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 DV DV = voltage = electric potential drop I = R I = current R = resistance Good conductors have low resistance. Unit of resistance : V / A = ohm (W) 4

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

Poll Question If you want to increase the current through a resistor, you need to Increase the resistance or voltage. Decrease the resistance or voltage. Increase the resistance or decrease the voltage. Decrease the resistance or increase the voltage.

Ohm’s Law Rearranged If you know two, you can find the third. DV DV DV = IR R = R I Good conductors have low resistance. I = current DV = potential R = resistance

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: 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

dissipated by a resistor Power dissipated by a resistor

Electric Power Power = VI Potential is energy per charge: V = DE / q Current is charge per time: I = q /Dt So, (potential times current) = (energy per time) = power Good conductors have low resistance. Power = VI

Group Work Power P = VI and V = IR. Using these, show that: P = I2R P = V2/R