Monday, February 17, 2014
Origin of resistivity
 ions are at rest, electrons flow
 ions vibrate, electrons collide with ions and stop
 Stopping of electron: stopping of a current.
 At higher T, ions vibrate with larger amplitude: higher
resistance
electrons
ions
A
2
Monday, February 17, 2014
Other Temperature Dependence
 At very low temperatures the
resistivity of some materials goes to
exactly zero (not just small, exactly
0). These materials are called
superconductors
resistivity of a superconductor
electrons do not collide with ions at all
 The resistance of some
semiconducting materials actually
decreases with increasing
temperature
number of conducting
electrons depends on T
Monday, February 17, 2014
42
Power dissipated by circuit with
resistively
 drifting electrons collide with ions and stop.
 energy of drift motion (energy of a flowing current)
is given to ions.
 Ions are heated
electrons
ions
A
 How much power a current dissipates?
4
Monday, February 17, 2014
Energy and Power in Electric Circuits (1)
 Consider a simple circuit in which a source of emf with voltage V
causes a current i to flow in a circuit.
 The work required to move a differential amount of charge dq around
the circuit is equal to the differential electric potential energy dU given
by
 So we can rewrite the differential electric potential energy as
 The definition of power P is
 Putting it together
65
Monday, February 17, 2014
Energy and Power in Electric Circuits (2)
 The power dissipated in a circuit or circuit element is given by the product of
the current times the voltage
 Using Ohm’s Law we can write equivalent formulations of the power
with
 The unit of power is the watt (W)
 Electrical devices are rated by the amount of power they consume in watts
 Electricity bill
is based
on how many
kilowatt-hours
of electrical energy you
kW
h = power
times
time
consume
 The energy is converted to heat, motion, light, …
1 kW h = 1000 W X 3600 s = 3.6 x 106 joules
66
Monday, February 17, 2014
 A resistor R is connected to a battery with fixed
voltage V. What needs to be done to increase the
dissipated power?
 A: Increase resistance
 B: Decrease resistance
Very small resistance: short circuit!
Only resistance of the battery.
7
Monday, February 17, 2014
 A resistor R is connected to a battery with fixed
voltage V. What needs to be done to increase the
dissipated power?
 A: Increase resistance
 B: Decrease resistance
Very small resistance: short circuit!
Only resistance of the battery.
Boom!
7
Monday, February 17, 2014
Electromotive Force
 To make current flow through a resistor one must
establish a potential difference across the resistor.
 This potential difference is termed an electromotive force, emf.
 A device that maintains a potential difference is called an emf device
and does work on the charge carriers
 The potential difference created by the emf device is termed Vemf
 We will assume that emf devices have terminals that we can connect
and the emf device is assumed to maintain Vemf between these
terminals
43
Monday, February 17, 2014
Ohm’s Law
 Consider a simple circuit of the form shown below
 Here a source of emf provides a voltage V across a resistor with
resistance R
 The relationship between the voltage and the resistance in this
circuit is given by Ohm’s Law
… where i is the current in the circuit
45
Monday, February 17, 2014
Two forms of Ohm’s law
L
V = iρ
A
V
i
=ρ
L
A
E = ρJ
10
Monday, February 17, 2014
Two forms of Ohm’s law
L
V = iρ
A
V
i
=ρ
L
A
E = ρJ
or
E = ρJ
10
Monday, February 17, 2014
Visualization of Ohm’s Law
 Now let’s visualize the same circuit in a different way, making it clearer
where the potential drop
happens and what part of the circuit is at
which potential
 The top part of this drawing is just
our original circuit diagram
 In the bottom part we show the same circuit,
but now the vertical dimension represents the
voltage drop around the circuit
 The voltage is supplied by the source of emf and
the entire voltage drop occurs across the single
resistor
46
