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Chapter 26: Problem 20

Two light bulbs for use at \(110 \mathrm{~V}\) are rated at \(60 \mathrm{~W}\) and \(100 \mathrm{~W}\), respectively. Which has the filament with lower resistance?

Short Answer

Expert verified
Answer: The 100 W light bulb has a lower resistance.

Step by step solution

01

Calculate resistance for the 60 W light bulb

To find the resistance of the first light bulb (\(60 \mathrm{~W}\)), we will rearrange the formula \(P = \frac{V^2}{R}\) to solve for resistance, giving us \(R = \frac{V^2}{P}\). Now, we will plug in the given data for the \(60 \mathrm{~W}\) light bulb: \(R_1 = \frac{(110 \mathrm{~V})^2}{60 \mathrm{~W}}\)
02

Calculate resistance for the 100 W light bulb

Similar to step 1, we will use the same formula to find the resistance for the second light bulb (\(100 \mathrm{~W}\)): \(R_2 = \frac{(110 \mathrm{~V})^2}{100 \mathrm{~W}}\)
03

Compare the resistances

Now that we have calculated the resistance for both light bulbs, we can compare them to see which one has the lower resistance. We have: \(R_1 = \frac{(110 \mathrm{~V})^2}{60 \mathrm{~W}}\) \(R_2 = \frac{(110 \mathrm{~V})^2}{100 \mathrm{~W}}\) Since the resistance is inversely proportional to the power, the light bulb with higher power will have lower resistance. So, the light bulb with \(100 \mathrm{~W}\) power rating has the lower resistance.

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