A light bulb is wired in series with a 142 resistor, and they are connected across a 120 V source. The power delivered to the light bulb is 22.3 W. What are the two possible resistances of the light bulb?

Answer 1

P = 142I^2 = Power to the resistor.
P = 22.3W. = Power to bulb.
P = 120I = Power from source.
The total power delivered is = to the
power from the source:

142I^2 + 22.3 = 120I,
142I^2 - 120I + 22.3 = 0,
Use Quadratic Formula and get:
I = 0.56914A; I = 0.27593A.

P = (0.56914)^2 * R = 22.3,
R = 22.3 / (0.56914)^2 = 68.8 Ohms.

R = 22.3 / (0.27593)^2 = 292.9 Ohms.