Monday, February 17, 2014
Resistances in Series
 Resistors connected such that all the current in a
circuit must flow through each of the resistors are
connected in series
 For example, two resistors R1 and R2 in series with one
source of emf with voltage Vemf implies the circuit shown
below
i
Charge
conservation:
i1=i2
47
Monday, February 17, 2014
Two Resistors in 3D
 To illustrate the voltage drops in this
circuit we can represent the same circuit
in three dimensions
 The voltage drop across resistor R1
is V1
 The voltage drop across resistor R2
is V2
 The sum of the two voltage drops
must equal the voltage supplied by
the battery
48
Monday, February 17, 2014
Resistors in Series
 The current must flow through all the elements of
the circuit so the current flowing through each
element of the circuit is the same
 For each resistor we can apply Ohm’s Law
… where
 We can generalize this result to a circuit with n
resistors in series:
49
Monday, February 17, 2014
Example: Internal Resistance of a Battery (1)
 When a battery is not connected in a circuit, the
voltage across its terminals is Vt
 When the battery is connected in series with a resistor with resistance R,
current i flows through the circuit
 When current is flowing, the voltage, V, across the terminals of the battery is
lower than Vt
 This drop occurs because the battery has an internal resistance, Ri, that can be
thought of as being in series with the external resistor
50
Monday, February 17, 2014
Example: Internal Resistance of a Battery (2)
battery
terminals of the
battery
 We can express this relationship as
 Consider a battery that has a voltage of 12.0 V when it is not
connected to a circuit
 When we connect a 10.0 Ω resistor across the terminals, the
voltage across the battery drops to 10.9 V
Question:
What is the internal resistance of the battery?
51
Monday, February 17, 2014
Example: Internal Resistance of a Battery (3)
Answer:
 The current flowing through the external resistor is
 The current flowing in the complete circuit must be the
same so
52
Monday, February 17, 2014
Resistances in Parallel (1)
 Instead of connecting resistors in series so that all the current
must pass through both resistors, we can connect the resistors in
parallel such that the current is divided between the two resistors
 This type of circuit is shown below
53
Monday, February 17, 2014
Resistance in Parallel (2)
 In this case, the voltage drop across each resistor is
equal to the voltage provides by the source of emf
 Using Ohm’s Law we can write the current in each resistor
 The total current in the circuit must equal the sum of
these currents
 Which we can rewrite as
54
Monday, February 17, 2014
Resistance in Parallel (3)
 We can then rewrite Ohm’s Law for the complete circuit as
Req
=
 .. where
 We can generalize this result for two parallel resistors to n
parallel resistors
55
Monday, February 17, 2014
Example: Network of Resistors (1)
 Consider the network of resistors shown below
 Calculate the current flowing in this circuit.
56
Monday, February 17, 2014
Example: Network of Resistors (2)
 First look for pairs of resistors: R3 and R4 are in series
 Now note that R34 and R1 are in parallel
57
Monday, February 17, 2014
Example: Network of Resistors (3)
 And now R2, R5, R6, and R134 are in series
58
Monday, February 17, 2014
Example 2: More Resistors (1)
 The figure shows a circuit containing one ideal 12-V battery (no internal
resistance) and 4 resistors with R1=20 ΩΩ, R2=20 ΩΩ, R3=30 ΩΩ, and R4=8 ΩΩ.
Question:
What is the current through the battery?
Answer:
Idea: Find the equivalent resistance and
use Ohm’s Law.
R2 and R3 are in parallel.
59
Monday, February 17, 2014
Example 2: More Resistors (2)
What is the current through the battery?
 R23=12 Ω
 R1, R23 and R4 are in series.
60
Monday, February 17, 2014
Example 2: More Resistors (3)
 The circuit contains one ideal 12 V battery (no
internal resistance) and 4 resistors with R1=20
ΩΩ, R2=20 ΩΩ, R3=30 ΩΩ, and R4=8 ΩΩ
Question:
What is the current i2 through R2?
