I am about to toss this book off the window

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SUMMARY

The discussion centers on calculating the percentage increase in power output of a 100W lightbulb when subjected to a voltage surge from 120V to 140V. The user initially calculated a power increase of 136.11% using the formula P = CΔV², but the correct percentage increase, as stated in the reference book, is 36.11%. This discrepancy arises from misunderstanding the definition of percentage increase, which is based on the ratio of the increase in power output to the initial power output.

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  • Ability to manipulate equations and understand constants in physics.
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Homework Statement



A 100W lightbulb connected to a 120V source experiences a voltage surge that produces 140V for a moment. By what percentage does its power output increase? Assume its resistance does not change

The Attempt at a Solution



(1) [tex]P_0 = C\Delta V_0 ^2[/tex]

(2) [tex]P = C\Delta V^2[/tex]

Note that C is a constant (the resistance)

Divide (1) by (2)

[tex]\frac{P_0}{P} = (\frac{\Delta V_0}{\Delta V})^2[/tex]

Plugging the numbers I get 1.36111...

Multiply it by 100 I get 136.11%

Now, the book says 36.11%

Why...?
 
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1. Behave yourself. You should give a apt title to your question.
2. You have been asked to calculate the percentage increase in power output. It is the percentage ratio of increase in power output to the initial power.
 
increase = final - initial = 36.11

so required percentage = 36.11%
 

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