Resistance and Power Probelm (again)

  • #1
A light bulb is wired in series with a 132 resistor, and they are connected across a 120 V source. The power delivered to the light bulb is 23.4 W. What are the two possible resistances of the light bulb?

a. lower value
b. higher value

relevant equations V=I/R and P=(I^2)(R)

attempt at soultion:

23.4=I^2(132)
=.4210

Wrong!


V=I/R
120=I/132
=15840

Wrong!
 

Answers and Replies

  • #2
mgb_phys
Science Advisor
Homework Helper
7,819
13
You have two resistors in series (the 'resistor' and the bulb )
What do you know about the current through these two resistors?
What equations do you know for the total resistance ?
 

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