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