# What is the internal resistance of the power supply

A 76 W light bulb with a resistance of 179 Ω is connected to a power supply with a 117 V. What is the internal resistance of the power supply? Express the answer with one decimal place.

So my attempt has been this but I'm not exactly sure whether or not so just making sure this would be the answer.

[117/(179+r)]^2(179)=76

sqrt(76/179)=0.652 A
(0.652 x 179)= 116.708
117-116.708=0.364
0.364/0.652=0.558 ohm

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I got the same answer as you to 2 significant figures

I got the same answer as you to 2 significant figures
So this would be the correct method to solve this question?

You got it right so you must have known what you were doing.
I prefer to break the question down into separate parts rather than produce 1 equation. What I did was:
76W, 179Ω so V^2/179 = 76
which gives V = 116.64 volts across the lamp, this means 117-116.64 volts = 0.36Vacross the internal resistance.
Also I^2.R =76W so this gives I = 0.652A flowing from the battery
So internal resistance = V/I = 0.36/0.652 = 0.552Ω
This is why I quoted my answer to 2 sig figs, I used V = 0.36 rather than the 0.364 that you used.