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1
Norah Ali Al moneef
Electron Configuration
 The Electron Configuration is the orbital description of
the locations of the electrons in an unexcited atom
 Electrons orbit in “SHELLS” or “Energy Levels”
 The higher the orbit, the higher the “Energy Level”
 Atoms react based on the Electron Configuration
 The outermost electron shell is the most important as far
as conductivity properties are concerned
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Norah Ali Al moneef
Bohr model of the aluminum atom
14 N
13 P
Protons = Electrons
13=13
Net charge is neutral or zero
Complete with 2
Complete with 8
+29
Complete with 18
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Norah Ali Al moneef
Incomplete with 1
Electrically Neutral
Electricity-the flow of free electrons
Valance electron
Positive source
Negative source
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Norah Ali Al moneef
Bound electron
Free electron
Atomic structure of conductors,
insulators, and semiconductors
Conductor - 1 to 3
valence electrons
Insulator - full
valence shell
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Norah Ali Al moneef
Semiconductor - 4
valence electrons
Conductors & Semiconductors
• In conductors, the valence band is only partially-full, so
electrons can easily move from being near one atom to being
near another
• In semiconductors and insulators, the valence band is
completely full, so electrons must gain extra energy to move
• In semiconductors, the band gap between the full valence band
and the empty conduction band is small, so electrons move
easily with only thermal energy
• In insulators, the band gap is larger, so electrons will not easily
move into the conduction band
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Norah Ali Al moneef
•
Conductors
&
Insulators
Electric current moves easily through some materials and less easily
through other materials
• Materials that have very “tightly bound” electrons have few free electrons
when an electric force is applied. These materials are insulators (e.g.
rubber, glass, dry wood)
• Materials that allow the movement of a large number of free electrons are
called conductors (e.g., silver, copper, aluminum)
– Electrical energy is transferred through a conductor by means of the
movement of free electrons that move from atom to atom
– Displaced electrons continue to “bump” each other
– The electrons move relatively slowly but this movement
creates
electrical energy throughout the conductor that is transferred almost
instantaneously throughout the wire
(e.g., billiard ball example,
wind vs. sound example)
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Norah Ali Al moneef
27.1 Electric Current
 Whenever electric charges move, an electric current is said to
exist
 The current is the rate at which the charge flows through a certain
cross-section
 For the current definition, we look at the charges flowing
perpendicularly to a surface of area A
Definition of the current:
Charge in motion through an area A. The time rate of the charge
flow through A defines the current (=charges per time):
I av 
Q
t
Units:1 C/s= 1 A
SI unit of the current: Ampere
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Norah Ali Al moneef
Electrical current
 If an electric field points from left to right, positive
charge carriers will move toward the
right
while negative charges will move toward the
left
 The result of both is a net flow of positive charge to the
right.
 Current is the net change in positive charge per time

