Question
An electric bulb marked 220V, 100W will get fused if it is made to consume 150W or more. What voltage fluctuation will the bulb withstand?

Answer

$\text{V}=220\text{v}$$\text{P}=100\text{w}$
$\text{R}=\frac{\text{V}^2}{\text{P}}$
$=\frac{220\times220}{100}=484\Omega$
$\text{P}=150\text{w}$
$\text{V}=\sqrt{\text{PR}}=\sqrt{150\times22\times22}$
$=22\sqrt{150}=269.4\approx270\text{v}$

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