A servo voltage stabiliser limits its output voltage to 220 V ± 1\% . When an electric bulb rated at 220 V , 100 W is connected to this stabiliser, determine the minimum and maximum power consumed by the bulb.

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 98 W and 102 W

The power consumed by the bulb is given by

P = \dfrac{V^2}{R}

.

For small variations, the fractional change in power is:

\dfrac{\Delta P}{P} = 2 \dfrac{\Delta V}{V}

Given that the voltage variation is

\pm 1\%

, we have:

\dfrac{\Delta P}{P} = \pm 2(1\%) = \pm 2\%

The nominal power of the bulb is

P = 100 \text{ W}

.

The variation in power is

\Delta P = \pm 2\%

of

100 \text{ W} = \pm 2 \text{ W}

.

Therefore, the minimum power consumed is

100 - 2 = 98 \text{ W}

and the maximum power consumed is

100 + 2 = 102 \text{ W}

.

A servo voltage stabiliser limits its output voltage to 220 V ± 1\%...

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