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Norah Ali Al-moneef1
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
Norah Ali Al-moneef2
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)
Norah Ali Al-moneef3
Electrons in an Electric Field
Norah Ali Al-moneef4
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/mAs 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 velocityv = at = –eEt/m
ELECTRON MOTION IN A CONDUCTOR WITH AND WITHOUT AN ELECTRIC FIELD
Norah Ali Al-moneef5
27.1 Electric Current
Norah Ali Al-moneef6
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
Charge in motion through an area A. The time rate of the charge flow through A defines the current (=charges per time):
Units:1 C/s= 1 A SI unit of the current: Ampere
Definition of the current:
t
QI av
Electrical current
Norah Ali Al-moneef7
If an electric field points from left to right, positive charge carriers will move toward therightwhile negative charges will move toward theleft
The result of both is a net flow of positive charge to the right.
Current is the net change in positive charge per time
t
QI av
Instantaneous current i = d q / d t
• Coulomb (C) – represents the total charge of approximately 6.25 x 1018 electrons
Norah Ali Al-moneef8
The direction of current flow is the direction positive charge would flowThis 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 carrierA charge carrier can be positive or negative
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 vdt from P2, where vd is the drift velocity (typically 10-4 m/s).
Norah Ali Al-moneef 9
Does the direction of the current depend on the sign of the charge? No!
(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.
Norah Ali Al-moneef 10
E
E
vd
vd
qvd
(-q)(-vd) = qvd
Charged particles move through a conductor of cross-sectional area A
n is the number of charge carriers per unit volume V (=“concentration”)
nAx=nV is the total number of charge carriers in V
Norah Ali Al-moneef 11
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]
Microscopic model of current
Norah Ali Al-moneef12
The drift speed, vd, is the speed at which the carriers movevd = Δx/Δt
Rewritten: ΔQ = (nAvdΔt)q
current, I = ΔQ/Δt = nqvdA
Δx
If the conductor is isolated, the electrons undergo (thermal) random motionWhen an electric field is set up in the conductor, it creates an electric force on the electrons and hence a current
Norah Ali Al-moneef13
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
A 1.5C/s 1.5s 2
C 3
t
QI
Example:
Example:
Norah Ali Al-moneef14
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 8.51028 per cubic meter. Find the magnitudes of (a) the current density
(b) the drift velocity. (a) A=d 2/4=(0.00102 m)2/4=8.210-7 m2
J=I /A=1.67 A/(8.210-7 m2)=2.0106 A/m2
(b) From J=I /A=nqvd
)C1060.1)(m105.8(
m/A100.219328
26
d
nq
Jv
vd=1.510-4 m/s=0.15 mm/s
Example:
15 Norah Ali Al-moneef
• 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 5.0min X 60s/1 min = 300s
A 0.80 s 300
C 240
t
Q I
Example:
16 Norah Ali Al-moneef
electrons 1061.2N 101.6
s 5835.0 t Q Q )
53.860 00.2
101.6 106.4
t
t
Q )
20.04
101.6 105.6
t
t
Q
19
19-
19-21
-1914
A
e
I
eNNeb
ANe
Ic
mANe
I
To the left
Example:
17 Norah Ali Al-moneef
electrons 1061.2N 101.6
s 5835.0 t Q Q )
19
19-
A
e
I
eNNeb
As
CIa 835.0
00.2
67.1
t
Q )
ANe
Ic 53.860 00.2
101.6 106.4
t
t
Q )
19-21
Example:
18 Norah Ali Al-moneef
II. Electric current1. Definition
t
QI
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.
As
C
t
qI 4
4
)10)(106.1( 2019
CsAItq 5.0)1)(5.0( CsAItq 30)60)(5.0(
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?
a)
b)
Norah Ali Al-moneef19
Example:
27.2 resistance
Norah Ali Al-moneef
Norah Ali Almoneef 20
• 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) :
• Ohm's Law: E = J J = σ E– = resistivity– = 1/ = conductivity– Good conductor: low and high
• Ohm's Law: – R = resistance Measured in Volt/Ampere = Ohm ()
dd nev
A
Avne
A
IJ
)(
20
In a homogeneous conductor, the current density is uniform over any cross section, and the electric field is constant along the length.
Norah Ali Al-moneef 21
a
b
V=Va-Vb=EL The ratio of the potential drop to
the current is called resistance of the segment:
Unit: 1V/A= 1ohm
Resistance in a circuit arises due to collisions between the electrons carrying the current with the fixed atoms inside the conductor
Ohm’s Law
Norah Ali Al-moneef22
V I V=const .I V=RIOhm’s Law is an empirical relationship that is
valid only for certain materialsMaterials that obey Ohm’s Law are said to be
ohmicI=V/R R, I0, open circuit; R0, I, short circuit
• The ratio of the potential drop to the current is called resistance of the segment:
Unit: V/A=ohm
I
VR
Resistivity and ResistanceJ = E/ρ where ρ is the resistivity
I = V/R
Norah Ali Al-moneef23
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.
