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PHYS2012
emp10_01
ELECTRIC CARS
Why do petrol cars dominate?
Why are hybrid petrol-electric cars becoming more popular?
Why do we not see battery powered electric cars on our roads?
What about capacitors as energy storage devices for electric cars?
What is the future of using ultracapacitors for electric cars?
Below are a few simple models which will give some insight into the above questions.
The models are not necessarily accurate ones, but do give “ball-park” results.
PETROL CARS – A LONG DISTANCE JOURNEY
Consider a petrol car going on a non-stop 250 km journey at a constant velocity. How
much energy and power does the petrol provide?
Parameters describing the simple model for the journey
Quantity
m
dx
dx
dt
dt
=
=
=
=
=
Value
1500
250
2.50E+05
4.00
1.44E+04
C
Vol
Vol

M
=
=
=
=
=
10
25
2.50E-02
750
19
u
u
U
e
=
=
=
=
46
4.60E+07
8.63E+08
0.30
MJ/kg
J/kg
J
=Mu
W
=
2.59E+08
J
=eU
P
P
=
=
1.80E+04
18
W
kW
= W / dt
Symbol
Mass of car
Distance travelled by car
Time for journey
Fuel consumption of car
Volume petrol used by car
Density of petrol
Mass of petrol used by car
Energy density of petrol
Energy used by car
Efficiency of heat engine
Energy expended as work done on
car
Average rate of energy usage, power
Unit
kg
km
m
h
s
L/100
km
L
m3
kg/m3
kg
Calculation
=  Vol
1L / 10 km
The amount of energy that is needed as useful work to be done on the car for the 250 km
journey is 2.6108 J and the average rate at which this energy must be supplied is 18 kW.
For this journey, only 19 kg of petrol is needed.
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1.1
PETROL CARS – A SHORT SPRINT
Consider a petrol car going starting at rest and accelerating to 100 km/h in 10 s.
How much energy and power does the petrol provide?
Parameters describing a simple model for the short sprint
Quantity
Mass of car
Initial speed of car
Final speed of car
Symbol
m
v0
v
Value
1500
0
100
27.8
10
Unit
kg
Calculation
Time to accelerate
dt
=
=
=
=
=
Work done = increase in KE
Efficiency of petrol motor
Energy needed from petrol engine
dK
e
QH
=
=
=
5.79E+05
0.30
1.93E+06
J
= 1/2 m v2
J
= dK/e
Power required in accelerating car
P
=
1.93E+05
193
W
kW
Energy density of petrol
Mass of fuel consumed
u
M
=
=
4.60E+07
0.042
J/kg
kg
km/h
m/s
s
= QH / dt
= QH / u
The amount of energy that is needed as
useful work to be done on the car
accelerating for a short period time is
1.9106 J and the average rate at
which this energy must be supplied is
190 kW. For this sprint, the petrol
needed is insignificant, 42 g.
Petrol is a fantastic fuel from the point of view of its energy density. A lot of energy is
released when a small amount of petrol is combusted. However, petrol is a nonrenewable resource and as a result of the combustion process, undesirable chemicals
(hydrocarbons, NOx, COx) are released into the atmosphere.
To have a better understanding of the claims made about hybrid and electric cars it is
necessary to know a little about the physical principles of heat engines used in a petrol
and diesel cars.
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1.2
HEAT ENGINES
A heat engine is a physical device that converts thermal energy to mechanical output. The
mechanical output is called work, and the thermal energy input is called heat. Heat
engines typically run on a specific thermodynamic cycle.
HEAT ENGINE
Cyclic process: W = |QH| - |QC|
Hot reservoir (heat source) TH
QH
Engine – working substance
W
QC
useful mechanical work output
Dissipative losses – friction, turbulence
Cold reservoir (heat sink) TC
A heat engine absorbs heat energy QH from the high temperature TH heat source,
converting part of it to useful work W and delivering the rest QC to the cold temperature
TC heat sink. In one cycle of a heat engine, there is zero change in the internal energy U
and using the First Law of Thermodynamics, the work and heat transfers are related by
U  Qnet  W  QH  QC  W  0
W  QH  QC
In general, the efficiency of a given heat engine is defined informally by the ratio of
"what you get out" to "what you put in." In the case of a heat engine, one desires to
extract work and put in a heat transfer. The efficiency e is given by
W
Q
e
 1 C
QH
QH
The theoretical maximum efficiency emax of any heat engine depends only on the
temperatures it operates between. This efficiency is usually derived using an ideal
imaginary heat engine such as the Carnot heat engine
T
emax  1  C
TH
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1.3
Is it possible to recover all the energy lost and convert it into useful mechanical energy?
Second Law of Thermodynamics: 100% of heat can’t be transformed into
mechanical energy  e < 1
e  1
QC
1 
QH
QC
0
QH

