Download - Resistance
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Resistance
and resistivity
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Current
• Current is sort of a vector• Direction is constrained by conductor• Restricted to forward or backward (+ or –)
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Resistance
• Current does not flow unhindered• Electrical resistance is analogous to
friction or drag• Expressed as potential needed to maintain
a current
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Ohm’s Law
I = V
R
I = currentV = voltage = electric potential dropR = resistance
Unit of resistance : V / A = ohm ()
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Voltage Causes Current
• Potential drop is the cause. • Current is the effect.• Resistance reduces the effect of potential.
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Does it Work?
• Approximation of varying utility:R is independent of V and I
• When true, the material is ohmic
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Poll Question
If you want to increase the current through a resistor, you need toA. 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.
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Ohm’s Law Rearranged
I = V
R
I = currentV = potentialR = resistance
If you know two, you can find the third.
R = V
IV = IR
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Example
A 1.5-V battery powers a light bulb with a resistance of 9 . What is the current through the bulb?Ohm’s Law I = V / RV = 1.5 V; R = 9 I = (1.5 V ) / (9 V/A) = 1/6 A
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Resistivity
For current through a cylinder:
• Longer L greater R.• Greater A smaller R.• More resistive material bigger R.
LA
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Resistivity
• R = L/A• is Resistivity• Unit: ohm·meter = m
• More or less constant depending on material, conditions
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Resistivity
• Intensive quantity• Does not depend on the amount of
material, only its conditions• Predictive value when mostly constant
(ohmic)
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Resistivities vary widely
Silver 1.59 10–8 m
Graphite 3.5 105 m
Quartz 75 1016 m
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Example
The resistivity of copper is 1.710–8 m. What is the resistance of a 100-km length of copper wire that is 1/4” in diameter?
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Classes of Conductors• How resistivity changes with temperature
= temperature coefficient of resistivity
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Classes of Conductors• How resistivity changes with temperature
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Power
dissipated by a resistor
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Electric PowerPotential is energy per charge:
V = E / qCurrent is charge per time:
I = q /tSo, (potential times current) =
(energy per time) = power
Power = VI
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Group Work
Power P = VI and V = IR. Using these, show that:
a. P = I2Rb. P = V2/R