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

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. 
Joe, check your 12411, 5:19pm Post.

((R1)(V^2))/((R1+R2)^2)=Power (watts)
in a ti89: solve(((x*120^2)/(X+133)^2)=22.5,x)