Transcript I. Electric Current

Electrodynamics #1
The Electric Current
HW #1 – 4 fits these notes.
I. Electric Current: {Start HW #1}
Introduction:
Our study of electricity so far has concentrated on static systems of
charges. In other words, the charges were at rest. This new unit
analyzes moving systems of charges.
The first task is to quantify the motion, or flow, of electric charge:
The flow of charge through some point in space is called the current.
The concept of electrical current is very similar to the flow of (and
currents in) water.
Current is represented by the letter I (or i).
Q
Q
I
or
t
t
The current is defined as the amount of charge passing by a point
in space divided by the elapsed time.
Units:
Current is given its own special unit name, the ampere (A). This is
defined as follows:
Q
coulomb
I
 ampere 
t
second
C
1A  1
s
Example #1: A wire carries a current of 2.50 mA. How many
electrons pass through the wire each second?

3 C   each electron 
2.50  10 A   2.50 10


19
1.6022

10
C
s



3
electrons 

 1.56  1016

s


Driving Force:
For a current to exist, there must be some kind of energy difference
between two points in space. Electric charges will move in response to
the energy difference. Rather than using potential energy to define the
motion of the charges, we use electric potential (V).
V = difference in electric potential between two points in space.
V = energy per unit charge available to push charges into motion,
usually provided by a battery or generator.
I
The difference in electric potential
pushes charges to flow. Direction of

flow is based on motion of positive
load
V
charges (Franklin’s convention).

Positive charges flow from high
potential (+) to low potential (–).
connecting wire
II. Response of Charge Flow: Resistance
When an electric potential is placed across some material, charges will
flow through the material.
How much charge moves depends on how
easily charge can move through the
substance. Define the resistance to
charge flow as follows:
V
resistance  R 
I
For a given electric potential, the greater
the current flow, the lower the resistance.
V = potential difference moving charges
I = current, flow of charges
high V
low V
R = relative difficulty for charge to flow
Units for Resistance:
Resistance is measured as electric potential per current:
V
resistance  R 
I
volts
V


amperes A
This is given its own special unit name,
called an ohm (W).
V
1W 
A
Georg
Simon
Ohm
Example #2: When a wire is connected to a 12.0 V battery, a current of
0.125 A is established in the wire. What is the resistance of the wire?
V 12.0V
R

I
0.125 A
 96.0 W
Example #3: A certain insulating material has a resistance of 15.0 MW.
When an electric potential (or “voltage”) of 75.0 kV is placed across
this device, how much current exists in the material?
V
V
R
 I
I
R
75.0 103V

15.0 106 W
I  5.00 103 A  5.00 mA
Ohm’s Law and Resistance:
In general, the resistance of a material is not constant, but varies with
temperature and current. A filament light bulb is an example of a device
whose resistance changes with temperature and with current carried
through the filament.
There are certain materials
whose resistance is stable over
current load and temperature,
and these materials are called
“ohmic”. Thus an ohmic
material has a constant
resistance. The equation for
resistance is usually written:
V  IR
resembles equation for line
Example #4: When an electric potential difference of 12.0 V is placed
across a given load, the load pulls 0.250 A of current. (a) What voltage
is required to put 3.25 A of current through the load?
V
12.0 V
R

I
0.250 A
 48.0 W
V  IR  3.25 A48.0 W
 156 V
(b) What current will the material carry if the electric potential is set to
420 volts?
V 420V
I

R
48.0 W
 8.75 A
Homework:
Read and Outline 18.1-18.3
Questions 2-5
Problems 1-11
Text book Check Friday.
III. Resistance and Resistivity: {Start HW #2}
In general, resistance is an extrinsic property. Resistance depends on
what material is being used to carry a current (for example, copper
versus glass). Resistance also depends on the amount and shape of the
material carrying the current. For example, five pieces of copper can
have very different resistances due to differing amounts and shapes.
Resistivity is an intrinsic property that relates to resistance. Resistivity
depends only on what type of material is used, and is not affected by
amount or shape.
Five pieces of copper have the same density (intrinsic property)
even though they may have different masses (extrinsic property).
The variable used to represent resistivity is the Greek letter “rho”:
  resistivity
Resistance is related to resistivity and the shape of the material.
L
I
A
high V
V
L
R
A
low V
V = voltage difference across the material.
L = length of the material.
RL
A = cross – sectional area of the material.
1
R
A
I = current = flow of charge through the material
The resistivity depends on what material is used, and the value is given
on a table on the homework sheet.
Units of Resistivity:
The units of resistivity are given as:
L
RA
R
 