QC  0
Some energy must be transferred to the cold reservoir to satisfy the Second Law
of Thermodynamics. Can’t convert 100% of thermal energy into work.
HEAT ENGINES – OTTO CYCLE
The four-stroke Otto cycle spark-ignition engine is the most common engine used for
automobiles today. The purpose of internal combustion engines is the production of
mechanical power from the chemical energy contained in the fuel. Each cylinder of the
engine contains a piston. The movement of the piston in response to the combustion of
the engine produces work. In a four stroke engine, during one cycle, the piston moves
through the cylinder four times, each of which is called a stroke. In the first stroke, the
intake stroke, the piston travels down the cylinder with the volume increasing and the
pressure decreasing which results in air being drawn into the cylinder, and then fuel is
added to the air. In the second stroke, the compression stroke, the piston travels up the
cylinder to its highest point (smallest volume) compressing the fuel-air mixture. Then
combustion occurs - a spark ignites the gases and changes the composition of the
mixture, raising the pressure and the temperature to their highest values. In the third
stroke, the expansion, or power stroke, the high pressure of the gases push the piston
down to its lowest point in the cylinder. This produces the work output of the cycle. In
the fourth stroke, the exhaust stroke, the exhaust valve is opened and the pressure within
the cylinder is higher than atmospheric pressure, so the exhaust gases leave the cylinder
as the piston moves from the bottom to top, pushing those gases out of the exhaust valve.
Otto Cycle
P
e  1
IGNITION
fuel
combustion
r
isothermals
adiabatic
3
1
 1
QH
power stroke
2
compression
stroke
W
Po
5
V2
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4
1
intake stroke
exhaust stroke
5 sep 10
QC
cooling of
exhaust gases
released to
surroundings
V
V1 = r V2
r = compression ratio
1.4
The theoretical efficiency of the Otto Cycle heat engine is given by
e  1
1
 1
r
C
where r is the compression ratio and  is the ratio of molar specific heats (   p )
CV
V
r  max
Vmin
Vmin and Vmax are the minimum and maximum volumes of the gas enclosed in the cylinder
by the movable piston.
Typical operating values for a Otto Cycle heat engine are:
r = 8  = 1.4 TC = 330 K TH = 1800 K
e = 0.56 emax = 0.82
The maximum efficiency for the heat engine based upon temperature difference between
the hot and cold reservoirs is 82% for a Carnot Cycle but for an ideal Otto Cycle it is only
56% efficient. However, in driving a real car on the road, the efficiency in delivering
useful work in moving the car is only between 20% to 30%.
HYBRID PETROL-ELECTRIC CARS
Why is a hybrid petrol-electric car more efficient than an equivalent petrol powered car?
A hybrid petrol-electric car has a higher efficiency than a petrol-only car because it
recovers some of the energy that would normally be lost as heat to the surrounding
environment in charging the batteries and energy to charge the batteries during breaking.
For example, if the efficiency of a typical petrol-only car engine is 20%, what efficiency
could be achieved if the amount of heat loss is halved?
The efficiency of a heat engine is given by
e
W
Q  QC
Q
 H
 1 C
QH
QH
QH
e  1
QC
 0.2 
QH
QC
 0.80
QH
Assuming the heat lost (QC) can be reduced by half, then
QC
 0.40  e  1  0.40  0.6
QH
The efficiency of the hybrid car is 60% compared to only 20% for the petrol. This
increase in efficiency is achieved by recovering 50% of the heat lost.The above is
only valid if the energy transferred to the cold reservoir is useful in generating
electricity to charge the batteries.
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1.5
Quotes from http://mb-soft.com/public/cars00.html
The electricity used in Hybrids is created by an alternator driven by a gasoline engine, using gasoline to do
it. Think about this! Instead of burning gasoline in the engine to turn the wheels directly, the gasoline is
burned in the engine to turn an alternator to produce electricity which is then put into batteries, where it
is then later used to send to electric motors, which then turn the wheels. Now, if all those mechanisms
and processes were perfectly efficient, fine, there would be no disadvantage of the complicated
sequence, but they all have efficiencies which are well below 100%! Adding in lots of extra processes is
NOT really beneficial! The very fact that the SAME gasoline is the initial source of energy, and the SAME
turning of the wheels is the final goal, makes this all somewhat bizarre. However, Hybrids DO have one
real advantage, in that the gasoline engine can run at essentially constant speed. Normal driving involves
the engine running at very different speeds, from idling to flat out, and the Physical laws do not permit
devices to have excellent efficiency over such wide ranges of speed. It is amusing that you will hear
people tell you a hundred different reasons of why they think Hybrids are spectacularly more efficient
than gasoline engines, and they are virtually all totally wrong! Just the fact that a gasoline engine has to
run in either case should be a clue. If it can always run at whatever speed where it has maximum
efficiency, fine, there can be a real advantage. That is rarely the case in what those "experts" describe!
And manufacturers and even the EPA seem willing to ONLY evaluate Hybrids WHILE the batteries are fully
charged and are discharging, where they then can claim really impressive mileage figures. I guess that is
not ABSOLUTE deception of potential buyers, but it certainly seems pretty close.
First, it will be WONDERFUL if and when battery-powered vehicles and/or hydrogen fuel-cell-powered
