21 I2 = 0
--eq. 3
Three equations, three unknowns; the rest is “algebra.”
118
Make sure to use voltages in V and resistances in . Then currents will be in A.
Collect our three equations:
I3 – I1 – I2 = 0
- 30 I1 + 45 - 41 I3 = 0
41 I3 -130 + 21 I2 = 0
Rearrange to get variables in “right” order:
– I1 – I2 + I3 = 0
- 30 I1 - 41 I3 + 45 = 0
21 I2 + 41 I3 -130 = 0
Use the middle equation to eliminate I1:
I1 = (41 I3 – 45)/(-30)
There are many valid sets of steps to solving a system of equations. Any that works is acceptable.
119
Two equations left to solve:
– (41 I3 – 45)/(-30) – I2 + I3 = 0
21 I2 + 41 I3 -130 = 0
Might as well work out the numbers:
1.37 I3 – 1.5 – I2 + I3 = 0
21 I2 + 41 I3 -130 = 0
– I2 + 2.37 I3 – 1.5 = 0
21 I2 + 41 I3 -130 = 0
Multiply the top equation by 21:
– 21 I2 + 49.8 I3 – 31.5 = 0
21 I2 + 41 I3 -130 = 0
120
Add the two equations to eliminate I2:
– 21 I2 + 49.8 I3 – 31.5 = 0
+ ( 21 I2 + 41 I3 -130 = 0 )
90.8 I3 – 161.5 = 0
Solve for I3:
I3 = 161.5 / 90.8
I3 = 1.78
Go back to the “middle equation” two slides ago for I1:
I1 = (41 I3 – 45)/(-30)
I1 = - 1.37 I3 + 1.5
I1 = - (1.37) (1.78) + 1.5
I1 = - 0.94
121
Go back two slides to get an equation that gives I2:
– I2 + 2.37 I3 – 1.5 = 0
I2 = 2.37 I3 – 1.5
I2 = (2.37) (1.78) – 1.5
I2 = 2.72
Summarize answers so your lazy professor doesn’t have to go
searching for them and get irritated (don’t forget to show
units in your answer):
I1 = - 0.94 A
I2 = 2.72 A
I3 = 1.78 A
Are these currents correct? How could you tell? We’d better
check our results.
122
I3 – I1 – I2 = 0
- 30 I1 + 45 - 41 I3 = 0
41 I3 -130 + 21 I2 = 0
I1 = - 0.94 A
I2 = 2.72 A
I3 = 1.78 A
1.78 – (-0.94) – 2.72 = 0

- 30 (-0.94) + 45 - 41 (1.78) = 0.22
?
41 (1.78) -130 + 21 (2.72) = 0.10
?
Are the last two indication of a mistake or just round off error? Recalculating
while retaining 2 more digits gives I1=0.933, I2=2.714, I3=1.7806, and the
last two results are 0.01 or less  round off was the culprit.
123
Today’s agendum:
RC Circuits.
You must be able to calculate currents and voltages in circuits containing both a resistor
and a capacitor. You must be able to calculate the time constant of an RC circuit, or use
the time constant in other calculations.
Leftovers.
Optional (not for test) material, if time permits.
124
RC Circuits
RC circuits contain both a resistor and a capacitor (duh).
Until now we have assumed that charge is
instantly placed on a capacitor by an emf.
Q
t
The approximation resulting from this
assumption is reasonable, provided the
resistance between the emf and the capacitor
being charged/discharged is small.
If the resistance between the emf and the
capacitor is not small, then the charge on the
capacitor does not change instantaneously.
Q
t
125
Charging a Capacitor
Switch open, no current flows.
I
Close switch, current flows.

C
Apply Kirchoff’s loop rule*
(green loop) at the instant
charge on C is q.
ε-
q
- IR = 0
C
This equation is
deceptively
complex because
I depends on q
and both depend
on time.
R
switch
t<0
t>0
*Convention for capacitors is “like” batteries: negative if going across from + to -.
126
Limiting Cases
q
ε - - IR = 0
C
When t=0, q=0 and I0=/R.
When t is “large,” the capacitor
is fully charged, the current
“shuts off,” and Q=C.
I
C

R
switch
127
Math:
ε-
q
- IR = 0
C
ε q
I= R RC
dq ε q
Cε q
Cε - q
= =
=
dt R RC RC RC
RC
dq
dt
=
Cε - q RC
dq
dt
=q- Cε
RC
128
More math:

q
0
t dt
dq
=-
0 RC
q- Cε
1 t
ln  q- Cε  0 = dt

0
RC
q
t
 q - Cε 
ln 
=

-C
ε
RC


t
q- Cε
= e RC
-Cε
q- Cε = -Cε e
-
t
RC
129
Still more math:
q = Cε - Cε e
-
t
RC
t