A
L
unit 
2
W
m
  
 m
 Wm
Example #5: (a) What is the resistance of a copper wire that is 0.100
mm in diameter and 12.0 m long?
Cu  1.69 108 W  m
L
L
R

A  r2
1.69 10


8
W  m  12.0 m 
 0.100 10 m 


2


R  25.8 W
3
2
(b) What voltage difference must be put across this wire to establish a
current of 0.340 mA?
V  IR   0.340 103 A   25.8 W 
V  8.78 103V  8.78 mV
Example #6: A parallel plate capacitor is made with two conducting
parallel plates that have a diameter of 2.00 cm. The parallel plates are
separated by a 0.0150 mm, and the space between the plates is filled
with fused quartz (an insulator). Determine the resistance of the fused
quartz.
quartz  75 1016 W  m
R
L
A

75 10

16

W  m 0.0150 103 m
 1.00 10 m
2
R  3.58 1016 W
2

Example #7: A wire has a given cross sectional area Ao, length Lo, and
resistance Ro. If this wire is cut into two identical lengths, and the two
pieces are connected side by side, what will be the new resistance of
this combination?
Lo
Ao
cut
L= ½Lo
L
R
A
R
A=2Ao
  Lo 2 

  1  Lo
2 Ao
4 Ao
1
R  Ro
4
Electrodynamics #1
The Electric Current
HW #1 – 4 fits these notes.
{day #2}
Example #8: A 1.00 gram lump of copper is pulled into a wire that has
a resistance of 0.100 W. Determine the length and diameter of this wire.
Cu  1.69 108 W  m
L
R
A
m  D Vol  DLA
mR  DLA
L
density  DCu  8.99 103 mkg3
L
A
 DL
1.00 10
L
2

kg 0.100 W
8.99 103 kg 1.69 108 W  m

3


mR
D
 0.811 m
Now solve for diameter:
L
r

R
L
R
A
 r  A
2
1.69 10

W  m 0.811 m 
 0.100 W 
8
r  2.09  10 4 m
diameter  2r  0.418 mm
L
R
IV. Temperature Dependence of Resistance and Resistivity:
{Start HW #3}
For most materials, heating the material causes the resistivity (and
resistance) of the material to increase. This effect can be modeled with
the following equation:
  o 1   T  To 
R  Ro 1   T  To 
 (or R) is the resistivity (or resistance) at some new temperature T
o (or Ro) is the resistivity (or resistance) at some starting
temperature To. Usually the starting temperature is 20.0 °C, and
the values given in the date table are calibrated to this starting
temperature.
 is the temperature coefficient of resistivity (or resistance)
Example #9: A copper wire has a resistance of 1.45 W at 20.0 °C. What
would be the resistance of this wire if it were heated to 165 °C?
 Cu  3.9 103 1C
R  Ro 1   T  To 
 


R  1.45 W  1  3.9 10 3 1C 165C  20.0C 
R  2.27 W
Example #10: A platinum wire has a resistance of 2.000 W at 20.0 °C.
What would be the temperature of the wire if the resistance was
measured only to be 0.145 W?
 Pt  3.92 10 3 1C
R  Ro 1   T  To 
R
 1   T  To 
Ro
R
 T  To    1
Ro

1 R
T  To    1
  Ro 

1 R
T  To    1
  Ro 

 0.145 W 
1

T  20.0C  
 1
3 1 
 3.92 10 C  2.000W 
T  217C
The resistivity of platinum
keeps a linear temperature
response over a wide range
of temperatures, making it
ideal for thermometer
work.
V. Power and Resistance: {Start HW #4}
When an electric current passes through an object, work must be done
to move the charged particles through the material.
As the charges move
through the material,
they collide with the
atoms of the material.
Each collision transfers
energy from the charges
to the atoms of the
material.
This transfer of energy
heats the resistive
materials.
The transfer of heat from the charges to the resistive material is given in
terms of power, or the rate at which energy is removed from the circuit.
V


R
The battery feeds power into the circuit,
driving the charges around the circuit.
This current passes through the resistive
load.
The resistor saps energy from the current. Overall, energy is
conserved. Energy in from the battery is taken out by the resistor.
The amount of power put in or taken out by
an element of the circuit is given as:
If Ohm’s law is substituted in,
there are two more versions of
the equation for power:
P  I 2R
P  I V 
2