vehicles become economically practical. Neither seems very likely during the next thirty or probably fifty
years, until and unless some great breakthroughs are found in energy production. Maybe YOU might
come up with such a concept some day, but in order to do that, you FIRST need to completely understand
the many subjects mentioned and discussed here.
It is certainly true that electric motors have some tremendous efficiency advantages over the common
21% (thermal) efficiency of most cars on the road with internal combustion engines! But, unfortunately,
the REST of the picture involves devices and technologies which are not as efficient as the electric motors
themselves, specifically the batteries and the methods of charging them.
Sadly, "battery-powered vehicles" and future Hydrogen-powered vehicles, will NOT be the wonderful
"energy solution" that people think they will be! People think they are "really efficient" because of no
exhaust, etc. That's true, IF you only consider the car itself! (This also applies to the electric operational
aspects of hybrid cars.)
People, including the so-called experts, seem to be overlooking a central concept! A battery does not
MAKE any electricity, it merely stores it. However much energy or work or power you want to get OUT of
a battery, must first get put INTO the battery! This is simply stating a long-known fact in science called the
Conservation of Energy! In other words, batteries are not FUEL like petroleum or natural gas or coal. They
actually have no fuel at all, and are instead simply STORAGE devices. Hydrogen is actually much the same,
as there is no existing supply of hydrogen gas; it must be produced, such as by the electrolysis of water
(which requires a LOT of electricity again, very similar to the battery situation). Where promotional
displays show the "simplicity" of plugging the car into house electricity, they neglect to note just how
much electricity that car is going to suck out of the house wiring!
Watch the TV commercials for the future Chevy Volt car. They brag that it will be able to go 40 miles
before ever needing its included gasoline engine to start up to recharge the batteries. They ONLY talk
about starting out with FULLY CHARGED batteries, and discharging them to provide the power to move
the vehicle that 40 miles. They NEVER mention the necessary fact that EXTERNAL power has to be
provided from somewhere to re-charge the batteries! In fact, in interviews, the Executives of General
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5 sep 10
1.6
Motors show amazing attitudes and apparent lack of knowledge regarding the central subjects! It was
NOT an Engineer that dreamed up the Volt, but an Executive with NO expertise in the needed areas! He
simply THOUGHT that since most drivers drive less than 40 miles in each round trip, he THOUGHT it would
be a great idea to create a vehicle that could go 40 miles without needing any fuel! His interviews seem to
sound like he believes in the Tooth Fairy to re-charge those batteries, and that the ONLY thing GM needs
to still do is to improve the performance of the batteries! The GM top executive even described their
advanced battery packs as having 16 kilowatts of capacity. Apparently, no GM Engineer ever told him that
CAPACITY (or energy) can only be described in kiloWatt-HOURS, which is a unit of energy. To describe ANY
battery as having a CAPACITY of 16 kiloWatts is simply meaningless and a statement that shows ignorance
of basic facts. (For their information, KiloWatts describes POWER, which is the RATE at which electricity or
other energy can be put into or taken out of a battery!) These are "facts" upon which GM is intending to
base the entire survival of their company! Wow!
But their commercials are like all others for electric/battery-powered vehicles and also hybrids, an
IMPLIED assumption of that Tooth Fairy to re-charge the batteries when no one is looking! As if FREE
energy and power can somehow be available by some hocus-pocus.
emp10_01
5 sep 10
1.7
ELECTRIC CARS – LEAD ACID BATTERIES
Can the energy stored in lead acid batteries be used as means to store energy for an
electric motor powered car?
Consider the electric car going on the non-stop 250 km journey at a constant velocity.
How many 12 V lead acids batteries are required?
Remember, the amount of energy needed as useful work to be done on the car for the 250
km journey is 2.6108 J and the average rate at which this energy must be supplied is
18 kW and only 19 kg of petrol was needed.
Can the batteries provide the necessary power for the sprint from 0 to 100 km.h-1?
The amount of energy that is needed as useful work to be done on the car accelerating for
a short period time is 1.9106 J and the average rate at which this energy must be
supplied is 190 kW.
emp10_01
5 sep 10
1.8
Parameters describing the simple model for the 250 km journey
maintenance free rechargeable sealed lead-acid
Quantity
Energy expended as work done on
car
Battery rating
Nominal battery voltage
Mass of battery
Volume of battery
Value
Symbol
Unit
Calculation
W
R
V
M
Vol
=
=
=
=
=
2.59E+08
8.00E+01
12
20
0.013
Rint
Imax
=
=
0.005
800
IC
upetrol
dt
dt
=
=
=
=
24
4.6×107
4.0
1.44.E+04
Average voltage
Vavg
=
11.0
V
Average current
Iavg
=
20.0
A
Average output power
Pavg
=
220
W
= Vavg I
Power dissipation - internal resistance
Plost
=
2.0
W
= Rint Iavg2
U
e
=
=
3.17E+06
0.8
J
= Pavg dt
WE
=
3.23E+08
J
=W /e
N
=
102
Pmax
=
22
Mbatteries
=
Internal resistance
Max discharge current (for 5 s)
Max charge current
Energy density - petrol
Time for journey
Energy supplied by 1 battery
Efficiency of electric motor
Energy required by electric motor
No. batteries required
Max power of batteries
Total mass of batteries

