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?
physics - bobpursley, Tuesday, January 25, 2011 at 3:12pm
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.
physics - Henry, Wednesday, January 26, 2011 at 3:46pm
Joe, check your 1-24-11, 5:19pm Post.
physics - Anonymous, Wednesday, February 11, 2015 at 11:57pm
in a ti89: solve(((x*120^2)/(X+133)^2)=22.5,x)