V 
P
R
Example #11: Show that the power equation does give units of power.
P  I V 
power  current voltage  amperevolt 
 C  J 
power    
 s  C 
J

s
 watt  W
Example #12: An incandescent light bulb is rated at 100 W when run
from a 120 V source. (a) What is the current through the light bulb?
P  I V 
100 W

120 V
P
I
V
 0.833 A
2

V 
P
(b) What is the resistance of the bulb?
R

V 
R
2
P

120 V 

2
100 W
 144 W
Example #13: An electric hot water heater is designed to pull 25.0 A
from a 220 V source. (a) What is the power output of this water heater?
P  I V   25.0 A220V 
 5,500W  5.50 kW
(b) How much time does it take to heat 200 kg of water from 20.0°C to
90.0°C?
energy mCT
P

time
t
J
C  specific heat  4186
kg C
mCT
t
P
mCT
P
t

J 
70.0C 
200 kg 4186
kg C 

t
5500W 

5.86 10 J 
t
7
5500W 
 1.07 10 4 s
 2.96 hours
(c) Energy is usually sold in units of kilowatt hours. Show that this is a
unit of energy.
1 kWh  1 kilowatt hour
1kWh  1000 watts3600 s 
1 kWh  3.6 106 J
(d) If electrical energy costs $0.500 per kilowatt hour, how much does it
cost to heat this volume of water?
energy  5.86 107 J
From part (b), the
energy needed is:
cost =
alternate =
 $0.500 


 1 kWh 
 $8.14
 $0.500 
5.50 kW  2.96 hours   1 kWh 


 $8.14
 1 kWh
5.86 10 J 
6
 3.6  10 J

7

Microscopic View of Electric Current
Superconductivity
Electrical Conduction in the Human Nervous
System
18.8 Microscopic View of Electric Current
Electrons in a conductor have large, random speeds just due to their temperature.
When a potential difference is applied, the electrons also acquire an average drift
velocity, which is generally considerably smaller than the thermal velocity.
18.8 Microscopic View of Electric Current
This drift speed is related to the current in the wire, and also to the number of electrons
per unit volume.
(18-10)
Example 18-14. Electron Speeds on a Wire.
A copper wire , 3.2 mm in diameter, carries a 5.0 Amp current. Determine the
drift speed of the electrons. Assume that one electron per Cu atom is free to
move.
“n” is the density of free electrons. This is equal to the density of Cu
atoms. A is the cross sectional area. Vd is the drift velocity. See text
book for complete calculation.
Calculation shows that the drift velocity of electrons is 4.7 X 10-5 m/s
If the distance bewtween the switch and the lights is 5 meters, how
long will it take the lights to come on?
Time = Distance/ Velocity = 29 hours! What is wrong with our model??
18.8 Microscopic View of Electric Current
29 hours is how long it would take for the electron at the switch to get to the light.
The electric field is established at the speed of light.
“The garden hose is full, so the water at the spigot is not the first water out the hose.”
18.9 Superconductivity
In general, resistivity decreases as temperature decreases. Some materials, however,
have resistivity that falls abruptly to zero at a very low temperature, called the critical
temperature, TC.
18.9 Superconductivity
Experiments have shown that currents, once started, can flow through these materials
for years without decreasing even without a potential difference.
Critical temperatures are low; for many years no material was found to be
superconducting above 23 K.
More recently, novel materials have been found to be superconducting below 90 K, and
work on higher temperature superconductors is continuing.
18.10 Electrical Conduction in the Human
Nervous System
The human nervous system depends on the flow of electric charge.
The basic elements of the nervous system are cells called neurons.
Neurons have a main cell body, small attachments called dendrites, and a long tail
called the axon.
18.10 Electrical Conduction in
the Human Nervous System
Signals are received by the dendrites,
propagated along the axon, and transmitted
through a connection called a synapse.
18.10 Electrical Conduction in the Human
Nervous System
This process depends on there being a dipole layer of charge on the cell membrane, and
different concentrations of ions inside and outside the cell.
18.10 Electrical Conduction in the Human
Nervous System
This applies to most cells in the body. Neurons can respond to a stimulus and conduct
an electrical signal. This signal is in the form of an action potential.
18.10 Electrical Conduction in the Human
Nervous System
The action potential propagates along the
axon membrane.
Problem Set: 49 – 53.
49. Assume one cubic meter of copper. Assume one electron per
atom of copper.
50. Same assumption as 49 for part c.
51. Use the drift velocity equation and sketch the motion of both
particles relative to the electric field.
52. Δ V = E Δx
53. Use the Definition of velocity.