A
A
J.kg-1
h
s
= WE / U
kW
= N Pavg
2042
kg
=NM
m3
= N Vol
J/kg
=U/M
Total volume of N batteries
Vbatteries
=
1.4
energy density of a battery
ubattery
=
1.58E+05
=
290
Energy density: petrol / battery
J
A.h
V
kg
m3
= upetrol / ubattery
Charging time for 1 battery
dtcharge1
=
3.1
h
= U /(V IC)
Charging time for N batteries
dtchargeN
=
13
days
= N dtcharge1
The energy density (J.kg-1) of the lead acid batteries is much less than that of petrol,
upetrol / ubattery ~ 290 as a consequence for the 250 km journey about 100 batteries are
required with a total mass of about 2000 kg which is greater than the mass of the car used
in the model. If the mass of the batteries is added to the car, then petrol is an even better
choice as an energy source for cars compared with lead acid batteries. The lead-acid
batteries can provide sufficient energy and provide the necessary power for the short
sprint lasting only 10 s. A battery can supply an enormous current (800 A for 5 s) but
only for a brief period of time. Another problem with batteries is that they can only be
charged and discharged only about 1000 times before they have to be replaced.
E333
emp10_01
5 sep 10
1.9
ELECTRICAL TERMINOLGY
In discussing the use of batteries and electric motors we come across the terms current,
voltage, resistance, etc, all the time but what do these terms really mean?
What is an electric current?
Often, electric current is simply defined as the movement of electric charge. But this
statement is imprecise. We have to be more careful in using scientific terminology.
The flow of charge gives rise to an electric current. The current is defined by the
equation
q
I
average current
t
where q is the amount of charge that passes a cross-section in the time interval t. A
still better definition is
dq
I
instantaneous current
dt
Consider a volume of electrons (charge e) moving through a wire of cross-sectional area
A, length dx and number density n. For the electrical conduction of electrons in a wire,
the electrons drift along at an average speed, vdrift = dx/dt. The current in the wire can be
expressed as
dq n e A dx
I

