Two bulbs consume 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
Medium
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(c) $P = \frac{{{V^2}}}{R}$ $\therefore R = \frac{{{V^2}}}{P}{\rm{or}}\,R \propto {V^2}$ i.e. $\frac{{{R_1}}}{{{R_2}}} = {\left( {\frac{{200}}{{300}}} \right)^2} = \frac{4}{9}$
When connected in series potential drop and power consumed are in the ratio of their resistances.
When connected in series potential drop and power consumed are in the ratio of their resistances.
So, $\frac{{{P_1}}}{{{P_2}}} = \frac{{{V_1}}}{{{V_2}}} = \frac{{{R_1}}}{{{R_2}}} = \frac{4}{9}$
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