I know itot
V
=
Rtot
itot
i2
i3
i2 + i3 = itot
V2 = V 3 = V
V = itot R23
R23
V
= itot
i2 =
R2
R2
61
Monday, February 17, 2014
Light Bulbs in Parallel and in Series
In parallel:
+12 V
Observation: Take
out one bulb,
nothing happens to
the others
- 12V
Assume: the bulbs are almost identical and
have the same resistance
63
Monday, February 17, 2014
Light Bulbs in Parallel and in Series
In series:
+12 V
- 12V
Observation: Taking
one bulb out breaks
the circuit. The
more bulbs we put in
series, the dimmer
they get!
Assume: the bulbs are almost identical and have
the same resistance
64
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (1)
67
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (1)
 A 100 W light bulb is connected to a source of emf with Vemf =
100 V. When the light bulb is operating, the temperature of
its tungsten filament is 2520 °C.
Question:
67
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (1)
 A 100 W light bulb is connected to a source of emf with Vemf =
100 V. When the light bulb is operating, the temperature of
its tungsten filament is 2520 °C.
Question:
What is the resistance of the light bulb at room temperature
(20 ºC)?
67
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (1)
 A 100 W light bulb is connected to a source of emf with Vemf =
100 V. When the light bulb is operating, the temperature of
its tungsten filament is 2520 °C.
Question:
What is the resistance of the light bulb at room temperature
(20 ºC)?
Answer:
67
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (1)
 A 100 W light bulb is connected to a source of emf with Vemf =
100 V. When the light bulb is operating, the temperature of
its tungsten filament is 2520 °C.
Question:
What is the resistance of the light bulb at room temperature
(20 ºC)?
Answer:
 Power when lighted
67
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (2)
68
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (2)
… so
68
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (2)
… so
The temperature dependence of the resistance
68
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (2)
… so
The temperature dependence of the resistance
68
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (2)
… so
The temperature dependence of the resistance
… solve for the resistance at room temperature, R0
68
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (2)
… so
The temperature dependence of the resistance
… solve for the resistance at room temperature, R0
68
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (2)
… so
The temperature dependence of the resistance
… solve for the resistance at room temperature, R0
68
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (2)
… so
The temperature dependence of the resistance
… solve for the resistance at room temperature, R0
68
Monday, February 17, 2014
Example: Temperature Dependence of the
Resistance of a Light Bulb (2)
… so
The temperature dependence of the resistance
… solve for the resistance at room temperature, R0
Look up the temperature coefficient for tungsten …
68
Monday, February 17, 2014
Example: Total Energy in a Flashlight Battery
 A standard flashlight battery can deliver about 2.0 Wh of
energy before it runs down
Question:
If a battery costs US $0.80, what is the cost of operating a
100 W lamp for 8.0 hours using standard batteries?
Answer:
69
Monday, February 17, 2014
Example: Total Energy in a Flashlight Battery
 A standard flashlight battery can deliver about 2.0 Wh of
energy before it runs down
Question:
If a battery costs US $0.80, what is the cost of operating a
100 W lamp for 8.0 hours using standard batteries?
Answer:
Energy provided by one battery: 2.0 W x 1 h
69
Monday, February 17, 2014
Example: Total Energy in a Flashlight Battery
 A standard flashlight battery can deliver about 2.0 Wh of
energy before it runs down
Question:
If a battery costs US $0.80, what is the cost of operating a
100 W lamp for 8.0 hours using standard batteries?
Answer:
Energy provided by one battery: 2.0 W x 1 h
Energy needed: 100 W x 8 h
69
Monday, February 17, 2014
Example: Total Energy in a Flashlight Battery
 A standard flashlight battery can deliver about 2.0 Wh of
energy before it runs down
Question:
If a battery costs US $0.80, what is the cost of operating a
100 W lamp for 8.0 hours using standard batteries?
Answer:
Energy provided by one battery: 2.0 W x 1 h
Energy needed: 100 W x 8 h
Number of batteries needed: 800Wh/2.0Wh=400
69
Monday, February 17, 2014
Example: Total Energy in a Flashlight Battery
 A standard flashlight battery can deliver about 2.0 Wh of
energy before it runs down
Question:
If a battery costs US $0.80, what is the cost of operating a
100 W lamp for 8.0 hours using standard batteries?
Answer:
Energy provided by one battery: 2.0 W x 1 h
Energy needed: 100 W x 8 h
Number of batteries needed: 800Wh/2.0Wh=400
400 batteries cost 400 x $0.80=$320
69
Monday, February 17, 2014
Direct Current Circuits
 Complex Circuits
 Time-dependent Currents
1
Monday, February 17, 2014
Circuits (1)
 We have been working with simple circuits containing either
capacitors or resistors.