 n e Avdrift
the concept of vdrift will be consider in more detail later
dt
dt
Often, a more useful quantity than current is the current density J
I
I   J dA
J
IJA
A
The number density n is given by the equation (assuming that each atom donates one
electron for conduction)
M 
m
N
N
n A
  total  mmolecule  n 

M
V
V
 NA 
where n is the number density [m-3],  is the density [kg.m-3], mtotal [kg] and V [m3] are
the mass and volume of the sample respectively, mmolecule is the mass of one molecule
[kg], M is the molecular mass [kg] and NA is Avogadro’s number.
What is causes an electric current?
When a comb is brushed through dry hair it crackles and the individual hairs repel each
other, nylon clothing crackles and tends to stick to the body. The comb after passing
through the hair will attract small pieces of paper. All these effects and countless others
are attributed to static electricity and the forces which act are called electrostatic forces.
The attractive or repulsive effect is because objects can possess a charge. Charge is an
intrinsic property of fundamental particles – electrons and protons. By convention, the
charge on an electron is negative and that on a proton is positive.
emp10_01
5 sep 10
1.10
All matter is composed of atoms and an atom has a structure with a central nucleus
containing protons (+) and neutrons (neutral). This positive nucleus is surrounded by an
“electron cloud”. Normally the number of electrons equals the number of protons in an
atom and it is electrically neutral. However, some atoms have a tendency to lose electrons
and form a positive ion (+) while other atoms can gain electrons to form a negative ion
(-). Objects can become charged when the atoms that make up the object gain or lose
electrons. For example, when glass is rubbed with a silk cloth, electrons are removed
from the glass rod and it becomes positively charged while the silk acquires a negative
charge. When a hard rubber rod is rubbed with wool it gains electrons to become negative
and the wool becomes positive.
rubber rod -
glass rod +
e-
e-
wool +
silk -
Since electrons repel electrons and protons repel protons but electrons attract protons 
objects with the same charge repel each other; objects with the opposite charge attract
each other; any charged object will attract a neutral object
charged
rod
+
neutral
+
same charges
repel
+
-
opposite charges
attract
neutral pieces
of paper
any charged object
will attract a neutral
object
Surrounding a charged object is an electric field. In this electric field other charges will
experience an electrical force. A field is a region of space
E
where some physical quantity has a define value at every point.
For a vector field, a definite direction is associated with the
physical quantity. Examples of fields: temperature (scalar),
+q
F
pressure (scalar), wind (vector), gravitational field (vector),
electric (vector), magnetic (vector).
An electric field is a region where a charge particle will experience an electrical force. If
E is the strength of the electric field at a point where a charge q is located then the
electrical force exerted on the charge q is
F = E q or F  q E
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5 sep 10
1.11
The electric field can be represented by electric field lines and the direction of the field is
defined to be the in the same direction as the force would act on a positive charge at any
point within the field. The electric field lines emerge from + charges and disappear into charges. The greater the density of the electric field lines then the greater the electric field
strength.
+
+
+
+
+
-
-
-
-
-
-
+
Electric field
NEGATIVE charge
Electric field
POSTIVE charge
Uniform electric field
The force between two point charges q1 and q2, a distance r apart is given by Coulomb’s
Law. For a medium separating the charges: permittivity  = r 0 where 0 is the
permittivity of free space and r is the relative permittivity or dielectric constant.
(The physics of dielectrics will be consider in detail later)
F
q1 q2
4  r  0 r 2
1
Therefore, a charge distribution gives rise to an electric field. This fact is expressed by
Gauss’s Law
 E dA 
 E
qenclosed
0

