Question

The corroded contacts in a lightbulb socket have 8.0 ? total resistance. How much actual power is dissipated by a 75 W (120 V) lightbulb screwed into this socket?

Answer 1

Assume that the resistance of the light bulb doesn't change with temperature (which is not true, but makes the problem simpler). Assume it is a 120VAC bulb, and that the voltage source is 120VAC.

The resistance of the light bulb is:
Rb = V^2 / P = (120V^2) / 75W = 192?

The contact resistance is:
Rc = 8.0?

The total circuit resistance is:
Rt = Rb + Rc = 192? + 8.0? = 200?

The current in the circuit is:
I = V / Rt = 120V / 200 = 0.6A

The power dissipated by the bulb is:
Pb = I^2 x Rb =( 0.6)²A x 200 ?
Pb = 72W