A) ratio of potential differences across them is 3/2
B) ratio of potential differences across them is 9/4
C) ratio of powers consumed across them is 4/9
D) ratio of powers consumed across them is 2/3
Correct Answer: C
Solution :
\[P=\frac{{{V}^{2}}}{R}\] \[\therefore \] \[R=\frac{{{V}^{2}}}{P}\] 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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