Resistance and Power Probelm (again)

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SUMMARY

The discussion focuses on calculating the possible resistances of a light bulb connected in series with a 132-ohm resistor across a 120 V source, delivering 23.4 W of power. The relevant equations include Ohm's Law (V=I/R) and the power formula (P=(I^2)(R)). The initial attempts at solving the problem were incorrect, highlighting the need to consider the total resistance in the circuit, which includes both the resistor and the light bulb.

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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!
 
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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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