A) 125 W
B) 100 W
C) 80 W
D) 64 W
Correct Answer: D
Solution :
Key Idea: The power consumption would be the average loss according to the relation \[P=\frac{{{V}^{2}}}{R}\]. Power consumed \[P=\frac{{{V}^{2}}}{R}\] and \[P=\frac{{{V}^{2}}}{R}\] Here. \[V=200\text{ }V.\text{ }V=160\text{ }V,P=100\text{ }W\] Hence, \[\frac{P}{P}={{\left( \frac{V}{V} \right)}^{2}}={{\left( \frac{160}{200} \right)}^{2}}={{(0.8)}^{2}}\] or \[P={{(0.8)}^{2}}\,P=0.64\times 100=64W\] NOTE: For a given voltage, the power consumed in bulb of higher resistance is less than in bulb of lower resistance.
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