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 ?
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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