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Norah Ali Al-moneef
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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•
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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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
 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 = at = –eEt/m
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ELECTRON MOTION IN A CONDUCTOR WITH AND
WITHOUT AN ELECTRIC FIELD
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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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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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 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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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
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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
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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
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 The drift speed, vd, is the speed at which the carriers move
vd = Δx/Δt
 Rewritten: ΔQ = (nAvdΔt)qΔx
 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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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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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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Example:
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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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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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Example:
a)
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 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
Conventional
current
Units: [ I ] = 1A = 1 C/s
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
t

4s
 4A
 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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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 density J) : J  I  ne(vd A)  nevd
A
• Ohm's Law: E =  J
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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Al-moneef
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 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
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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
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Resistivity and Resistance
J = E/ρ where ρ is the resistivity
Consider a bar or wire of crosssection 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:
I = V/R
also called Ohm’s Law.
∆V = IR
I = V/R and R = ρL/A is the resistance of the bar.
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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
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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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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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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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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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R
(Ohms)
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Example:
What voltage is required to produce 2a though
a circuit with a 3 resistor.
3
V = IR = 2A x 3 = 6v
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V
I = 2a
2. Ohm’s Law
Nonohmic device
I
I
V
V
Resistance
V
R
I
V
I
R
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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

18

10

2

3
A
p (10 m)

Temperature dependence of resistivity
T  0 1   (T  T0 )
T   0   0 (T  T0 )
   0 (T  T0 )
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

 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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Drift Velocity
• In a conductor the free electrons are moving very
fast in random directions (v ~ 106 m/sec)
• They collide with the atoms of the lattice and are
scattered in random directions
• If an electric field is present, there is a slow net drift
of electrons in the direction opposite the electric
field
• vDRIFT ~ mm/sec
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• Amperes: the measure of the rate of current flow.
– 6.24 × 1018 electrons passing a point per second is equal to
one amp.
• A current occurs whenever there is a source of
electricity, conductors and a complete circuit.
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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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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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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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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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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
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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
4. 3
A
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  
L
A
A2  A1 / 4
L2  2 L1
R
r2  r1 / 2
A  pr 2
I 2 / I1  ?
C. I1 > I2
B. I1 = I2
A. I1 < I2
I1
V
L2
2 L1
R2  

 8 R1
A2
A1 / 4
V
V
1
I2 

 I1
R2 8 R1 8
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I2
Voltage and Current Relationship for
Linear Resistors
Current (A)
Voltage versus Current
for a 10 ohm Resistor
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
Voltage (V)
Voltage and current are linear when resistance is held constant.
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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)
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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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B
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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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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• 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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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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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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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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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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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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B
Norah Ali Al-moneef
• 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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Norah Ali Al-moneef
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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Norah Ali Al-moneef
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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Norah Ali Al-moneef
If an electric fire uses 1.8 MJ of energy in a time of 10
minutes, calculate the power output of the fire.
E = 1.8 MJ = 1.8x106 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:
P = IV
• Examples of resistance loads are heaters and
incandescent lamps.
P = Watts
example
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.Norah
PbAli> 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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
Norah Ali Al-moneef
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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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