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?

Answer 1

Given that the operating voltage is V and power consumed is P.
Therefore, the resistance of the bulb,
$R=\frac{V^2}{P}=\frac{(220\times 220)}{100}=484\mathrm{\Omega }$
The power fluctuation, p = 150 W. So, the voltage fluctuation that the bulb can withstand,
$$\begin{gathered} v=\sqrt{pR}=\sqrt{150\times 484} \\ =269.4\mathrm{V}=270\mathrm{V} \end{gathered}$$
The bulb will withstand up to 270 V.