Q
t
I
Instantaneous current
av
i=dq/dt
• Coulomb (C) – represents the total charge of approximately 6.25 x
1018 electrons
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Norah Ali Al moneef
 The direction of current flow is the direction positive
charge would flow
 This is known as conventional (technical) current flow, i.e., from
plus (+) to minus (-)
 However, in a common conductor, such as copper, the current is due to
the motion of the negatively charged electrons
 It is common to refer to a moving charge as a mobile
charge carrier
 A charge carrier can be positive or negative
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Norah Ali Al moneef
Charge Carrier Motion in a Conductor
The electric field force F
imposes a drift on an electron’s
random motion (106 m/s) in a
conducting material. Without
field the electron moves from P1
to P2. With an applied field the
electron ends up at P2’; i.e., a
distance vdt from P2, where vd
is the drift velocity (typically
10-4 m/s).
Norah Ali Al moneef
11
Does the direction of the
current depend on the
sign of the charge? No!
qvd
E
vd
(a) Positive charges moving in
the same direction of the field
produce the same positive
current as (b) negative charges
moving in the direction
opposite to the field.
E
vd
Norah Ali Al moneef
(-q)(-vd) = qvd
12
Microscopic model of current
 Charged particles move through a
conductor of cross-sectional area A
 n is the number of charge carriers per
unit volume V (=“concentration”)
 nAx=nV is the total number of charge
carriers in V
The total charge is the number of carriers times the
charge per carrier, q (elementary charge)
ΔQ = (nAΔx)q [unit: (1/m3)(m2 m)As=C]
Norah Ali Al moneef
13
 The drift speed, vd, is the speed at which the carriers move
Δx
vd = Δx/Δt
 Rewritten: ΔQ = (nA vdΔt)q
 current, I = ΔQ/Δt = nqvdA
If the conductor is isolated, the electrons undergo (thermal)
random motion
When an electric field is set up in the conductor, it creates an
electric force on the electrons and hence a current
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Norah Ali Al moneef
Example:
coulombs of charge pass a point in a wire every two
seconds. Calculate current.
 Coulomb (C) – represents the total charge of
approximately 6.25 x 1018 electrons
 Unit of Current – Ampere (A) = 1coulomb/second
Q 3C
I 
 1.5 C/s  1.5 A
t
2s
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Norah Ali Al moneef
Example:
An 18-gauge copper wire (diameter 1.02 mm) carries a constant
current of 1.67 A to a 200 W lamp. The density of free electrons is
8.51028 per cubic meter. Find the magnitudes of (a) the current
density
(b) the drift velocity.
(a) A=d 2p/4=(0.00102 m)2p/4=8.210-7 m2
J=I /A=1.67 A/(8.210-7 m2)=2.0106 A/m2
(b) From J=I /A=nqvd
J
2.0  10 6 A / m 2
vd 

nq (8.5  10 28 m 3 )(1.60  10 19 C)
vd=1.510-4 m/s=0.15 mm/s
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Norah Ali Al moneef
Example:
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Norah Ali Al moneef
Example:
• If 240 C of charge pass a point in a conductor
in 5 min, what is the current through that point
in the conductor?
Convert 5 min to seconds
Q 240 C
I 
 0.80 A
t
300 s
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Norah Ali Al moneef
5.0min X 60s/1 min = 300s
Example:
 Q Ne 5.6 1014  1.6 10-19
I


 2mA
t t
0.04
Q
Ne 6.4 1021  1.6 10-19
c)
I


 8.53 A To the left
t
t
2.00  60
 Q I  t 0.835 A  5 s
b)
 Q  Ne
 N


e
e
1.6 10-19
N  2.611019 electrons
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Norah Ali Al moneef
Example:
a)
20
 Q 1.67C
I 

