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Resistance
and resistivity
Current
• Current is sort of a vector
• Direction is constrained by conductor
• Restricted to forward or backward (+ or –)
Resistance
• Current does not flow unhindered
• Electrical resistance is analogous to
friction or drag
• Expressed as potential needed to maintain
a current
Ohm’s Law
I=
DV
R
I = current
DV = voltage = electric potential drop
R = resistance
Unit of resistance : V / A = ohm (W)
Voltage Causes Current
• Potential drop is the cause.
• Current is the effect.
• Resistance reduces the effect of potential.
Does it Work?
• Approximation of varying utility:
R is independent of DV and I
• When true, the material is ohmic
Circuit symbol
• zigzag
• straight line is a perfect conductor
Poll Question
If you want to increase the current through a
resistor, you need to
A. Increase the resistance or voltage.
B. Decrease the resistance or voltage.
C. Increase the resistance or decrease the
voltage.
D. Decrease the resistance or increase the
voltage.
Ohm’s Law Rearranged
If you know two, you can find the third.
DV
I=
R
DV = IR
I = current
DV = potential
R = resistance
R=
DV
I
Example
A 1.5-V battery powers a light bulb with a
resistance of 9 W. What is the current
through the bulb?
Ohm’s Law I = V / R
V = 1.5 V; R = 9 W
I = (1.5 V ) / (9 V/A) = 1/6 A
Resistivity
For current through a cylinder:
A
L
• Longer L  greater R.
• Greater A  smaller R.
• More resistive material  bigger R.
Resistivity
• R = r L/A
• r is Resistivity
• Unit: ohm·meter = Wm
• More or less constant depending on
material, conditions
Resistivity
• Intensive quantity
• Does not depend on the amount of
material, only its conditions
• Predictive value when mostly constant
(ohmic)
Resistivities vary widely
Silver
1.59  10–8 Wm
Graphite
3.5  105 Wm
Quartz
75  1016 Wm
Example
The resistivity of copper is 1.710–8 Wm.
What is the resistance of a 100-km length of
copper wire that is 1/4” in diameter?
Classes of Conductors
• How resistivity changes with temperature
a = temperature coefficient of resistivity
Classes of Conductors
• How resistivity changes with temperature
Power
dissipated by a resistor
Electric Power
Potential is energy per charge:
V = DE / q
Current is charge per time:
I = q /Dt
So, (potential times current) =
(energy per time) = power
Power = VI
Group Work
Power P = VI and V = IR. Using these,
show that:
a. P = I2R
b. P = V2/R