Homework Help: physics

Posted by joe on Tuesday, January 25, 2011 at 2:06pm.

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.

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