 0.835 A
t
2.00 s
 Q I  t 0.835 A  5 s


e
e
1.6 10-19
N  2.611019 electrons
b)
 Q  Ne
c)
 Q Ne 6.4 1021 1.6 10-19
I


 8.53 A
t t
2.00  60
Norah Ali Al moneef
 N
Example:
II. Electric current
1. Definition
Q
I
t
Units: [ I ] = 1A = 1 C/s
Conventional
current
Electron
flow
1020 electrons passed through the electric conductor during 4
seconds. Find the electric current through this conductor.
q (1.6 1019 C )(1020 )
I 
 4A
t
4s
 Example:The electric current of 0.5 A is flowing through the electric
conductor. a) What electric charge is passing through the conductor
during each second
b) What electric charge will pass through the
conductor during 1 minute?
q  It  (0.5 A)(1 s)  0.5 C
a)
q  It  (0.5 A)(60 s)  30 C
b)
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Norah Ali Al moneef
27.2 resistance
• I = n q vd A
– n = number of free charge carriers/unit volume
• Current density
• (The current per unit cross-section is called the current
I ne(vd A)
density J) :
J 
 nev
A
• Ohm's Law: E =  J
d
A
J=σE
–  = resistivity
–  = 1/ = conductivity
– Good conductor: low  and high 
• Ohm's Law:
– R = resistance } Measured in Volt/Ampere = Ohm (){
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Ali Almoneef
Norah Ali AlNorah
moneef
22
 In a homogeneous conductor, the current
density is uniform over any cross section, and
the electric field is constant along the length.
b
a
The ratio of the potential drop to the
current is called resistance of the
segment:
Unit: 1V/A= 1 (ohm)
V=Va-Vb=EL
Resistance in a circuit arises due to collisions between the
electrons carrying the current with the fixed atoms inside the
conductor
Norah Ali Al moneef
23
Ohm’s Law
 V  I  V=const .I  V=RI
 Ohm’s Law is an empirical relationship that is valid only for certain
materials
 Materials that obey Ohm’s Law are said to be ohmic
 I=V/R
 R, I0, open circuit; R0, I, short circuit
•
The ratio of the potential drop to the current is called
resistance of the segment:
V
R 
I
Unit: V/A= (ohm)
Norah Ali Al moneef
24
Resistivity and Resistance
J = E/ρ where ρ is the resistivity
Consider a bar or wire of cross-section
A and length L, carrying current I and
with potential difference V =Vb - Va
between the ends.
We know E = V/L so I/A = J = V/Lρ.
Thus:
also called Ohm’s Law.
I = V/R
∆V = IR
I = V/R and R = ρL/A is the resistance of the bar.
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Norah Ali Al moneef
Ohm’s Law, final
Ohmic
Plots of V versus I for (a)
ohmic and (b) nonohmic
materials. The resistance
R=V/I is independent of I
for ohmic materials, as is
indicated by the constant
slope of the line in (a).
Nonohmic
Norah Ali Al moneef
26
Ohmic Resistors
• Metals obey Ohm’s Law linearly so long as their
temperature is held constant
• Their resistance values do not fluctuate with temperature
• i.e. the resistance for each resistor is a constant
• Most ohmic resistors will behave non-linearly outside of a
given range of temperature, pressure, etc.
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Norah Ali Al moneef
On What Does Resistance Depend?
• If I increase the length of a wire, the current flow
decreases because of the longer path
• If I increase the area of a wire, the current flow
increases because of the wider path
R =  L/A
• If I change to a material with better conductivity, the
current flow
increases because charge carriers move better
• If I change the temperature, the current flow
changes
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Norah Ali Al moneef
More on Resistance
I = V/R
Units of Resistivity : ρ is in Ω-m (ohm-meters),
so R is in (Ω-m)(m)/(m2)
= Ω (ohms) = V/A (volts/ampere)
Resistivity ρ depends only upon the material (copper,
silver…).
Resistance R depends upon the material and also upon
the dimensions of the sample (L, A).
- R = ρL/A
Note: Some devices (e.g.semiconductor diode) do not obey Ohm’s law!
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Norah Ali Al moneef
Resistors are designed to have a specific resistance to reduce
the amount of current going to a specific part of a circuit
To obey Ohm’s law means a conductor has a constant resistance
regardless of the voltage.
V
(Volts)
A
(Amps)
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Norah Ali Al moneef
R
(Ohms)
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Norah Ali Al moneef
Example:
What voltage is required to produce 2a though
a circuit with a 3 resistor.
V = IR = 2A x 3 = 6v
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Norah Ali Al moneef
V
3
I = 2a
2. Ohm’s Law
Nonohmic device
I
I
V
V
Resistance
V
R
I
V
I
R
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Norah Ali Al moneef
Units: [ R ] = 1Ω = 1 V/A
V  IR
Ohm’s Law:
R  const
Resistivity
I
Definition:
A
L
L
R
A
A
R
L
Example:What is the resistance of 1 m of nichrome wire of 2 mm diameter ?
L
1m
6
3
R    10   m

3

10

2
2
A
p (
10
 (T  T m
) )
0
0
Temperature dependence of resistivity
T  0 1   (T  T0 )
T   0   0 (T  T0 )
   0 (T  T0 )
34


