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(C) Given that both bulbs consume the same power $P$ at their respective rated voltages $V_1 = 200\, V$ and $V_2 = 300\, V$.Using the formula $P = \fr

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Ratio of power consumed across them is $4/9$

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(C) Given that both bulbs consume the same power $P$ at their respective rated voltages $V_1 = 200\, V$ and $V_2 = 300\, V$.Using the formula $P = \frac{V^2}{R}$,we have $R = \frac{V^2}{P}$.Since $P$ is constant for both,$R \propto V^2$.Therefore,the ratio of their resistances is $\frac{R_1}{R_2} = \left( \frac{200}{300} \right)^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$.When connected in series,the current $I$ flowing through both bulbs is the same.The potential difference across each bulb is $V' = I \cdot R$,so $\frac{V'_1}{V'_2} = \frac{R_1}{R_2} = \frac{4}{9}$.The power consumed by each bulb in series is $P' = I^2 R$,so the ratio of power consumed is $\frac{P'_1}{P'_2} = \frac{R_1}{R_2} = \frac{4}{9}$.

Two bulbs consume the same power when operated at $200\, V$ and $300\, V$ respectively. When these bulbs are connected in series across a $D.C.$ source of $500\, V$,then:

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