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

R

I = currentV = voltage = electric potential dropR = 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

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.

Ohm’s Law Rearranged

I = V

R

I = currentV = potentialR = resistance

If you know two, you can find the third.

R = V

IV = IR

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

Resistivity

For current through a cylinder:

• Longer L greater R.• Greater A smaller R.• More resistive material bigger R.

LA

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 3.5 105 m

Quartz 75 1016 m

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?

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 PowerPotential is energy per charge:

V = E / qCurrent is charge per time:

I = q /tSo, (potential times current) =

(energy per time) = power

Power = VI

Group Work

Power P = VI and V = IR. Using these, show that:

a. P = I2Rb. P = V2/R