 0 T
   0 (T  T0 )
Norah Ali Al moneef

T
 The drift speed is much smaller than the average speed
between collisions
 When a circuit is completed, the electric field travels with a
speed close to the speed of light
 Although the drift speed is on the order of 10-4 m/s the
effect of the electric field is felt on the order of 108 m/s
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Norah Ali Al moneef
Example
What is the current flow in a circuit with a voltage of 120 volts and a
resistance of 0.23 ?
V = IR
V 120 V
I=
=
= 521.7 A
R 0.23 
Example
With the increase in the length of the wire, the current increases.
A. True
B. False
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Norah Ali Al moneef
Example
A 4v battery is placed in a series circuit with a 2 resistor.
What is the total current that will flow through the circuit?
V = IR
2
4v = I x 2 
4v
I = 2A
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Norah Ali Al moneef
I=?
Example
What voltage is required to produce 2A though a circuit with
a 3 resistor.
V= IR
3
V = 2A x 3
V
V = 6v
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Norah Ali Al moneef
I = 2A
Example
What resistance is required to limit the current
to 4 A if a 12 V battery is in the circuit?
V = IR
3
12 = 4 x R
R = 3
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Norah Ali Al moneef
12v
I = 4a
Example
A cylindrical copper rod has resistance R. It is reformed
to twice its original length with no change of volume.
Its new resistance is:
1. R
2. 2R
3. 4R
4. 8R
5. R/2
Norah Ali Al moneef
40
Two conductors are made of the same material and
have the same length. Conductor A is a solid wire of
diameter 1 mm. Conductor B is a hollow tube of inside
diameter 1 mm and outside diameter 2 mm. The ratio
of their resistances RA/RB is
1. 1/2
2. 1
3. 2
A
4. 3
5. 4
B
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Norah Ali Al moneef
Two cylinders are made of the same material and have
the same length but different diameters. They are
joined end-to-end and a potential difference is
maintained across the combination. Which of the
following quantities is the same for the two cylinders?
1. the potential difference
2. the current
3. the current density
4. the electric field
5. none of the above
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Norah Ali Al moneef
Example
Two cylindrical resistors, R1 and R2, are made of identical
material. R2 has twice the length of R1 but half the radius of
R1. They are connected to a battery V as shown. Compare
the currents flowing through R1 and through R2.
 2  1  
L2  2 L1
r2  r1 / 2
A  pr 2
I 2 / I1  ?
C. I1 > I2
B. I1 = I2
A. I1 < I2
L
A
A2  A1 / 4
R
V
L2
2 L1
R2  

 8 R1
A2
A1 / 4
V
V
1
I2 

 I1
R2 8 R1 8
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Norah Ali Al moneef
I1
I2
Which one of the following graphs correctly represents
Ohm's law, where V is the voltage and I is the current?
(a)
(b)
(c)
(d)
(e)
44
A
B
C
D
A and C
Norah Ali Al moneef
If a piece of wire has a certain resistance, which wire made
of the same material will have a lower resistance?
A )a hotter wire
B ) a thicker wire
C ) a longer wire
D) a thinner wire
ANS:
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Norah Ali Al moneef
B
27.3
A model for electrical conduction
Electrons in an Electric Field
 Conduction electrons move randomly in all directions in the
absence of a field.
 If a field is applied, the electric force results in acceleration in
a particular direction:
F=ma= –eE  a = –eE/m
•This acceleration, which occurs for only a short time
interval between collisions, enables the electron to acquire a
small drift velocity.
•the motion of free electrons in a conductor. In the absence
of an electric field, there is no net displacement after many
collisions An electric field E modifies the random motion
and causes the electrons to drift in a direction opposite that
of E
Norah Ali Al moneef
46
ELECTRON MOTION IN A CONDUCTOR WITH AND
WITHOUT AN ELECTRIC FIELD
we assume that the motion of an electron after a collision is independent of its
motion before the collision. We also assume that the excess energy acquired by
the electrons in the electric field is lost to the atoms of the conductor when
the electrons and atoms collide. The energy given up to the atoms increases
their vibrational energy, and this causes the temperature of the conductor to
increase. The temperature increase of a conductor due to resistance
the free electrons drift slowly in a direction opposite that of the electric
field, with an average drift speed vd that is much smaller (typically 10-4 m/s)
than their average speed between collisions (typically 106 m/s).
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Norah Ali Al moneef
As the charges accelerate, the potential energy stored in the
electric field is converted to kinetic energy which can be
converted into heat and light as the electrons collide with
atoms in the wire This acceleration produces a velocity
v
f
 vi  at  vi 
the average value of vi is zero
qE
t
me
The term (qE/me)τ is the velocity change of the electron due to
the electric field during one trip between atoms
The average value of the second term of is (qE/me)τ, where τ is
the average time interval between successive collisions. Because the
average value of vf is equal to
the drift velocity,
qE
Drift velocity in terms of
v
f
 vd 
m
e
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Norah Ali Al moneef

