Question

A servo voltage stabiliser restricts the voltage output to 220 V ± 1%. If an electric bulb rated at 220 V, 100 W is connected to it, what will be the minimum and maximum power consumed by it?

Answer 1

Output voltage, V = 220 V ± 1% = 220 V ± 2.2 V

The resistance of a bulb that is operated at voltage V and consumes power P is given by

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

\[ \Rightarrow R = \frac{48400}{100} = 484  \Omega\]

(a) For minimum power to be consumed, output voltage should be minimum. The minimum output voltage,

V' = (220 − 2.2) V

= 217.8 V

The current through the bulb,

\[i' = \frac{V'}{R} = \frac{217 . 8}{484} = 0 . 45  A\]

Power consumed by the bulb, P' = i' × V'

                                                    = 0.45 × 217.8 = 98.0 W

(b) For maximum power to be consumed, output voltage should be maximum. The maximum output voltage,

V" = (220 + 2.2) V

     = 222.2 V

The current through the bulb,

\[i''  = \frac{V''}{R} = \frac{222 . 2}{484} = 0 . 459  A\]

Power consumed by the bulb,

P" = i" × V"

    = 0.459 × 222.2 = 102 W