Two bulbs are rated (40 ; W, 200 ; V) and (10 | vedclass

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(C) The power rating of an electrical appliance is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.For the

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$R_{40} > R_{100}$

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(C) The power rating of an electrical appliance is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage and $R$ is the resistance.For the first bulb $(40 \; W, 200 \; V)$:$R_{40} = \frac{V^2}{P_1} = \frac{(200)^2}{40} = \frac{40000}{40} = 1000 \; \Omega$.For the second bulb $(100 \; W, 200 \; V)$:$R_{100} = \frac{V^2}{P_2} = \frac{(200)^2}{100} = \frac{40000}{100} = 400 \; \Omega$.Comparing the two values,we find that $1000 \; \Omega > 400 \; \Omega$,which implies $R_{40} > R_{100}$.

Two bulbs are rated $(40 \; W, 200 \; V)$ and $(100 \; W, 200 \; V)$. What is the correct relation for their resistances?

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