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(A) The power dissipated in an electric bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage across the bulb and $R$ is its resis

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$2:1$

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(A) The power dissipated in an electric bulb is given by the formula $P = \frac{V^2}{R}$,where $V$ is the voltage across the bulb and $R$ is its resistance.Since the bulbs are connected in parallel to a constant voltage source,the voltage $V$ across each bulb is the same.Therefore,the power dissipated is inversely proportional to the resistance: $P \propto \frac{1}{R}$.Given the ratio of resistances is $\frac{R_1}{R_2} = \frac{1}{2}$,the ratio of power dissipated is $\frac{P_1}{P_2} = \frac{R_2}{R_1} = \frac{2}{1}$.Thus,the ratio of power dissipated is $2:1$.

Two electric bulbs whose resistances are in the ratio of $1:2$ are connected in parallel to a constant voltage source. The ratio of the power dissipated in them is:

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