r 0

q free  qbound
0

q free
r 0
q free  q f
qbound  qb
 charge density [C.m-3]
0 = 8.85×10-12 C2.N-1.m-2 or F.m-1
A charge experiences a force in an electric field. Therefore, we can define a potential
energy U. The potential energy per unit charge at a point in an electric field is called the
electric potential V at that point
U
V
q
The electric potential difference V between any two points 1 and 2 in an electric field
is
U U
U
V  V2  V1  2  1 
q
q
q
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5 sep 10
1.12
The change in potential energy is the negative of the work
W
V  V2  V1  
q
Alternatively, when a charge moves from one point to another in an electric field then
work is done on or by the charge
W = q V
If we set U1 = 0 at infinity as our reference point, the electric potential must also be zero
there, then we can define the potential at any point in an electric field to be
W
V  
q
where W is the work done by the electric field on a charged particle as the particle
moves in from infinity to that point.
Considering moving a charge q from point 1 to point 2 in an electric field
Incremental work done
dW  F dl  qE dl
Work done
W  q  E dl
Potential difference
V  
Electric field
E  V
2
1
2
W
   E dl
1
q
Dot product of two vectors A B  A B cos   Ax Bx  Ay By  Az Bz
 angle between the two vectors when place tail-to-tail
    
Vector differential operator   grad   , , 
 x y z 
Around a closed loop
 E dl
0
electric field is conservation
We can set up a simple model to describe the current through a section of wire.
length L
E
Electrons move slowly through
the conductor with an average
velocity called the drift velocity
vdift.
+
area
A
resistance R
resistivity 
conductivity 
number density n
I
_
v
electron
J
_
V
e-
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5 sep 10
e-
1.13
A current I flows because of the potential difference V = V across the ends of the wire
and the current is limited by the resistance R such that
V
I
R
LE
1
 E  E
I = J A and | V | = E L  J 
AR

L

Conductivity  [-1.m-1]
AR
AR
L

R
Resistivity  [.m]
L
A
As the electrons move through the wire they collide with the lattice and lose energy. A
simple model is that an electron losses all its energy in a collision and then is accelerated
from rest until the next collision. The time between the collisions is known as the
relaxation time  and the distance travelled between collisions is called the mean free
path . Therefore, we can model the electron as moving through the wire with a constant
drift velocity vdrift  vd.
Accelerating force
Drag force
FE  e E
Fdrag   m vd
FE  Fdrag
Current density
Resistivity
Conductivity
(impulse = change in momentum)
vd 
eE

m
n e2 E  1
J  n e vd 
 E
m

m
 2
ne 

n e2
m
This relationship is useful in showing that the conductivity of a metal depends upon the
density of conduction carriers, their mass, charge and relaxation time. However, this
model is not satisfactory. For example, some metals become superconducting (R = 0) at
very low temperatures approaching absolute zero.
J   E implies that the current is due to the electric field created by a voltage source
known as an emf (electromotive force). When the lead acid battery is connected to a
circuit the, emf is due to chemical reactions taking place to transfer electrons to make one
terminal positive and the other negative. The nominal battery emf is 12 V. However, a
battery has an internal resistance Rint. When a current is drawn from the battery the
potential across the terminals of the battery decreases with increasing current
Vbattery  emf  I Rint
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1.14
A generator can be used to create an emf by moving a conductor through a magnetic
field. The induced emf is determined by the time rate of change of the magnetic flux
d
emf   B
B   B dA
Faraday’s Law
dt
When the current flows through a resistance, energy is dissipated as thermal energy
dW qV
I2
V
P

