Question

A servo stabilizer restricts voltage output to $220V\pm 1$. If an electric bulb 100 W/ 20 V is connected to it, find the maximum power consumed by it in W.

Answer 1

In this question we have been asked to calculate the maximum power consumed by the bulb when connected to the given servo stabilizer. To solve this question, we shall first calculate the resistance of the bulb. Later, we shall calculate the maximum Voltage of the bulb. Using these parameters, we shall calculate our using the Joules heating effect, which states the relation between power, resistance and voltage.

Formula used:
$R=\dfrac{{{V}^{2}}}{P}$

Complete step by step answer:
It is given that an electric bulb of 100 W/ 20 V is connected to a servo stabilizer.
From the Joule's heating effect we know,
$R=\dfrac{{{V}^{2}}}{P}$
Therefore, resistance of bulb ${{R}_{B}}$is given by,
${{R}_{B}}=\dfrac{{{20}^{2}}}{100}$
Therefore,
${{R}_{B}}=4\Omega$ ………………….. (1)
Now, the range of servo stabilizer is given as $220V\pm 1$
Therefore,
Minimum voltage will be,
$220-\dfrac{220}{100}=217.8$V
Similarly, maximum voltage will be
$220+\dfrac{220}{100}=222.2$V ……………. (2)
Therefore, using ohm's law,
$I=\dfrac{V}{R}$
From (1) and (2)
We get,
$I=\dfrac{222.2}{4}$
Therefore,
$I=55.55A$…………… (3)
Now, we know that power is given by the equation,
$P={{I}^{2}}R$
Therefore, from (1) and (3)
We get,
$P={{55.55}^{2}}\times 4$
Therefore,
$P=123432.21W$
It can also be written as,
$P=123.4kW$
Therefore, the maximum power will be 123.4 kW.

Note:
When an electric current is passed through an electrical conductor, the temperature of the conductor rises. This happens due to the resistance applied by the conductor material to the current flowing. This effect of heat generation is known as Joule's heating effect. It is also known as the heating effect of current.