 Capacitors wired in parallel
 Capacitors wired in series
2
Monday, February 17, 2014
Circuits (2)
 Resistors wired in parallel
 Resistors wired in series
 We dealt with circuits containing either capacitors or
resistors that could be resolved in systems of parallel or
series components
3
Monday, February 17, 2014
Complex Circuits
 Some circuits can be constructed that cannot be resolved
into series or parallel systems of capacitors or resistors
 Examples:
4
Monday, February 17, 2014
Kirchhoff’s Rules for Multi-loop Circuits
 To handle these types of circuits, we must apply
Kirchhoff’s Rules.
 Kirchhoff’s Rules can be stated as
• Kirchhoff’s Junction Rule
• The sum of the currents entering
a junction must equal the sum of
the currents leaving a junction
• Kirchhoff’s Loop Rule
• The sum of voltage drops
around a complete circuit
loop must sum to zero
5
Monday, February 17, 2014
Kirchhoff’s Junction Rule
 Kirchhoff’s Junction Rule is a direct consequence of the
conservation of charge.
 In a conductor, charge cannot be created or destroyed.
 At a junction: all charges streaming into the junction must
also leave the junction
6
Monday, February 17, 2014
Kirchhoff’s Loop Rule
 Kirchhoff’s Loop Rule is a direct
consequence of the conservation
of electric potential energy
 Suppose that this rule was not valid
• We could construct a way around a loop in such a way that each turn
would increase the potential of a charge traveling around the loop
• We would always increase the energy of this charge, in obvious
contradiction to energy conservation
 Kirchhoff’s Loop Rule is equivalent to the law of energy
conservation
7
Monday, February 17, 2014
EMF Devices - Directions
 An emf device (e.g., a battery) keeps the positive
terminal (labeled +) at a higher electrical potential
than the negative terminal (labeled -)
 When a battery is connected in a circuit, its internal
chemistry causes a net current inside the battery:
positive charge carriers move from the negative to the
positive terminal, in direction of the emf arrow
 Or: Positive charge carriers move from a region of low
electric potential (negative terminal) to a region of
high electric potential (positive terminal)
 This flow is part of the current that is set up around
the circuit in that same direction
8
Monday, February 17, 2014
Single Loop Circuits (1)
 We begin our study of more
complicated circuits by analyzing
circuits with several sources of emf
and resistors connected in series
in a single loop
 We will apply Kirchhoff’s Rules to
these circuits
 To apply these rules we must establish conventions for determining the
voltage drop across each element of the circuit depending on the assumed
direction of current and the direction of the analysis of the circuit
 Because we do not know the direction of the current in the circuit before
we start, we must choose an arbitrary direction for the current
9
Monday, February 17, 2014
Single Loop Circuits (2)
 Choose a direction for the current
 We can determine if our assumption for the direction of the
current is correct after the analysis is complete
 Resulting current positive
• Current is flowing in the same direction as we had chosen
 Resulting current negative
• Current is flowing in the direction opposite to what we had chosen
 We can also choose the direction in which we analyze the
circuit arbitrarily
• Any direction we choose will give us the same information
10
Monday, February 17, 2014
Single Loop Circuits (3)
 If we move around the circuit in the
same direction as the current, the
voltage drop across a resistor
will be negative
 V=iR
 V- i R =0
11
Monday, February 17, 2014
Single Loop Circuits (3)
 If we move around the circuit in the
same direction as the current, the
voltage drop across a resistor
will be negative
 If we move around the circuit in the
direction opposite to the current, the voltage drop across the resistors
will be positive
 If we move around the circuit and encounter a source of emf pointing in
the same direction, we say this source of emf contributes a positive
voltage
 If we encounter a source of emf pointing in the opposite direction, we
consider that component to contribute a negative voltage
11
Monday, February 17, 2014
Single Loop Circuits (1)
 We have studied circuits with various networks of resistors but only one
source of emf
 Circuits can contain multiple sources of emf as well as multiple resistors
 We begin our study of more complicated circuits by analyzing a circuit
with two sources of emf (Vemf,1 and Vemf,2) and two resistors (R1 and R2)
connected in series in a single loop
 We will assume that the two sources of emf have opposite polarity
13
Monday, February 17, 2014
Single Loop Circuits (2)
 We assume that the current is flowing around the circuit in a clockwise direction
 Starting at point a with V = 0, we analyze around the circuit in a clockwise direction
 Because the components of the circuit are in series, all components have the same
current, i
14
Monday, February 17, 2014
Single Loop Circuits (3)
 Start at point a.
