Ohm's Law ---------------------------- Joseph F. Alward, PhD     Department of  Physics   University of the Pacific

Georg Simon Ohm (1787-1854)

   

"Electromotive":  relating to something which moves electricity "Electromotive Force (EMF)":  is not a force       The source of an EMF is any device or mechanism which maintains a potential difference between two points.     EMF is measured in volts, not newtons.  EMF is a potential difference, not a force.

  Batteries Provide Electromotive Force      

 
   Inside the Dry Cell Battery                                         

The source of the emf in this  case is chemical energy.

A battery ripped open.

The electric field of a battery.

  Defining Current                                      

Each second, 15 coulombs of charge cross the plane.  The current is I = 15 amperes (15 A). One ampere is one coulomb per second.

Andre M. Ampere  1775-1836

 Conventional Current Direction                                        

In a wire, electrons are the only charged particles moving in an electrical current.

Conventional current is the direction along which imaginary positive charge carriers may be imagined to flow. At the left, negative charges moving to the left is equivalent to positive charges moving to the right.

   Electron Current:  Microscopic Picture                     

Positive terminal of battery (not shown) is at the right. ---------------------------------------- Electrons collide with metal ions as they move toward the positive terminal. ----------------------------------------- Electron current is to the right, but conventional current I is to the left.

   Drift Speed of Electrons in a Wire                                                .

Speed between collisions is  106 m/s.  Because of direction  changes, the electron "drifts"  slowly down the wire:  Drift speed = 0.01 cm/sec  ----------------------------------------------  If the drift speed is so small, why  does the light bulb go on almost  instantaneously when the switch is  thrown?

    Ohm's Law                                                                                                                                      

V = 10 volts (V)          R = 4 ohms  (W)   I = 10/4 = 2.5 amperes (A)

Filament provides resistance  to the flow of electrons.

Georg Simon Ohm (1787-1854)

   Calculating V from Ohm's Law        

Ohm's Law:  V = IR -------------------------- V = (6A)(3 W)     = 18 volts (V) -------------------------- Battery is assumed to be ideal (have zero internal resistance).

    Calculating I from Ohm's Law  

Ohm's Law:  I = V/R --------------------------- I = (12 V) /(3 W)   = 4 amperes (A) -------------------------- Battery is assumed to be ideal (have zero internal resistance).

     Calculating R from Ohm's Law    

Ohm's Law: R = V/I -------------------------- R = (36 V) /(6 A)     = 6 W -------------------------- Battery is assumed to be ideal (have zero internal resistance).

  Resistors and Resistance                                         

  Resistor Color Code                                       

Orange (3)-Black (0)-Brown (1)              30 x 101= 300 W

Resistor Color Code Black     0              Blue       6 Brown   1              Violet     7 Red       2              Gray       8 Orange  3              White     9 Yellow   4              Gold      5 % Green     5             Silver   10 %

   Resistance and Resistivity                                                         

A cylindrical resistor of  length L and area A

R = rL/A r = resistivity Units:  ohms-meter (W-m) ------------------------------------ Example:  Aluminum wire r = 3 x 10-8 W-m L = 20 m     A = 5 x 10-6 m2 R = 3 x 10-8 (20)/ 5 x 10-6      = 0.12 W

Table of Resistivities       x 10-8 W-m ------------------------------- Copper           1.72 Aluminum        2.82 Gold                2.44 Tungsten            5.6 Carbon           3500 Wood         3 x 1010 Rubber       1 x 1024

   Resistance of Wires                                                                           

R = rL/A -------------------------------- The longer the wire, the larger its resistance. The larger the area, the smaller the resistance.

For wires with a circular cross section, A = pr2

R = rL/A    = rL/pr2 --------------------------------- The thicker the wire, the less its resistance.

  Resistance and Cross-Sectional Area A       

Which has the greater resistance:  The copper cord, or the tungsten filament? --------------------------------------- R = rL/A                     A = pr2

If the length and diameter of a copper wire are each doubled, what happens to the resistance?

   Medical Application                                                 

Measuring venous thrombosis (clotting). Greek:  plethysmos, increase

Pressure shuts off venous blood flow back to heart, but permits arterial flow.  Cross sectional area of calf calf increases, causing the resistance to decrease. When pressure is released, resistance rises quickly if venous blood clotting is low.

 Temperature Coefficient of Resistivity    

R = rL / A Resistivity r is a function of temperature ----------------------------------------------   r0 = resistivity at temperature T0     r      = r0 [1 + a(T - T0)]        = resistivity at  temperature T a =  temperature coefficient         of resistivity

  Temperature Coefficient of Resistivity Example            

Problem:  The resistance R0 of a length of wire is 20 W at 300 C.  The temperature coeffficient of resisitivity is a = 7 x 10-3 W/C. What will be the resistance R at T = 500 C?

Solution: The resistivity increases by a factor of   r/r0 = [1 + a(T - T0)]         = [1 + 7 x 10-3(500-300)]         = 2.4 New resistance will be 2.4 times 20 W, or 48 W.

  Measuring Current and Voltage                                      

Voltmeter measures the potential difference across the filament.   By Ohm's Law, this voltage is V = IR. Ammeter measures current I through the battery, the filament, and itself.

This arrangement measures the voltage across the battery.

 Power Dissipated in a Resistor:  Three Ways to Calculate It         

P = I2R             = (0.5)2 100             = 25 watts

P = (V/R)2 R             = V2/R             = (50)2 /100             = 2500/100             = 25 watts

P = I2(V/I)            =  IV            = (0.50)50            = 25 watts

   Power Dissipated by a Light Bulb                        
    (Conceptual Question 2)

When the light bulb is turned on, its tungsten filament heats up.   Assuming the temperature coefficient of tungsten is positive, what happens to the power as time passes? --------------------------------------------- P = I2R P = V2/R P = IV

  Power Dissipated by a Light Bulb                                       
    (Conceptual Question 6)

Which bulb has the greater filament resistance? P = V2/R V and R are independently fixed. Can't use P = I2R because I will not be the same for each bulb; I depends on R.

  Output Power of a Battery                          

60 watts is output by this battery (assumed to be ideal).   60 joules of energy per second will appear as heat and light in the bulb.