A) 25 W
B) 50 W
C) 75 W
D) 90 W
Correct Answer: B
Solution :
From Joule's law, the power consumed by bulb of resistance R is\[P=\frac{{{V}^{2}}}{R}.\]where V is potential difference. Given, V= 220 volt, P=200W \[\therefore \] \[R=\frac{{{V}^{2}}}{P}=\frac{220\times 220}{200}=242\,\Omega \] When, V = 110 volt, R = 242 Q \[P=\frac{110\times 110}{242}=50\,W\]You need to login to perform this action.
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