 The first circuit component is a source of emf Vemf,1, which
produces a positive voltage gain of +Vemf,1
 Next we find resistor R1, which produces a voltage drop V1
given by -iR1
 Continuing around the circuit we find resistor R2, which
produces a voltage drop V2 given by -iR2
 Next we meet a second source of emf, Vemf,2
 This source of emf is wired into the circuit
with a polarity opposite that of Vemf,1
 We treat this component as a voltage drop
of -Vemf,2 rather than a voltage gain
 We now have completed the circuit
and we are back at point a
15
Monday, February 17, 2014
Single Loop Circuits (4)
 Kirchhoff’s loop rule for the
voltage drops states
Vemf,1 − Vemf,2
i=
R1 + R 2
 Generalization: the voltage drops across
components in a single loop circuit must sum to
zero
 This statement must be qualified with
conventions for assigning the sign of the
voltage drops around the circuit
16
Monday, February 17, 2014
Single Loop Circuits (5)
 Now let’s analyze the same circuit in the counterclockwise direction starting at point a
 The first circuit element is Vemf,2 which is a positive
voltage gain
 The next element is R2
17
Monday, February 17, 2014
Single Loop Circuits (6)
 Because we have assumed that the
current is in the clockwise direction
and we are analyzing the loop in the
counter-clockwise direction, this
voltage drop is +iR2
Proceeding to the next element in the loop, R1, we use a
similar argument to designate the voltage drop as +iR1
 The final element in the circuit is Vemf,1, which is aligned in a
direction opposite to our analysis direction, so the voltage
drop across this element is -Vemf,1
18
Monday, February 17, 2014
Single Loop Circuits (7)
 Kirchhoff’s Loop Rule then gives us
 Comparing this result with the result we obtained by
analyzing the circuit in the clockwise direction
… we see that they are equivalent.
 The direction that we choose to analyze the circuit does not
matter.
19
Monday, February 17, 2014
Multi-Loop Circuits (1)
 To analyze multi-loop circuits, we must apply both the Loop
Rule and the Junction Rule
 To analyze a multi-loop circuit, identify all complete loops and
all junction points in the circuit and apply Kirchhoff’s Rules to
these parts of the circuit separately
 At each junction in a multi-loop circuit, the current flowing
into the junction must equal the current flowing out of the
circuit
20
Monday, February 17, 2014
Multi-Loop Circuits (2)
 Assume we have a junction point a
 We define a current i1 entering junction a and two currents i2
and i3 leaving junction a
 Kirchhoff’s Junction Rule tells us that
21
Monday, February 17, 2014
Multi-Loop Circuits (3)
 By analyzing the single loops in a multi-loop circuit with
Kirchhoff’s Loop Rule and the junctions with Kirchhoff’s
Junction Rule, we can obtain a system of coupled equations in
several unknown variables
 These coupled equations can be solved in several ways
• Solution with matrices and determinants
• Direct substitution
 Next: Example of a multi-loop circuit solved with Kirchhoff’s
Rules
22
Monday, February 17, 2014
Example: Kirchhoff’s Rules (1)
 The circuit here has three resistors, R1, R2, and R3 and two sources of
emf, Vemf,1 and Vemf,2
 This circuit cannot be resolved into simple series or parallel structures
 To analyze this circuit, we need to assign currents flowing through the
resistors
 We can choose the directions of these currents arbitrarily
23
Monday, February 17, 2014
Example: Kirchhoff’s Rules (1)
 The circuit here has three resistors, R1, R2, and R3 and two sources of
emf, Vemf,1 and Vemf,2
 This circuit cannot be resolved into simple series or parallel structures
 To analyze this circuit, we need to assign currents flowing through the
resistors
 We can choose the directions of these currents arbitrarily
23
Monday, February 17, 2014
Example: Kirchhoff’s Laws (2)
 At junction b the incoming current must equal the outgoing
current
 At junction a we again equate the incoming current and the
outgoing current
 But this equation gives us the
same information as the
previous equation!