RC
q = Cε 1- e 


t


RC
q  t  = Q 1- e 


dq Cε - RCt
Cε - RCt
ε - RCt
It =
=
e =
e = e
dt RC
RC
R
RC is the “time constant” of the circuit; it tells us “how fast”
the capacitor charges and discharges.
130
Charging a capacitor; summary:
t


RC
q  t  = Q 1- e 


ε - RCt
It = e
R
Charging Capacitor
0.01
0.05
0.008
0.04
0.006
0.03
I (A)
q (C)
Charging Capacitor
0.004
0.002
0.02
0.01
0
0
0
0.2
0.4
0.6
t (s)
0.8
1
0
0.2
0.4
0.6
0.8
1
t (s)
Sample plots with =10 V, R=200 , and C=1000 F.
RC=0.2 s
131
In a time t=RC, the capacitor charges to Q(1-e-1) or 63% of its
capacity…
…and the current drops to Imax(e-1) or 37% of its maximum.
Charging Capacitor
0.01
0.05
0.008
0.04
0.006
0.03
I (A)
q (C)
Charging Capacitor
0.004
0.002
0.02
0.01
0
0
0
0.2
0.4
0.6
0.8
1
t (s)
0
0.2
0.4
0.6
0.8
1
t (s)
RC=0.2 s
=RC is called the time constant of the RC circuit
132
Discharging a Capacitor
Capacitor charged, switch
open, no current flows.
Close switch, current flows.
Apply Kirchoff’s loop rule*
(green loop) at the instant
charge on C is q.
q
- IR = 0
C
I
C
+Q
+q
-q
-Q
R
switch
t<0
t>0
*Convention for capacitors is “like” batteries: positive if going across from - to +.
133
Math:
q
- IR = 0
C
IR =
q
C
I=
dq
dt
-R
negative because
charge decreases
dq q
=
dt C
dq
dt
=q
RC
134
More math:
t dt
dq
1 t
Q q = - 0 RC = - RC 0 dt
q
1 t
ln  q  Q = dt

0
RC
q
t
q
ln   = RC
Q
q(t) = Q e
-
t
RC
t
dq Q - RCt
I(t) = - =
e = I0 e RC
dt RC
same equation
as for charging
135
Disharging a capacitor; summary:
q(t) = Q e
-
t
RC
ε - RCt
It = e
R
Discharging Capacitor
0.01
0.05
0.008
0.04
0.006
0.03
I (A)
q (C)
Discharging Capacitor
0.004
0.02
0.002
0.01
0
0
0
0.2
0.4
0.6
t (s)
0.8
1
0
0.2
0.4
0.6
0.8
1
t (s)
Sample plots with =10 V, R=200 , and C=1000 F.
RC=0.2 s
136
In a time t=RC, the capacitor discharges to Qe-1 or 37% of its
capacity…
…and the current drops to Imax(e-1) or 37% of its maximum.
Discharging Capacitor
0.01
0.05
0.008
0.04
0.006
0.03
I (A)
q (C)
Discharging Capacitor
0.004
0.02
0.002
0.01
0
0
0
0.2
0.4
0.6
0.8
1
t (s)
0
0.2
0.4
0.6
0.8
1
t (s)
RC=0.2 s
137
Today’s agendum:
RC Circuits.
You must be able to calculate currents and voltages in circuits containing both a resistor
and a capacitor. You must be able to calculate the time constant of an RC circuit, or use
the time constant in other calculations.
Leftovers.
Optional (not for test) material, if time permits.
138
Real quickly… some leftover circuits material…if time permits…
EMF’s in Series and in Parallel: Charging a Battery
If you put batteries in series the “right way,” their voltages add:
+
6V
=
3V
9V
If you put batteries in series the “wrong way,” their voltages
add algebraically:
+
=
magnitudes only 
6V
3V
chosen loop direction
-6 V
+3 V
3V
-3 V  algebraically,
using chosen loop
139
direction
Why would you want to connect emf’s in series?
More voltage! Brighter flashlights, etc. Chemical reactions in
batteries yield a fixed voltage. Without changing the chemical
reaction (i.e., inventing a new battery type), the only way to
change voltage is to connect batteries in series.
Why would you want to connect batteries in series the
“wrong” way?
You could connect a source of emf – like the alternator in your
car – so that it charges a battery.
Rechargeable batteries use an ac to dc converter as a source of
emf for recharging.
140
Go to www.howstuffworks.com to see how batteries work.
They even expose the secret of the 9 volt battery!
Click on the picture above only if you are mature enough to
handle this graphic exposé.
141
Go to www.howstuffworks.com to see how batteries work.
They even expose the secret of the 9 volt battery!
Shocking!
Six 1.5 V batteries in series!
142
Could you connect batteries (or sources of emf) in parallel?
Sure!
a
b
3V
3V
You would still have a 3 V voltage drop across your resistor,
but the two batteries in parallel would “last” longer than a
single battery.
You could use Kirchhoff’s rules to analyze this circuit and show
that Vab = 3 V.
143
d.
Energy storage in magnetic fields
i) RL-circuit
EMF  IR  EMFL  0
dI
0
dt
EMF
L dI
I 
0
R
R dt
EMF  IR  L
I
EMF 
1  e
R 
R
t
L
I  EMF  I 2 R  LI
dU m
dI
 LI
dt
dt
Um
I
0
0
 dU m   LIdI
Um 




