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General Physics (PHY 2140)
Lecture 9
 Electrodynamics
Electric current
 temperature variation of resistance
 electrical energy and power
Chpter 17-18
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1
Lightning Review
Last lecture:
1. Current and resistance



Current and drift speed
Resistance and Ohm’s law
• I is proportional to V
Resistivity
• material property
Q
I
t
I  nqvd A
V  IR
RA

l
Review Problem: Consider two resistors wired one after another. If
there is an electric current moving through the combination, the
current in the second resistor is
a. equal to
b. half
c. smaller, but not necessarily half
the current through the first resistor.
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I
a
R1
b
R2
c
2
17.4 Resistivity - Example
(a) Calculate the resistance per unit length of a 22-gauge
nichrome wire of radius 0.321 m.
Cross section:


2
A   r   0.32110 m  3.24 107 m2
2
3
Resistivity (Table): 1.5 x 106 Wm.
6
R

1.5

10
Wm
Resistance/unit length:
W
 

4.6
m
7 2
l A 3.24 10 m
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17.4 Resistivity - Example
(b) If a potential difference of 10.0 V is maintained across a
1.0-m length of the nichrome wire, what is the current?
V 10.0V
I

 2.2 A
R
4.6W
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17.4 Temperature Variation of Resistance
- Intro
• The resistivity of a metal depends on many
(environmental) factors.
• The most important factor is the temperature.
• For most metals, the resistivity increases with
increasing temperature.
• The increased resistivity arises because of larger
friction caused by the more violent motion of the
atoms of the metal.
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For most metals, resistivity increases
approx. linearly with temperature.

  o 1   T  To  
•
T
Metallic Conductor
 is the resistivity at temperature T (measured in Celsius).
o is the reference resistivity at the reference temperature To
(usually taken to be 20 oC).
•  is a parameter called temperature coefficient of resistivity.
•
For a conductor with fixed cross section.
R  Ro 1   T  To  
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
T
Superconductor
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17.5 Temperature Variation of Resistance Example
Platinum Resistance Thermometer
A resistance thermometer, which measures temperature by measuring the
change in the resistance of a conductor, is made of platinum and has a
resistance of 50.0 W at 20oC. When the device is immersed in a vessel
containing melting indium, its resistance increases to 76.8 W. Find the melting
point of Indium.
Solution:
Using =3.92x10-3(oC)-1 from table 17.1.
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Platinum Resistance Thermometer
A resistance thermometer, which measures temperature by measuring the change in the
resistance of a conductor, is made of platinum and has a resistance of 50.0 W at 20oC.
When the device is immersed in a vessel containing melting indium, its resistance
increases to 76.8 W. Find the melting point of Indium.
Solution:
Using =3.92x10-3(oC)-1 from table 17.1.
Ro=50.0 W.
To=20oC.
R  Ro
76.8W  50.0W
T  To 

R=76.8 W.
1
 Ro
 
3.92 103 o C

 50.0W

 137o C
T  157 C
o
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Appendix: Superconductivity
1911: H. K. Onnes, who had figured out how to
make liquid helium, used it to cool mercury to 4.2
K and looked at its resistance:
At low temperatures the resistance of some
metals0, measured to be less than 10-16•ρconductor
(i.e., ρ<10-24 Ωm)!
–Current can flow, even if E=0.
–Current in superconducting rings can flow for years with no
decrease!
1957: Bardeen (UIUC!), Cooper, and Schrieffer (“BCS”) publish theoretical
explanation, for which they get the Nobel prize in 1972.

It was Bardeen’s second Nobel prize (1956 – transistor)
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17.7 Electrical energy and power
In any circuit, battery is used to induce electrical current

chemical energy of the battery is transformed into kinetic energy
of mobile charge carriers (electrical energy gain)
Any device that possesses resistance (resistor) present
in the circuit will transform electrical energy into heat

kinetic energy of charge carriers is transformed into heat via
collisions with atoms in a conductor (electrical energy loss)
C
D
+ -
I
B
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V = IR
A
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Electrical energy
Consider circuit on the right in detail
AB: charge gains electrical energy
form the battery
E  Q  V
(battery looses chemical energy)
B
A
C
D
CD: electrical energy lost (transferred
into heat)
Back to A: same potential energy
(zero) as before
Gained electrical energy = lost
electrical energy on the resistor
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Power
Compute rate of energy loss (power dissipated on the
resistor)
E Q
P

