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
$$\begin{gathered} R=\frac{V^2}{P}=\frac{(220{)}^2}{100} \\ \Rightarrow R=\frac{48400}{100}=484\mathrm{\Omega } \end{gathered}$$

(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,
$i\text{'}=\frac{V\text{'}}{R}=\frac{217.8}{484}=0.45\mathrm{A}$
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,
$i"=\frac{V"}{R}=\frac{222.2}{484}=0.459\mathrm{A}$
Power consumed by the bulb,
P" = i" × V"
= 0.459 × 222.2 = 102 W