A light bulb is wired in series with a 133-Ω resistor, and they are connected across a 120-V source. The power delivered to the light bulb is 22.5 W. What are the two possible resistances of the light bulb?

Answer 1

Power to bulb= I^2 R

but I= 120/(R+133)

22.5= 120^2/(R+133)^2 * R

solve for R. Notice it is a quadratic.

Answer 2

Joe, check your 1-24-11, 5:19pm Post.

Answer 3

((R1)(V^2))/((R1+R2)^2)=Power (watts)

in a ti89: solve(((x*120^2)/(X+133)^2)=22.5,x)