How Do You Calculate the Resistance of a Light Bulb in a Series Circuit?

AI Thread Summary
To calculate the resistance of a light bulb in a series circuit with a 144 Ohm resistor and a 120.0-V source, the power delivered to the bulb is given as 23.4W. The relationship between power, current, and resistance is used, leading to the equation I = (V / (Rb + Rr))^2 x Rb. The discussion focuses on rearranging the equation to fit the standard quadratic form ax^2 + bx + c = 0. Participants suggest expanding and simplifying the equation to isolate Rb, emphasizing the need for algebraic manipulation to solve for the bulb's resistance. The conversation highlights the importance of careful mathematical steps to arrive at the correct solution.
rcrx
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Homework Statement


A light bulb is wired in series with a 144Ohm resistor, and they are connected across a 120.0-V source. The power delivered to the light bulb is 23.4W. What is ther resistance of the light bulb? Note that there are two possible answers.


Homework Equations


P = I^2R and I = V/R


The Attempt at a Solution


Solve for I to find Rbulb. I = (V / Rb + Rr)^2 x Rb.

Rearranged 23.4(Rb^2 + 288Rb + 144^2) = 120Rb

This is where I'm stuck, I need to rearrange that so I can get ax^2 + bx + c = 0

My math skills are a little weak since I have been out of school for a few years.

Thanks a lot!:smile:
 
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Expand the brackets and subtract 120Rb from each side...
 
So, pardon me, but I'd get the following?

23.4Rb^2 + (23.4)288Rb + (23.4)144^2 - 120Rb = 0
 
rcrx said:
So, pardon me, but I'd get the following?

23.4Rb^2 + (23.4)288Rb + (23.4)144^2 - 120Rb = 0

Yes. Now, since you are dealing with numbers, you can simplify 23.4*288 and subtract 120 to obtain an equation in your required form.
 

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