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An electric bulb has its specification 60 W, 220 volt. The resistance of the bulb is :

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708 $Ω$

807 $Ω$

4 $Ω$

8070 $Ω$

answer is C.

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Detailed Solution

$\begin{array}{rcl} 60 & = & \frac{V^{2}}{R} = \frac{220^{2}}{R} \\ R & = & \frac{220^{2}}{60} = 806.66 Ω \end{array}$

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