Two bulbs consume same power when operated at | vedclass

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(c) $P = \frac{{{V^2}}}{R}$ $	herefore R = \frac{{{V^2}}}{P}{m{or}}\,R \propto {V^2}$ i.e. $\frac{{{R_1}}}{{{R_2}}} = {\left( {\frac{{200}}{{300}}}

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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