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Posted by 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 - , 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 - , Wednesday, January 26, 2011 at 3:46pm

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

  • physics - , Wednesday, February 11, 2015 at 11:57pm

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

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

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