A bulb is rated 220;V and 100;W. What is the | vedclass

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(D) The power rating of the bulb is given by $P = 100\;W$ and the voltage rating is $V = 220\;V$.We know the relationship between power,voltage,and re

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$484\;\Omega$

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(D) The power rating of the bulb is given by $P = 100\;W$ and the voltage rating is $V = 220\;V$.We know the relationship between power,voltage,and resistance is given by the formula $P = \frac{V^2}{R}$.Rearranging the formula to solve for resistance $R$,we get $R = \frac{V^2}{P}$.Substituting the given values: $R = \frac{(220)^2}{100}$.$R = \frac{48400}{100}$.$R = 484\;\Omega$.Therefore,the resistance of the bulb is $484\;\Omega$.

$A$ bulb is rated $220\;V$ and $100\;W$. What is the resistance of this bulb?

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