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Question
An electric bulb marked 220 V, 100 W will get fused if it is made to consume 150 W or more. What voltage fluctuation will the bulb withstand?
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Solution
Given that the operating voltage is V and power consumed is P.
Therefore, the resistance of the bulb,
\[R = \frac{V^2}{P} = \frac{(220 \times 220)}{100} = 484 \Omega\]
The power fluctuation, p = 150 W. So, the voltage fluctuation that the bulb can withstand,
\[v = \sqrt{pR} = \sqrt{150 \times 484}\]
\[ = 269 . 4 V = 270 V\]
The bulb will withstand up to 270 V.
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