A) 25 W
B) 50 W
C) 100 W
D) 200 W
Correct Answer: A
Solution :
Resistance of the bulb, \[(R)=\frac{{{V}^{2}}}{P}\] \[=\frac{220\times 220}{100}=484\Omega \] \[\therefore \] Power consumed by the bulb \[P=\frac{V{{'}^{2}}}{R}=\frac{110\times 110}{484}\] \[25W\]
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