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
ARatio of potential difference across them is $3/2$
BRatio of potential difference across them is $9/4$
CRatio of power consumed across them is $4/9$
DRatio of power consumed across them is $2/3$
Medium
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CRatio of power consumed across them is $4/9$
c (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.
So, $\frac{{{P_1}}}{{{P_2}}} = \frac{{{V_1}}}{{{V_2}}} = \frac{{{R_1}}}{{{R_2}}} = \frac{4}{9}$
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