Resistance and Power Probelm (again)

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The discussion focuses on calculating the resistance of a light bulb connected in series with a 132-ohm resistor across a 120 V source, delivering 23.4 W of power. Initial attempts to solve the problem using the equations V=I/R and P=(I^2)(R) were incorrect. The current through both resistors must be the same, and the total resistance can be expressed as the sum of the individual resistances. The correct approach involves using the power equation and the relationship between voltage, current, and resistance in series circuits. Ultimately, the problem requires determining both the lower and higher possible resistances of 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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