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General Physics (PHY 2140)
Lecture 5
 Electrodynamics
Electric current
• temperature variation of resistance
• electrical energy and power
Direct current circuits
• emf
• resistors in series
http://www.physics.wayne.edu/~alan/2140Website/Main.htm
Chapter 17-18
5/23/2017
1
Department of Physics and Astronomy
announces the Spring-Summer 2007 opening of
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on Monday, May 21 in
Room 172 of Physics Research Building.
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The Center will be open:
5/23/2017
Monday, May 21 to Thursday, August 2, 2007.
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Exam 1 this Wednesday 5/23
12-14 questions
You must show your work for credit.
Closed book.
You may bring an 8 ½ x 11 inch page of notes.
Bring a calculator and a pen/pencil (I don’t have extras).
5/23/2017
4
Suggested Course Sequence
FRESHMAN YEAR
Fall
Physics 2130/2131 or 2170/2171
Mathematics 1800
English (BC)
UGE 1000
General Education Req.
Winter
JUNIOR YEAR
Fall
Physics 5340/41: Optics
Chemistry 2220/2230 or 2280/2290
Biology Elective
College Foreign Lang. II
Winter
Physics 5620: Electronics
Physics 6Z00: Biomedical Seminar#
Physics 2140/2141 or 2180/2181
Elective
College Foreign Lang. III
Chemistry 1220/1230
Mathematics 2010
English (IC)
SENIOR YEAR
Fall
SOPHOMORE YEAR
Fall
Physics 3X00:
Application of Mathematics in Biomed.
Chemistry 1240/50
Mathematics 2020
Biology 1500
Winter
Physics 4X00:
Introduction to Biomedical Physics
Biology 1510
College Foreign Lang. I
General Education Req.
5/23/2017
New Degree Program in
BIOMEDICAL PHYSICS
College Group Req.
Physics 6X00: Biological Physics
Elective
Elective
General Education Req.
Winter
Physics/Radiology 6Y00:
Physics in Medicine
Physics 6W00: Biomedical Research#
College Group Req.
General Education Req.
#These will also be offered in the summer
For more info, please contact Prof. Peter
Hoffmann at [email protected]
5
Lightning Review
1
Q2 1
U  QV 
 CV 2
2
2C 2
Last lecture:
1.
2.
Capacitance and capacitors
 Capacitors with dielectrics (C↑ if  ↑)
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
A
C   0 , C   C0
d
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
6
17.5 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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17.5 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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17.6 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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17.6 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.
R  Ro 1   T  To  
Solution:
Using =3.92x10-3(oC)-1 from table 17.1.
Ro=50.0 W.
To=20oC.
R  Ro
76.8W  50.0W
R=76.8 W.
T  To 

1
 Ro
 
3.92 103 o C

 50.0W

 137o C
T  157 C
o
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17.7 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 (1’st: 1956 – transistor)
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17.8 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
5/23/2017


1 kWh  103W  3600 s   3.60 106 J
16
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=?
5/23/2017
P  I R  1000 A   49.6 W   49.6 106 W
2
2
17
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, and its resistance is R0.
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 V 2 / R
Thus, there is a power spike just after the switch is thrown, which burns
the light bulb.
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Chapter 18:
Direct Current Circuits
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Introduction: elements of electrical circuits
A branch: A branch is a single electrical element or device (resistor, etc.).





A circuit with 5 branches.
A junction: A junction (or node) is a connection point between two or more
branches.





A circuit with 3 nodes.
If we start at any point in a circuit (node), proceed through connected
electric devices back to the point (node) from which we started, without
crossing a node more than one time, we form a closed-path (or loop).
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18.1 Sources of EMF
 Steady current (constant in magnitude and direction)
• requires a complete circuit
• path cannot consist only of resistances:
cannot have 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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18.1 Sources of EMF
(continued)
Important note: in our complete circuit, conservation of energy
requires that as we travel around the circuit we should encounter
a number of voltage drops and increases so that we get back to
the same potential we started with!
V = IR
I
+ E
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V = IR = E
Start with 0V here
and move clockwise
22
EMF
Each real battery has some
internal resistance
AB: potential increases by E,
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)
5/23/2017
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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
5/23/2017
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
24
Measurements in electrical circuits
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
5/23/2017
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Direct Current Circuits
Two Basic Principles:

Conservation of Charge

Conservation of Energy
a
I
Resistance Networks
Vab  IReq
Req
5/23/2017
Vab

I
Req
b
26
18.2 Resistors in series
A
+
v2 _
B
R2
+
v _
+
Ii1
R1
1. Because of the charge conservation, all
charges going through the resistor R2 will
also go through resistor R1. Thus, currents
in R1 and R2 are the same,
I1  I 2  I
v1
_
C
2. Because of the energy conservation, total
potential drop (between A and C) equals to
the sum of potential drops between A and B
and B and C,
V  IR1  IR2
By definition,
V  IReq
Thus, Req would be
Req 
V IR1  IR2

 R1  R2
I
I
Req  R1  R2
5/23/2017
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Resistors in series: notes
Analogous formula is true for any number of resistors,
Req  R1  R2  R3  ...
(series combination)
It follows that the equivalent resistance of a series
combination of resistors is greater than any of the
individual resistors.
5/23/2017
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Resistors in series: example
In the electrical circuit below, find voltage across the resistor R1 in terms of
the resistances R1, R2 and potential difference between the battery’s
terminals V.
Energy conservation implies:
A
+
v2 _
with
V  V1  V2
V1  IR1 and V2  IR2
v1
Then,
V  I  R1  R2  , so I 
C
Thus,
B
R2
+
v _
+
Ii1
R1
_
V1  V
V
R1  R2
R1
R1  R2
This circuit is known as voltage divider.
5/23/2017
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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

I
I
V1 V2 V3 V1 V2 V3
     
I I
I I1 I 2 I 3
Req 
R2
V2=I2R2
R3
V3=I3R3
Req  R1  R2  R3
b
Voltage Divider:
5/23/2017
V1
R1

Vab Req
30
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
1
I I1  I 2  I 3


Req Vab
Vab

I1 I 2 I 3 I1 I 2 I 3
 
  
Vab Vab Vab V1 V2 V3
1
1 1 1
  
Req R1 R2 R3
Current Divider:
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I1 Req

I
R1
31
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
E=18V
+
R1=4W
R2=3W
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R3=6W
32