Two electric bulbs A and B are marked 220, V, | vedclass

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(A) The resistance $R$ of an electric bulb is calculated using the formula $R = \frac{V^2}{P}$,where $V$ is the voltage and $P$ is the power.For bulb

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$A$ has greater resistance

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(A) The resistance $R$ of an electric bulb is calculated using the formula $R = \frac{V^2}{P}$,where $V$ is the voltage and $P$ is the power.For bulb $A$: $R_A = \frac{220^2}{60} \approx 806.67\, \Omega$.For bulb $B$: $R_B = \frac{220^2}{100} = 484\, \Omega$.Since the voltage $V$ is constant for both bulbs,the resistance is inversely proportional to the power $(R \propto \frac{1}{P})$.Therefore,the bulb with the lower power rating $(60\, W)$ will have a higher resistance.Thus,bulb $A$ has greater resistance.

Two electric bulbs $A$ and $B$ are marked $220\, V, 60\, W$ and $220\, V, 100\, W$ respectively. Which one of the two has greater resistance?

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