Current and Resistance

currmetalAni.gif (94996 bytes)

Current flow through a metal.
(Note: The electrons are colored differently to make tracking easier.)

currvacuumAni.gif (18091 bytes) Current flow through a vacuum.
(Note: The electrons are colored differently to make tracking easier.)

Previously Asked Questions

Q: "    A 9.0 volt battery is connected across a light bulb (R = 3.0 omega.gif (841 bytes))
a. How many electrons pass through the resistor in one minute?
b. How many joules of energy are generated in one minute?"

A:
a
.  The resistance, R, of a conductor is the voltage, V, applied across the conductor divided by the current, I, through the conductor:

R = V / I .

Current, delta - delta.gif (839 bytes)I, is the time rate of flow of charge, delta - delta.gif (839 bytes)Q, through an area:

I = delta - delta.gif (839 bytes)Q / delta - delta.gif (839 bytes)t.

The amount of charge, delta - delta.gif (839 bytes)Q, is simply a number of particles, z, multiplied by the charge per particle, q:

(delta - delta.gif (839 bytes)Q = z q)

Using these facts we can derive and equation for the number of particles, z, passing through a conductor with resistance, R, in an amount of time, t:

R = V / I      ->     I = V / R =

I = V / R = delta - delta.gif (839 bytes)Q / delta - delta.gif (839 bytes)t  = ( z q ) / ( delta - delta.gif (839 bytes)t )  =  V / R       ->

     z = ( V * delta - delta.gif (839 bytes)t ) / (R * q)

Now that we have a formula we can solve the problem.
V = 9.0 V
t = 1 minute = 60 s
R = 3.0
omega.gif (841 bytes)
q = the charge of an electron = 1.60217733 x 10-19 C / electron

Solve for z:

z  = ( 9.0 V * 60 sec) / (3.0 omega.gif (841 bytes) * -1.6021917 x 10^-19 C / electron)
    = 1.123 x 1022 (V sec)/ (
omega.gif (841 bytes) C / electron)
    = 1.123 x 1022 ( (J/C) sec) / ( (J sec/C2) C / electron)
    = 1.123 x 1022 ( J sec / C) / ( (J sec/ C) / electron)
    = 1.123 x 1022 electrons

1.123 x 1022 electrons pass through the resistor in one minute.

b.  The power, P, loss in a conductor is the square of the current, I 2,   multiplied by the resistance, R:

P = I 2 R.

Using the fact that I = V / R we can rewrite this as:

P = V 2 / R .

Power, P, is energy, E, divided by time, t:

(P = E / t ).

Rewrite the equations:

P = V 2 / R as  E / t = V 2 / R.

This yields

E = V 2 t / R.

Now we can find the energy lost through the light bulb.

E  = V 2 t / R
    = ( 9.0 V)2 * 60 sec / 3.0
omega.gif (841 bytes)
    = 4860 V2 sec / 3.0
omega.gif (841 bytes)
    = 1620 V2 sec /
omega.gif (841 bytes)
    = 1620 ((J/C)2 sec) / ( J sec/C2)
    = 1620 (J2 sec / C2) / (J sec/C2)
    = 1620 J

The energy dissipated by the light bulb in 1 minute is 1620 Joules.

Q:     Do all conductors obey Ohm's Law?

A:    Yes, conductors are also called "Ohmic materials."

Q:     Two light bulbs use the standard 110 V light socket.  One is rated at 60 Watts and the other at 100 Watts.
a)  Which bulb carries the greater current?
b)  Which bulb has a higher resistance?
c)  Which bulb is brighter?

A:    The formula for power is P = I * V = (V/R) * V = V^2/R.
a)  Power is proportional to current, therefore the bulb with higher power carries greater current.
b)  Power is inversely proportional to resistance, therefore the bulb with lower power has a higher resistance.

r_squared.gif (10451 bytes) c)  Brightness is synonymous with intensity, I, and intensity is defined as "the portion of the power emitted by a point source, that reaches a unit area at a distance r from the source."  The total power of the source is uniformly spread over the area of sphere with radius r.   This is the very important and very well known law usually referred to in physics as "the one over r squared law."  It results that the bulb with higher power also provides higher intensity (or brightness) at a distance from it.

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References

Definitions

Semiconductors: materials with few conduction electrons but with available conduction-level states that are close, in energy to their valance bands.  These materials become conductors when they are doped with other atoms that contribute electrons to the conduction band.

Superconductors: materials that lose all electrical resistance at low temperatures.

Facts

The SI unit of electric current is the ampere (A):   1 A = 1 C/s.
The SI unit of resistivity is the ohm-meter (omega.gif (841 bytes) m).

Ohm's Law:  for many materials (including most metals), the ratio of the current density to the electric field is a constant, sigma2.gif (831 bytes), that is independent of the electric field producing the current.

Equations

Electric current (i) 27-1.gif (146 bytes)
Current density (J) 27-4.gif (170 bytes)
Drift speed of the charge carriers 27-7.gif (149 bytes)
Resistance of a conductor R = V / i
Resistivity (rho2.gif (832 bytes)) and conductivity (sigma2.gif (831 bytes)) 27-12.gif (168 bytes)
Ohm's Law J = sigma2.gif (831 bytes) E
Resistance of a conductive wire of length L and uniform cross section 27-16.gif (148 bytes)
Change of resistivity (rho2.gif (832 bytes)) with temperature rho2.gif (832 bytes) - rho2.gif (832 bytes)0 = rho2.gif (832 bytes)0 alpha2.gif (828 bytes) (T - T0)
Resistivity of metals 27-20.gif (168 bytes)
Power (rate of energy transfer) P = iV
Resistive dissipation 27-22.gif (208 bytes)

 

Color Code For Resistors

  Number Multiplier Tolerance (%)
Black 0 1  
Brown 1 101  
Red 2 102  
Orange 3 103  
Yellow 4 104  
Green 5 105  
Blue 6 106  
Violet 7 107  
Gray 8 108  
White 9 109  
Gold   10-1 5%
Silver   10-2 10%
Colorless     20%

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List of Topics

Measurements Electric Potential Magnetism Electrical Circuits (AC) Optical Instruments: Mirrors and Lenses
Electrostatics Capacitance Sources of Magnetic Fields Maxwell's Equations Interference
Electric Fields Current and Resistance Magnetism in Matter Electromagnetic Waves Diffraction
Electric Flux Electrical Circuits (DC) Electromagnetic Induction Interaction of Radiation with Matter: Reflection, Refraction, Polarization  

 

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