V  I V
t
t
Use Ohm’s law
P  I V  I
2
V 

R
2
R
Units of power: SI:
watt
delivered energy: kilowatt-hours
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

1 kWh  103W  3600 s   3.60 106 J
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Example
Power Transmission line
A high-voltage transmission line with resistance of 0.31 W/km carries 1000A ,
starting at 700 kV, for a distance of 160 km. What is the power loss due to
resistance in the wire?
Given:
V=700000 V
=0.31 W/km
L=160 km
I=1000 A
Observations:
1. Given resistance/length, compute total resistance
2. Given resistance and current, compute power loss
R   L   0.31 W km160 km  49.6 W
Now compute power
Find:
P=?
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P  I R  1000 A   49.6 W   49.6 106 W
2
2
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Mini-quiz
Why do the old light bulbs usually fail just after you turn
them on?
When the light bulb is off, its filament is cold, so its resistance is large.
Once the switch it thrown, current passes through the filament heating it
up, thus increasing the resistance,
R  Ro 1   T  To  
This leads to decreased amount of power delivered to the light bulb, as
P  I 2R
Thus, there is a power spike just after the switch is thrown, which burns
the light bulb.
Resume: electrical devices are better be turned off if there is a power loss
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Direct Current Circuits
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18.1 Sources of EMF
 Steady current (constant in magnitude and direction)
• requires a complete circuit
• path cannot be only resistance
cannot be only potential drops in direction of current flow
 Electromotive Force (EMF)
• provides increase in potential E
• converts some external form of energy into electrical energy
 Single emf and a single resistor: emf can be thought of as a
“charge pump” V = IR
I
+ -
V = IR = E
E
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EMF
Each real battery has some
internal resistance
AB: potential increases by E
on the source of EMF, then
decreases by Ir (because of
the internal resistance)
Thus, terminal voltage on the
battery V is
V  E  Ir
B
C
r
R
E
A
D
Note: E is the same as the
terminal voltage when the
current is zero (open circuit)
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EMF (continued)
Now add a load resistance R
Since it is connected by a
conducting wire to the battery →
terminal voltage is the same as
the potential difference across the
load resistance
V  E  Ir  IR, or
E  Ir  IR
Thus, the current in the circuit is
I
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E
Rr
B
C
r
R
E
A
D
Power output:
I E  I 2r  I 2 R
Note: we’ll assume r negligible unless otherwise is stated
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Measurements
Voltmeters measure Potential Difference (or voltage)
across a device by being placed in parallel with the
device.
V
Ammeters measure current through a device by being
placed in series with the device.
A
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Direct Current Circuits
Two Basic Principles:
Conservation of Charge
Conservation of Energy
Resistance Networks
a
Vab  IReq
I
V
Req  ab
I
Req
b
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Resistors in series
Conservation of Charge
I = I 1 = I2 = I 3
Conservation of Energy
Vab = V1 + V2 + V3
I
a
R1
V1=I1R1
Vab V1  V2  V3
Req 

I
I
V1 V2 V3 V1 V2 V3
 

 

I
I
I
I1 I 2 I 3
Req  R1  R2  R3
Voltage Divider:
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R2
V2=I2R2
R3
V3=I3R3
b
V1
R1

Vab Req
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Resistors in parallel
Conservation of Charge
R1
I = I 1 + I 2 + I3
V1=I1R1
Conservation of Energy
Vab = V1 = V2 = V3
I
a
R2
V2=I2R2
R3
V3=I3R3
b
I1  I 2  I 3
1
I


Req Vab
Vab
I1
I2
I3
I1 I 2 I 3



  
Vab Vab Vab V1 V2 V3
1
1
1
1



Req R1 R2 R3
I 1 Req
Current Divider:

I
R1
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Example:
Determine the equivalent resistance of the circuit as shown.
Determine the voltage across and current through each resistor.
Determine the power dissipated in each resistor
Determine the power delivered by the battery
R1=4W
E=18V
R2=3W
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R3=6W
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