I am about to toss this book off the window

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A 100W lightbulb connected to a 120V source experiences a voltage surge to 140V, prompting a calculation of the percentage increase in power output. The initial power is calculated using the formula P = CΔV², leading to a result of 136.11% when comparing the new power to the original. However, the correct interpretation focuses on the actual increase in power, which is 36.11% relative to the original power. The confusion arises from misunderstanding how to calculate the percentage increase versus the total power output. The final answer confirms that the percentage increase in power output is indeed 36.11%.
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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) P_0 = C\Delta V_0 ^2

(2) P = C\Delta V^2

Note that C is a constant (the resistance)

Divide (1) by (2)

\frac{P_0}{P} = (\frac{\Delta V_0}{\Delta V})^2

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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