 We need more information
to determine the three currents – 2 more independent
equations
24
Monday, February 17, 2014
Example: Kirchhoff’s Laws (3)
 To get the other equations we must apply Kirchhoff’s Loop
Rule.
 This circuit has three loops.
• Left
• R1, R2, Vemf,1
• Right
• R2, R3, Vemf,2
• Outer
• R1, R3, Vemf,1, Vemf,2
25
Monday, February 17, 2014
Example: Kirchhoff’s Laws (3)
 To get the other equations we must apply Kirchhoff’s Loop
Rule.
 This circuit has three loops.
• Left
• R1, R2, Vemf,1
• Right
• R2, R3, Vemf,2
• Outer
• R1, R3, Vemf,1, Vemf,2
25
Monday, February 17, 2014
Example: Kirchhoff’s Laws (3)
 To get the other equations we must apply Kirchhoff’s Loop
Rule.
 This circuit has three loops.
• Left
• R1, R2, Vemf,1
• Right
• R2, R3, Vemf,2
• Outer
• R1, R3, Vemf,1, Vemf,2
25
Monday, February 17, 2014
Example: Kirchhoff’s Laws (3)
 To get the other equations we must apply Kirchhoff’s Loop
Rule.
 This circuit has three loops.
• Left
• R1, R2, Vemf,1
• Right
• R2, R3, Vemf,2
• Outer
• R1, R3, Vemf,1, Vemf,2
25
Monday, February 17, 2014
Example: Kirchhoff’s Laws (4)
26
Monday, February 17, 2014
Example: Kirchhoff’s Laws (4)
 Going around the left loop counterclockwise starting at point
b we get
26
Monday, February 17, 2014
Example: Kirchhoff’s Laws (4)
 Going around the left loop counterclockwise starting at point
b we get
 Going around the right loop clockwise starting at point b we
get
26
Monday, February 17, 2014
Example: Kirchhoff’s Laws (4)
 Going around the left loop counterclockwise starting at point
b we get
 Going around the right loop clockwise starting at point b we
get
26
Monday, February 17, 2014
Example: Kirchhoff’s Laws (4)
 Going around the left loop counterclockwise starting at point
b we get
 Going around the right loop clockwise starting at point b we
get
 Going around the outer loop
clockwise starting
at point b we get
26
Monday, February 17, 2014
Example: Kirchhoff’s Laws (4)
 Going around the left loop counterclockwise starting at point
b we get
 Going around the right loop clockwise starting at point b we
get
 Going around the outer loop
clockwise starting
at point b we get
26
Monday, February 17, 2014
Example: Kirchhoff’s Laws (4)
 Going around the left loop counterclockwise starting at point
b we get
 Going around the right loop clockwise starting at point b we
get
 Going around the outer loop
clockwise starting
at point b we get
 But this equation gives us no new
information!
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Monday, February 17, 2014
 General rule:
 Current conservation: # of nodes -1
 Voltage: # of loops -1.
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Example: Kirchhoff’s Laws (5)
 We now have three equations
 And we have three unknowns i1, i2, and i3
 We can solve these three equations in a variety of ways
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