microscopic quantities
Drift velocity= Acceleration x Mean free time(τ )
The average time taken between collisions is called
F

vd 
as relaxation time, (or mean free time)
m
e
the magnitude of the current density is
J  nq vd 
2
nq E
me

Current density in terms
of microscopic quantities
where n is the number of charge carriers per unit volume.
Ohm’s law,
J=σE
The relationships for conductivity and resistivity of
a conductor:
 

49
nq
2

me
1



Norah Ali Al moneef
me
n q2
Conductivity in terms of
Microscopic quantities
Resistivity in terms of
microscopic quantities
According to this classical model, conductivity and resistivity do not
depend on the strength of the electric field. This feature is
characteristic of a conductor obeying Ohm’s law.
The average time interval τ between collisions is related to
the average distance between collisions (that is, the mean free
path)  and the average speed v through the expression

 
v
The average distance taken between collisions is
called as mean free path, 
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Norah Ali Al moneef
Since there is no applied field, the movement of the charge
carriers will be completely random. This randomness result no
net current flow. As a result of thermal energy there are almost an
equal number of carriers moving right as left, in as out or up as
down
How long does a carrier move in time before collision ?
The average time taken between collisions is called as
relaxation time, (or mean free time)
How far does a carrier move in space (distance) before
a collision?
The average distance taken between collisions is called as
mean free path, .
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Norah Ali Al moneef
27.4 Resistance and Temperature
•Resistance (R) is proportional to resistivity ():
R=L/A
The resistivity () depends on temperature and the physical properties of
the material, so it has a different value for each material
 The resistivity (and hence resistance) varies
with temperature.
 For metals, this dependence on temperature is
linear over a broad range of temperatures.
 An empirical relationship for the temperature
dependence of the resistivity of metals is given
by
   0 [1   (T  T0 )]
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Norah Ali Al moneef
Copper
•  is the resistivity at temperature T
• 0 is the resistivity at some standard temperature T0
•  is the “temperature coefficient” of electric resistivity
for the material under consideration
• Some materials, when very cold, have a resistivity
which abruptly drops to zero. Such materials are called
superconductors.
T  0  0 (T  T0 )
T  0 1   (T  T0 )
   0 (T  T0 )
T 

 0
• The temperature coefficient of resistivity can be expressed as.