V I V 2 / R 
I
V IR
dt
t
R
R
For lead acid-batteries, the internal resistance is very low (< 5 m), so even for large
currents drawn from a battery very little energy is dissipated and the temperature of a
battery will not increase significantly. However, when a battery provides a large current
to drive an electrical motor, the resistance in the electric motor windings could cause
excessive heating of the motor.
E221
The relaxation time has another equivalent interpretation. It measures the rate at which
the current or current density changes when the applied field is suddenly changed, e.g.,
turned on or off. For example, if the field is suddenly turned off, the current follows the
exponential law
I  I o e  t /
current decreases exponentially with time
Most electric motors are ac. Therefore, the dc current from a battery needs to be
converted to ac. However, for ac currents in a wire, the energy flow is due to
electromagnetic field surrounding the wire
 electromagnetic field drives the currents
 currents radiates the field
While propagating along the wire, the electromagnetic fields penetrate about one skin
depth  into the wire ( << a, a wire diameter). The current is not uniformly distributed
across the wire. Most of the wire is unused as the current is confined to a thickness of
about one skin depth. The wire might as well be hollowed out.
1
f
frequency of ac current
 2
f
 ac resistance is greater than the DC resistance and increases with frequency.
 Energy does not flow along the inside of the wire.
 Energy undergoes ohmic dissipation by the current flow into the wire from the
outside.
 The energy that flows is parallel to the wire in the space surrounding the wire.
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For ac currents, the charges must distribute themselves on the surface and this takes time
and this is determined by the relaxation time . As the frequency of the source driving the
ac current increases, the resistance of the wire increases because:


the effective cross sectional area for the conduction decreases
the rate at which the charges re-distribute themselves can’t match the
frequency of the applied fields i.e. the charges can no longer follow the field.
Copper is not such a good conductor at frequencies above 1010 Hz (microwave
frequencies))
Copper:
f = 50 Hz
 = 8.5 mm
f = 1010 Hz
 = 6.610-5 m
Can a domestic powerpoint be used to provide the electrical energy for an electric car
in a reasonable time frame?
A domestic powerpoint provides a voltage of 240 Vrms and a maximum current of
15 Arms.
%% power point
close all; clear all; clc;
Vrms = 240;
Irms = 15;
dW = 2.59e8;
P = Vrms * Irms;
dt = dW / P;
dt = dt / 3600;
% rms voltage
% rms current
% work need to be done on car for 250 km journey
% average power
% time need to provide the electrical energy [s]
% time need to provide the electrical energy [h]
fprintf('Power rating of powerpoint, P = %6.0f W\n',P)
fprintf('Time need to provide the electrical energy, dt = %4.2g h\n',dt)
Power rating of powerpoint, P = 3600 W
Time need to provide the electrical energy, dt = 20 h
The electric car would have to be connected to a normal powerpoint for about 20 h to
provide sufficient energy for the 250 km journey.
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1.16
NEXT STEP
It is quite clear that lead acid battery powered cars can’t replace existing petrol powered
ones. Another way in which electrical energy can be stored is by using capacitors. A
capacitor is a component that stores charge on two conductive plates that are separated by
an insulating medium called a dielectric. Our next step is to look the physics of
capacitors from a macroscopic view and develop simple models to judge whether
capacitors can be used as an energy source for electric motor powered cars. It may be
possible that newer super (ultra) capacitors which have capacitance values of 1000’s
farads may be the answer. However, before looking at how the energy stored by
capacitors can be dramatically increased with the presence of a dielectric filling the space
between the plates it is a necessary to have a good understanding of dielectrics from a
fundamental microscopic view.
E221
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