A) Ratio of potential difference across them is 3/2
B) Ratio of potential difference across them is 9/4
C) Ratio of power consumed across them is 4/9
D) Ratio of power 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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