 0 T
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Norah Ali Al moneef
• In everyday applications we are interested in the
temperature dependence of the resistance of various
devices.
• The resistance of a device depends on the length and the
cross sectional area.
• These quantities depend on temperature
• However, the temperature dependence of linear expansion is
much smaller than the temperature dependence of resistivity
of a particular conductor.
• So the temperature dependence of the resistance of a
conductor is, to a good approximation,
R  R0  1   (T  T 0) 
where R0 and T0 are the resistance and temperature at a standard
temperature, usually room temperature or 20o C.
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Norah Ali Al moneef
Reminder: Battery as a “ski lift for charges”:
Ski lift raises objects to higher potential energy
- flow may vary, but potential energy difference fixed
Battery also fixed potential diff. , but current may vary
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Norah Ali Al moneef
27.6 Electrical Power
• The chemical energy of the battery is
converted to U, electrical potential
energy: Echem  U
• The resulting electric field causes the
electrons to accelerate: UK
• Collisions in the lattice structure transfer
the energy to the lattice as thermal
energy: KEth
• Thermal energy is a dissipative energy
(i.e. can’t be recovered like mechanical
energy.
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Norah Ali Al moneef
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Norah Ali Al moneef
the rate at which the system loses electric potential energy as the charge
Q passes through the resistor:
du d
dQ
 ( Q V ) 
V  I V
dt dt
dt
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Norah Ali Al moneef
•The system regains this potential energy when the charge
passes through the battery,
• Since a resistor obeys Ohm’s Law:
PR = I2R = (∆VR)2/R
Electrical Energy = Voltage x Electrical Current x Time Interval
energy =
V
x I (amps) x
t (sec)
E = V x I x t
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Norah Ali Al moneef
How is Electrical Power calculated?
Electrical Power is the product of the current (I) and the
voltage (v)
The unit for electrical power is watt (W)
Example
How much power is used in a circuit which is 110 volts and has
a current of 1.36 amps?
P = IV
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Norah Ali Al moneef
Power = (1.36 amps) (110 V) = 150 W
electrical energy: Electrical energy is a measure of the
amount of power used and the time of use.Electrical energy is
the product of the power and the time.
Electrical Energy = Voltage x Electrical Current x Time Interval
energy =
V
x I (amps) x
t (sec)
E = V x I x t
Example
energy = Power X time
P = (2A) (120 V) = 240 W
P = IV
E = (240 W) (4 h) = 960Wh = 0.96 kWh
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example
A 9-volt battery drives an electric current through a circuit
with 4-ohm resistance. What is the electric current
running through the circuit?
0.44 A
b. 2.25 A
c. 5 A
d. 36 A
a.
ANS:
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• Joule’s Law
– States that the rate at which heat
produced in a conductor is directly
proportional to the square of the current
provided its resistance is constant
– i.e. P = I 2R
In order to prevent power lines from
overheating, electricity is
transmitted at a very
high voltage
From Joule’s law the larger the current the more
heat produced hence a transformer is used to
increase voltage and lower current
i.e. P = V I
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Power dissipated by a bulb relates to the brightness of the bulb.
The higher the power, the brighter the bulb.
For example, think of the bulbs you use at home. The 100W
bulbs are brighter than the 50W bulbs.
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example
If an electric fire uses 1.8 MJ of energy in a time of 10
minutes, calculate the power output of the fire.
Energy = 1.8 MJ = 1.8x106 J
t=10 minutes = 600 s
Power = Energy / time
p = 1.8x106 J / 600 =3 10 3 watt
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example
Calculate the power of a vacuum cleaner if the
operating voltage is 120v, and the current flowing
through it when it is used is 7.90A.
P=VxI
P = 120V x 7.9A
P = 948 W
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example
Calculate the voltage of a computer that has
600W of power and 1.9A flowing into the
monitor?
V= P
I
V = 600W
1.9A
V = 316V
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example

If a 500 watt speaker need 10 amps to operate, what is the
voltage requirement?
500
V 
10
V  50V
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Example
• How much would you be charged for using a 60 Watt
light bulb for 10 hours if electricity costs 0.07 $per
kWh?
• E = PT= 0.06kW x 10h = 0.6kWh
• Cost = 0.6kWh x 0.07 $/kWh= 0.04$
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If an electric fire uses 1.8 MJ of energy in a time of 10
minutes, calculate the power output of the fire.
6
1.8x10
E = 1.8 MJ =
J
t=10 minutes = 600 s
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Power
• Power is the rate of doing work.
• Electrical power is usually expressed in watts or kilowatts
• In DC and AC circuits, with resistance loads, power can
be determined by:
• Examples of resistance loads are heaters and incandescent lamps.
example
P = IV
P = Watts
I = Amps
V = Volts
• Determine the power consumed by a resistor in a 12 volt
system when the current is 2.1 amps.
P = IV = 2.1 A x 12 V = 25.2 W
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ELECTRIC POWER
When there is current in a circuit as a result of a voltage, the electric
power delivered to the circuit is:
P  IV
SI Unit of Power: watt (W)
Many electrical devices are essentially resistors:
P  I (IR )  I R
2
V2
V 
P   V 
R
R
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Rank in order, from largest to smallest, the powers
Pa to Pd dissipated in resistors a to d.
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1. Pb > Pa = Pc = Pd
2. Pb = Pc > Pa > Pc
3. Pb = Pd > Pa > Pc
4. Pb > Pc > Pa > Pd
5.NorahPAlib Al>moneef
Pd > Pa > Pc
Example
• Determine the amount of energy a 100 Watt light bulb
will use when operated for 8 hours.
Energy = Power x Time
= 100 watts x 8 hour
= 800 wh
• What will it cost to operate the light bulb if the electrical energy
costs 0.12 $/kWh?

