Chapter 18 Lecture Notes

Formulas: 

I =q/t
V = IR
R = rL/A              
r = r_o[1+a(T-T_o)]
P = IV = I²R = V²/R

Demonstrations:

Light Bulb and Power.

EMF and Current

     Alessandro Volta (1800): Invented the electric battery and was able to produce a steady flow of electric charge. A battery provides a potential difference called an electromotive force or 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 that can cause the charges to flow around a circuit.  So in a circuit, where 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? I = q/t = (1´1012 electrons)(1.60´10-19 C/electron)/0.5 s = 3.2´10-7A   = .32 mA

Ohm's Law

     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 the ohms, and we use the Greek letter omega, W. 

     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 SF = 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.

PROBLEM: A hair dryer draws 13 Amps when plugged into a 120 V line. What is the resistance of the hair dryer? R = (Voltage)/(Current)= 120 V/13 A = 9.2 W.

     If two resistances 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.

V₁ = IR₁      V₂ = IR₂

Resistance and Resistivity

     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.   

     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=rL/A

where L = length, A = cross sectional area and r= 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 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 10m, and aluminum has a cross sectional area of 1mm and copper 0.5mm, what is the ratio of the resistance of the copper to the aluminum? r for copper is 1.72´10-8W-m and for aluminum 2.82 ´ 10-8W-m.  What is the ratio of the resistance? ratio = R(copper)/R(aluminum) = (rCLC/AC)/ (rALA/AA) = rCAA/rAAC (1.72´10-8W-m)(.001) / (2.82´10-8W-m)(.0005) = 1.22

For many metals resistivity is a function of temperature and is given by

r = r_o[1+a(T-T_o)].

Then for material with the same L and A,

R = R_o[1+a(T-T_o)].

We will do problems with this in a moment. 

Electrical Power

Recall that Energy/time = power and that Energy = qV. 
Then;

(Energy)/(time) = qV/t =V(q/t)= IV = P

P = IV, and since V=IR,

P = IV  = I²R = V²/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, 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? A computer plus screen draws about 4 amps at 120 volts.  P = IV = 480 watts. So we have $ = (.48kW)(30days)(24 hours/day)($.10/kW-hr) $ = $34.60

PROBLEM/DEMONSTRATION:  A light bulb filament is made of tungsten which has a coefficient of resistivity (a) of 0.0045 (C°)-1.  At room temperature of 20° C the filament has a resistance of 10 W.  (a) What is the average power of the light bulb immediately after it has been plugged into a 120 V ac outlet? P = V2/R = (120) 2/10 W = 1,440 W (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? R = V/I = (120 V)/(0.833 A) = 144 W (c) What is the average power of the light bulb when it is glowing brightly as in part (b)? P = V2/R = (120) 2/144 W = 100 W (d) What is the temperature of the filament when it is glowing brightly? R = Ro[1+a(T-To)] R/Ro = 1+a(T-To) R/Ro - 1 =a(T-To) R/aRo - 1/a =T –To T = R/aRo - 1/a  +To T  = (144 W)/(.0045)(10W) - 1/0.0045 + 20° = 3000° C

Alternating Currents

     An alternating current is a current that changes direction, going first one way and then the other in a circuit. 

If I were to plot the value of the current, as a function of time, I would see that it is changing in a sinusoidal way. 

This curve is given by the equation I=I_osin(2pft) 

I₀ 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_osin(2pft)

where V_o 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_o/Ö2 or for voltage V_RMS = V_o/Ö2

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.

In the U.S. f=60 Hz (cycles/second), and V_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_RMSR or V_o =I_oR

and   = V_RMSI =I²_RMSR

PROBLEM:  A vacuum cleaner has a rating of 11 amps.  (Ratings are RMS currents).  What is the maximum power consumed? Io= Ö2IRMS = Ö2(11 A) = 15.6 A Vo=Ö2VRMS = Ö2(120 V) = 170 V Po=VoIo= (170 V)(15.6 A) = 2600 W

Drift Velocity

     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 drift velocity by:

I = DQ/Dt = 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.