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,

\[R = \frac{V^2}{P} = \frac{(220 \times 220)}{100} = 484  \Omega\]

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.