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 = VV R I = current  V = voltage = electric potential drop R= resistance Unit of resistance : V / A = ohm (  )

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  V 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 I = VV R I = current  V = potential R= resistance If you know two, you can find the third. R = VV I  V = I R

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

Resistivity For current through a cylinder: Longer L  greater R. Greater A  smaller R. More resistive material  bigger R. L A

Resistivity R =  L/A  is Resistivity Unit: ohm·meter =  m 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  m Graphite 3.5  10 5  m Quartz 75   m

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?

Classes of Conductors How resistivity changes with temperature  = 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 =  E / q Current is charge per time: I = q /  t So, (potential times current) = (energy per time) = power Power = VI

Group Work Power P = VI and V = IR. Using these, show that: a. P = I 2 R b. P = V 2 /R