Chapter 18 Lecture Notes
Physics 2424 - Strauss
Formulas:
I =q/t
V = IR
R =
ρ
L/A
ρ
=
ρ
0
[1+
α
(T-T
P = IV = I 2 R = V
0
)]
2 /R
1.
EMF AND C URRENT
A battery provides a potential difference called an electromotive force o r
EMF. This is not truly a force, but rather a potential difference. It is called an emf for historical reasons. There is a gravitational analogy to this process. Raising the potential in a circuit (what the battery does) is like raising a ball to a higher potential. The battery provides a potential which can cause the charges to flow around a circuit. So in a circuit, when you connect wires to a battery, you get charges moving. If I stand in one place and watch the charge that goes by per unit time, I get a current.
- +
I = q/t
SI Units of current is Coulombs/Second = Amps
Standard current is defined as the direction positive charge would move around the circuit.
Problem: Suppose I have one trillion electrons flowing past a point in one half second. What is the current?
2.
O HM ’ S L AW
Resistance in a circuit is like friction on an object moving. It impedes the flow of charge. Ohm’s law tells us how much the resistance impedes the flow.
V = IR.
The SI unit of resistance is given by ohms, and we use the Greek letter omega,
Ω
.
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What does V = IR this mean? For a given voltage, if I increase the resistance, the current drops. This is a well known relationship that you may be expected to know for MCAT, etc. However, Ohm’s law is not truly a fundamental law like Newton’s second law,
Σ
F = ma. Ohm’s law is a relationship that works only for metals in a limited temperature range.
We draw a resistor in a circuit with a jagged line like this.
Battery
Resistor
Problem: A hair dryer draws 13 A when plugged into a 120 V line. What is the resistance of the hair dryer?
If two resistance are linked together with a single wire as in the diagram, the current through each resistor is the same. The potential drop across each resistor is equal to the current times the resistance of that resistor. A gravitational analogy is that the resistors are like water wheels which can use the power from the current flow.
V
1
= IR
1
V
2
= IR
2
3.
R ESISTANCE AND R ESISTIVITY
All material has a resistivity. For some it is very low (conductors), for others it is quite high (insulators). For semi-conductors it is somewhere in between. See Table 18-1.
The resistivity is a property of the material. The resistance is a property of the resistivity, plus how the material is constructed. For many materials we find
R=
ρ
L/A where L = length, A = cross sectional area and
ρ
=resistivity. If I want to have less resistance, I can get a material with a lower resistivity, or I can use the same material and get a larger cross sectional area. Wire that has
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to carry a lot of current or is used across a large potential difference is often very thick so that it has a low resistance.
Problem: If an aluminum wire and a copper wire each have a length of 10 meters, and aluminum has a cross sectional area of 1 mm and copper 0.5
mm, (a) what is the ratio of the resistance of the copper to the aluminum?
ρ
for copper is 1.72
×
10 -8
Ω
-m and for aluminum 2.82
×
10 -8
Ω
-m.
For many metals resistivity is a function of temperature and is given by
ρ
=
ρ
0
[1+
α
(T-T
0
)].
Then for material with the same L and A,
R = R
0
[1+
α
(T-T
0
)].
We will do a problems with this in a moment.
4.
E LECTRICAL P OWER
Recall that Energy/time = power and that electrical energy = qV. Then
Energy/t = qV/t = IV = P
P = IV, and since V=IR,
P = IV = I 2 R = V 2 /R
As we have said before, the SI unit of power is watts. Energy is Power multiplied by time and is often expressed in kw-hrs (kilowatt-hours) although this is not an SI unit. So if a hair dryer uses 1000 watts for 10 minutes, it uses
(1kw)(10 minutes)(1 hour/60 minutes) = .17 kw-hours at about $0.1 per kilowatt hour or about 2 cents to run.
Problem: How much does it cost to keep your computer on all the time for one month?
Problem/Demonstration: A light bulb filament is made of tungsten which has a coefficient of resistivity (
α
) of 0.0045 (C
°
) -1 . At room temperature of 20
°
C the filament has a resistance of 10
Ω
.
(a) What is the average power of the light bulb immediately after it has been plugged into a 120 V ac outlet?
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(b) After a very brief time, the light bulb filament has changed temperature and is glowing brightly, the rms current is measured to be
0.833 amp. What is the resistance of the light bulb now?
(c) What is the average power of the light bulb when it is glowing brightly as in part (b)?
(d) What is the temperature of the filament when it is glowing brightly?
5.
A LTERNATING C URRENTS
An alternating current is a current which changes direction, going first one way and then the other.
If I was to plot the value of the current, as a function of time, I would see that it is changing in a sinusoidal way.
t
This curve is given by the equation
I=I
0
sin 2
π ft
I
0
is the maximum height of the current, and f is the frequency, which is how long it takes for the current to make one complete cycle.
The voltage follows the same curve and is given by
V=V
0
sin 2
π
ft. where V
0
is the maximum voltage.
Suppose I want to tell you what the voltage is for alternating current.
What should I use. I can’t really use the average voltage, because that is zero. I could use the maximum voltage. But what is actually done is that when I am describing an alternating circuit I use Root Mean Square (RMS) values. RMS is always related to the maximum for sinusoidal curves by the formula
I
RMS
= I
0
/
√
2 or for voltage V
RMS
= V
0
/
√
2
So if I talk about average current or average voltage (or just use the terminology of voltage or current) for an alternating current, I mean RMS values. If I talk about “peak” voltage or current, I mean their maximum values.
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In the U.S. f=60 Hz (cycles/second), andV
RMS
= 120 V.
All the formulas we have learned for dc currents work for ac currents, either their average values or their RMS values. So
V
RMS
=I
RMS
R or V
0
=I
0
R and
P = V
RMS
I = I 2
RMS
R
Problem: A vacuum cleaner has a rating of 11 amps. (Ratings are RMS
I currents). What is the maximum power consumed?
V
0
0
=
√
2I
RMS
=
√
2V
=
√
2(11 A) = 15.6 A
RMS
=
√
2(120 V) = 170 V
P
0
= V
0
I
0
= (170 V)(15.6 A) = 2600 W
6.
D RIFT V ELOCITY
Electrons in a wire do not move individually as fast as the electric current moves through the wire. They actually bounce around and drift at a much slower velocity than the current in the wire. Electricity in a wire usually moves almost the speed of light in a vacuum, but the electrons drift at speeds that are thousands of times slower.
We find that the current through a wire is related to the drift velocity by
I =
∆
Q/
∆
t = nev d where n is the number of free electrons per unit volume that can drift, e is the electric charge, and v d
is the average drift velocity of each electron.
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