dI
0
dt
Specifically for a coil,
L   o n 2 A
, and inside the coil (a long coil)
B   o nI
1 2
LI
2
. This allows us to substitute in Um for the I in terms of B.
Um
1 2
B2
LI 
A
2
2 o
The energy density is the energy divided by the volume enclosed inside the coil.
um 
Um
B2

 B2
A 2 o
144
ii) LC-circuit
We assume that there is no resistance in the circuit. Initially, the capacitor is charged with Qmax.
2
Qmax
The total energy in the circuit is U = UC + Um. At t = 0, I = 0 and
.
U
2C
When the switch is closed, the capacitor begins to discharge.
The current increases to a maximum Imax at t = t’, at which time U
Q
dI
L
0
C
dt
Q
d 2Q
L 2 0
C
dt
2
d Q
1

Q
2
LC
dt
 t 
Q  Qmax cos

 LC 
I
Q
 t 
dQ
  max sin 

dt
LC
 LC 

1
LC
145
iii) RLC-circuit
dU
 I 2 R
dt
d 2Q
dQ Q
R
 0
2
dt C
dt
1


 Rt
 1  R 2  2 
2L


Q  Qmax e cos
 
t

 
  LC  2 L   


L
This is exactly the same equation we obtain for the damped harmonic
oscillator—a harmonic oscillator with a dissipative force, such as friction.
146
a.
Alternating current & voltage
1.5
1
current
v  Vmax sin  t 
v V
i   max sin  t   I max sin  t 
R
R
P  i R  RI
2
2
max
I max
Vrms 
Vmax
0
-0.5
-1
sin  t 
2
2
 P  R  I max
sin 2  t  
I rms 
0.5
-1.5
2
max
0
RI
2
100
200
300
400
500
600
700
800
time
2
2
i) reactance
Consider an inductor circuit.
Similarly for a capacitor, we define a capacitive reactance,
XC 
1
C
Q  CVmax sin  t 
i
dQ


 CVmax  cos t   CVmax  sin   t  
dt
2

vL
di
0
dt
Vmax sin  t   L
vC  I max X C sin  t 
VC  I max X C
VL  I max X L
VR  I max R
XL 
di
0
dt
Vmax
 L
I max
v L  I max X L sin  t 
t
Vmax
sin  t dt
L
0
 di  
V
V


i L   max cos t    max sin   t  
L
L 
2
147
ii) impedance
Consider an RLC circuit, with an alternating EMF.
v  Vmax sin  t
v L  I max X L sin  t
IL 
Vmax


sin   t  
XL
2



v R  Vmax sin   t  
2

 

vC  I max X C sin   t   
2 2

2
Vmax
 VL  VC   VR2
2
Vmax  VR2  VL  VC   I max R 2   X L  X C   I max Z
2
2
c.
Transformers
Consider a circuit having two coils, a source EMF, and a load resistance as shown.
v1 = input voltage (primary)
v2 = output voltage (secondary)
N1 = number of turns (loops) on primary coil
N2 = number of turns on secondary coil
R = load resistance
v1   N 1
d
dt
v2   N 2
v  N
d
  N 2  1   2 v1  i2 R
dt
 N1  N1
2
N 
N
v
N
v N
v1  i1 1 R 1  i1 1 R 1 1  i1  1  R  i1 Req
N 2 v2
N 2 v1 N 2
 N2 
148