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Published byAlfred Todd Modified over 9 years ago
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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 = VV R I = current V = voltage = electric potential drop R= 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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Circuit symbol zigzag straight line is a perfect conductor
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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.
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Ohm’s Law Rearranged I = VV R I = current V = potential R= resistance If you know two, you can find the third. R = VV I V = I R
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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 / R V = 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. L A
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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 10 5 m Quartz 75 10 16 m
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Example The resistivity of copper is 1.7 10 – 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 Power Potential is energy per charge: V = E / q Current is charge per time: I = q / t So, (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 = I 2 R b. P = V 2 /R
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