I = V/R
and R = ρL/A is the resistance of the bar.
also called Ohm’s Law.
We know E = V/L so I/A = J = V/Lρ. Thus:
∆V = IR
Ohm’s Law, final
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). Norah Ali Al-moneef 24
Ohmic
Nonohmic
Norah Ali Al-moneef25
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.
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 flowincreases because of the wider path
R = L/A• If I change to a material with better
conductivity, the current flowincreases because charge carriers move better
• If I change the temperature, the current flowchangesNorah Ali Al-moneef26
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!
Norah Ali Al-moneef27
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)
R(Ohms)
28 Norah Ali Al-moneef
29 Norah Ali Al-moneef
V = IR = 2A x 3 = 6v
What voltage is required to produce 2a though a circuit with a 3 resistor.
V
3
I = 2a
Example:
30 Norah Ali Al-moneef
Resistance
I
V
I
V
Nonohmic device 2. Ohm’s Law
I
VR
IRV
Units: [ R ] = 1Ω = 1 V/A
constR Ohm’s Law:
R
VI
31 Norah Ali Al-moneef
Resistivity
L
AI
A
LR
L
AR
Definition:
Temperature dependence of resistivity
)(1 00 TTT
Example: What is the resistance of 1 m of nichrome wire of 2 mm diameter ?
)(
)(
00
000
TT
TTT
T
0
323
6 1018310
110
m
mm
A
LR
T
)( 00 TT 32 Norah Ali Al-moneef
Norah Ali Al-moneef33
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
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/secNorah Ali Al-moneef34
• 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.
Norah Ali Al-moneef35
Example
Norah Ali Al-moneef36
What is the current flow in a circuit with a voltage of 120 volts and a resistance of 0.23 ?
A 521.7= 0.23
V 120=
R
V=I
IR=V
ExampleWith the increase in the length of the wire, the current increases.A.TrueB.False
V = IR
4v = I x 2
I = 2A
A 4v battery is placed in a series circuit with a 2 resistor.
What is the total current that will flow through the circuit?
4v
2
I = ?
Example
37 Norah Ali Al-moneef
V= IR
V = 2A x 3
V = 6v
What voltage is required to produce 2A though a circuit with a 3 resistor.
V
3
I = 2A
Example
38 Norah Ali Al-moneef
V = IR
12 = 4 x R
R = 3
What resistance is required to limit the current to 4 A if a 12 V battery is in the circuit?
12v
3
I = 4a
Example
39 Norah Ali Al-moneef
Example
Norah Ali Al-moneef40
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-moneef41
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
5. 4B
A
Norah Ali Al-moneef42
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
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.
A. I1 < I2 B. I1 = I2 C. I1 > I2
?/
2/
2
12
2
12
12
12
II
rA
rr
LL
4/12 AAA
LR
11
1
2
22 8
4/
2R
A
L
A
LR
VI1 I2
112
2 8
1
8I
R
V
R
VI
Norah Ali Al-moneef43
Example
Voltage and Current Relationship for Linear Resistors
Norah Ali Al-moneef44
Voltage versus Current for a 10 ohm Resistor
00.10.20.30.40.50.6
0 1 2 3 4 5 6
Voltage (V)
Cu
rre
nt
(A)
Voltage and current are linear when resistance is held constant.
Which one of the following graphs correctly represents Ohm's law, where V is the voltage and I is the current?
(a)A(b)B(c)C(d)D(e)A and C
45 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: B
46 Norah Ali Al-moneef
Norah Ali Al-moneef47
27.4 Resistance and Temperature
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
Copper
)](1[ 00 TT
•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
Norah Ali Al-moneef48
• Some materials, when very cold, have a resistivity which abruptly drops to zero. Such materials are called superconductors.
)(1 00 TTT )( 000 TTT
T
0
)( 00 TT
0
T
• 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
• The temperature coefficient of resistivity can be expressed as.
Norah Ali Al-moneef49
• 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,
)(1 00 TTRR
where R0 and T0 are the resistance and temperature at a standard temperature, usually room temperature or 20o C.
Reminder: Battery as a “ski lift for charges”:
Ski lift raises objects to higher potential energy - flow may vary, but potential energy difference fixedBattery also fixed potential diff. , but current may vary
50 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: UK
• Collisions in the lattice structure transfer the energy to the lattice as thermal energy: KEth
• Thermal energy is a dissipative energy (i.e. can’t be recovered like mechanical energy.
51 Norah Ali Al-moneef
52 Norah Ali Al-moneef
the rate at which the system loses electric potential energy as the charge Q passes through the resistor:
V I V Q
) V Q ( dt
d
dt
d
dt
du
53 Norah Ali Al-moneef
Electrical Energy = Voltage x Electrical Current x Time Interval
energy = V x I (amps) x t (sec)E = V x I x t
•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
54 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)
How much power is used in a circuit which is 110 volts and has a current of 1.36 amps?
