A) 80 W
B) 67 W
C) 76 W
D) 65 W
Correct Answer: B
Solution :
Key Idea: Effective resistance in series is sum of individual resistances. Power consumed by a bulb, potential difference across which is V, is \[P=\frac{{{V}^{2}}}{R}\] where, R is resultant resistance. In series combination, \[R={{R}_{1}}+{{R}_{2}}\] \[\therefore \] \[P=\frac{{{V}^{2}}}{{{R}_{1}}+{{R}_{2}}}=\frac{{{V}^{2}}}{\frac{{{V}^{2}}}{{{P}_{1}}}+\frac{{{V}^{2}}}{{{P}_{2}}}}\] \[\Rightarrow \] \[\frac{1}{P}=\frac{1}{{{P}_{1}}}+\frac{1}{{{P}_{2}}}\] \[=\frac{1}{200}+\frac{1}{100}\] \[\Rightarrow \] \[P=67\,W\]
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