$
1 kW
$ = 0.12
x 800 W x
x 8 h = 0.77 $
kWh
1, 000 W
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Energy Use Calculations
How much electrical energy will an electric blanket use per
month if it is used 8 hours a day? The blanket is on a 120 V
circuit and draws 1.5 amp.
8 h 30 day
1 kW
Energy (kWh) = (120 V x 1.5 A) W x
x
x
= 43.2 kWh
day month 1,000 W
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Energy = Power x Time
E = (100 W) (300 s)
E = 30,000 J
E = 30 kJ
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Energy = Power x Time
= (4.2 kW) x (20 h)
= 84 kWh
Cost = Energy x rate per kWh
= (84 kWh) x ($0.12)
Rated for 4.2 kW
Used 20 h/month
Cost of 12 $ per kWh
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= $10.08
Example
Given copper wire 1mm diameter . 100m long has a
potential diffidence of 12 V Find
a) resistance, b) current in wire, c) current density,
d) electric field in wire, e) concentration of electrons
(assuming 1electron / atm), f) drift velocity,
g) amount of electric charge flowing in 1 minute
Resistivity ρ = 1.72x10-8 Ohm-m
Density D = 8.9 E 3 kg/m3
molecular weight M = 63.546 g/mole
Avogadro's #
6.022x10 23
electric charge
e = 1.6x10-19 C
r = 5x10-4m radius, L=100m, t=60sec
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Equations:
a)
b)
c)
d)
e)
f)
g)
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Answers:
R= ρL/A, A=πr2
A=7.85x10-7, R=2.19 Ω
V=IR
I=V/R I=5.48 A
J=I/A
J=6.977x106 A/m2
E = ρJ
0.12 V/m
n =D Na /M
8.434E28 e/m3
I = n q vd A
vd = I/nqA = 5.17x10-4 m/s
I= dQ/dt
Q = It = 329 C
Norah Ali Al moneef
Example
 100 W light bulb connected to 110V what is
 a) current
b) resistance c) at 10cents/kwhour how
much to illuminate for a year, d) how many can be
connected to a 15 ampere circuit breaker,
e) how
much electric power consumed by all these bulbs,
f)
if the temperature is 4500K and made from tungsten (α =
0.0038/K) what is the room temperature resistance at 300K
Given P=100 W, V= 100V Imax = 15A
price = 0.1 $/kW h T=4500 K To = 300 K
α = 0.0038/K
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Equations
Answers
a) P = IV
I=P/V = 0.909 A
b) V = IR
R=V/I = 121 Ω
c) cost = ($0.1)(.1KW)(24 x 365) = $87.60
d) Imax> Nmax I
Nmax = 16
e) Pmax = Nmax P
Pmax = 1600W
f)
R=Ro(1 + α (T-To)) =
Ro = 7.13 Ω
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summary
OHM’S LAW FORMULAS
Find current: I=ΔQ/Δt
I=nqAvd
Find Current
Find Voltage
Find Resistance
V
I=
R
V=IxR
V
R=
Current equals
voltage divided
by resistance
Voltage equals
current multiplied
by resistance
Resistance equals
voltage divided
by current
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I
Unit of Measure
Quantity
Name
Symbol
Voltage
V, emf
or E
Current
I
Resistance
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R
Name
Voltage
Ampere
Ohm
Symbol
Function
V
Pressure which
makes current
flow
A
Rate of flow
of electrons

Opposition to
current flow
Resistance related to physical parameters
The dimensions and geometry of the resistor as well as the particular
material used to construct a resistor influence its resistance. The
resistance is approximately given by
R
L
A
T  0 1   (T  T0 )
R  R0  1   (T  T 0)

P  IV
PI R
2
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V2
P
R