P = I V Power = (1.36 amps) (110 V) = 150 W
Example
55 Norah Ali Al-moneef
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.
Example
energy = Power X time
P = I V
P = (2A) (120 V) = 240 W
E = (240 W) (4 h) = 960Wh = 0.96 kWh
Electrical Energy = Voltage x Electrical Current x Time Interval
energy = V x I (amps) x t (sec)E = V x I x t
56 Norah Ali Al-moneef
a. 0.44 Ab. 2.25 Ac. 5 Ad. 36 A
ANS:
B
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?
57 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 58 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.
59 Norah Ali Al-moneef
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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.
Energy = 1.8 MJ = 1.8x106 Jt=10 minutes = 600 sPower = Energy / time p = 1.8x106 J / 600 =3 10 3 watt
example
61 Norah Ali Al-moneef
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 = V x IP = 120V x 7.9AP = 948 W
example
62 Norah Ali Al-moneef
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
example
63 Norah Ali Al-moneef
If a 500 watt speaker need 10 amps to operate, what is the voltage requirement?
example
VV
V
5010
500
64 Norah Ali Al-moneef
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 $/kWh= 0.04$
65 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 Jt=10 minutes = 600 s
66 Norah Ali Al-moneef
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.
Volts=V
Amps=I
Watts=P
IV=P
example
• Determine the power consumed by a resistor in a 12 volt system when the current is 2.1 amps.
W 25.2=V 12A x 2.1=IV=P
Norah Ali Al-moneef67
IVP
ELECTRIC POWER
When there is current in a circuit as a result of a voltage, the electricpower delivered to the circuit is:
SI Unit of Power: watt (W)
Many electrical devices are essentially resistors:
RIIRIP 2
R
VV
R
VP
2
Norah Ali Al-moneef68
Rank in order, from largest to smallest, the powers Pa to Pd dissipated in resistors a to d.
1. Pb > Pa = Pc = Pd 2. Pb = Pc > Pa > Pc 3. Pb = Pd > Pa > Pc 4. Pb > Pc > Pa > Pd 5. Pb > Pd > Pa > Pc
Norah Ali Al-moneef69
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?
$ =0.12 $
kWh x 800 W x
1 kW
1,000 W x 8 h =0.77 $
Norah Ali Al-moneef70
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.
kWh 43.2= W1,000
kW 1 x
month
day 30 x
day
h 8 W xA) 1.5 x V (120=(kWh)Energy
Norah Ali Al-moneef71
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Energy = Power x Time
E = (100 W) (300 s)
E = 30,000 J
E = 30 kJ
Norah Ali Al-moneef73
Rated for 4.2 kWUsed 20 h/month
Cost of 12 $ per kWh
Energy = Power x Time
= (4.2 kW) x (20 h)
= 84 kWh
Cost = Energy x rate per kWh
= (84 kWh) x ($0.12)
= $10.08
ExampleGiven 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-mDensity D = 8.9 E 3 kg/m3
molecular weight M = 63.546 g/moleAvogadro's # 6.022x10 23
electric charge e = 1.6x10-19 Cr = 5x10-4m radius, L=100m, t=60sec
74 Norah Ali Al-moneef
Equations: Answers:a) R= ρL/A, A=πr2 A=7.85x10-7, R=2.19
Ωb) V=IR I=V/R I=5.48 Ac) J=I/A J=6.977x106 A/m2
d) E = ρJ 0.12 V/me) n =D Na /M 8.434E28 e/m3
f) I = n q vd A vd = I/nqA = 5.17x10-4 m/s
g) I= dQ/dt Q = It = 329 C
75 Norah Ali Al-moneef
Example100 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 Kα = 0.0038/K
76 Norah Ali Al-moneef
Equations Answersa) P = IV I=P/V = 0.909 Ab) V = IR R=V/I = 121 Ωc) cost = ($0.1)(.1KW)(24 x 365) = $87.60d) Imax> Nmax I Nmax = 16e) Pmax = Nmax P Pmax = 1600Wf) R=Ro(1 + α (T-To)) = Ro =
7.13 Ω
77 Norah Ali Al-moneef
Norah Ali Al moneef78
OHM’S LAW FORMULAS
Current equalsvoltage dividedby resistance
Voltage equalscurrent multiplied
by resistance
Resistance equalsvoltage divided
by current
Find Current Find Voltage Find Resistance
summaryFind current: I=ΔQ/Δt I=nqAvd
Norah Ali Al moneef79
Quantity Unit of Measure
FunctionName NameSymbol Symbol
Voltage V, emfor E
Voltage VPressure whichmakes currentflow
Current I Ampere A Rate of flowof electrons
Resistance R Ohm Opposition tocurrent flow
Resistance related to physical parameters
The dimensions and geometry of the resistor as well as the particularmaterial used to construct a resistor influence its resistance. Theresistance is approximately given by
A
LR
Norah Ali Al moneef80
)(1 00 TTRR
)(1 00 TTT
IVP
RIP 2
R
VP
2
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