Two electric bulbs whose resistances are in t | vedclass

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(C) When two resistors are connected in parallel to a constant voltage source $V$,the power dissipated in each resistor is given by the formula $P = \

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$2:1$

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(C) When two resistors are connected in parallel to a constant voltage source $V$,the power dissipated in each resistor is given by the formula $P = \frac{V^2}{R}$.Since the voltage $V$ is constant for both bulbs,the power dissipated is inversely proportional to the resistance,i.e.,$P \propto \frac{1}{R}$.Given the ratio of resistances is $\frac{R_1}{R_2} = \frac{1}{2}$.Therefore,the ratio of powers dissipated is $\frac{P_1}{P_2} = \frac{R_2}{R_1} = \frac{2}{1}$.Thus,the ratio of powers 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 powers dissipated in them have the ratio:

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