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.
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?
3 answers
Joe, check your 1-24-11, 5:19pm Post.
((R1)(V^2))/((R1+R2)^2)=Power (watts)
in a ti89: solve(((x*120^2)/(X+133)^2)=22.5,x)
in a ti89: solve(((x*120^2)/(X+133)^2)=22.5,x)