Question

An electric bulb is rated $220$ volt and $100$ watt, Power consumed by it when operated on $110$ volt is :
(A) $50\;{\text{W}}$
(B) $85\;{\text{W}}$
(C) $90\;{\text{W}}$
(D) $25\;{\text{W}}$

Answer 1

Hint
The power rating of an electric appliance is the highest input we can flow through it. It would depend on the resistance and voltage rating of the electric appliance.

Complete step by step answer
Given, the voltage rated at the bulb is $V = 220\;{\text{V}}$ and power rated is $P = 100\;{\text{W}}$.
The power rating will help the users to use the electric appliance not giving the input greater than the rating. Hence it is for more safety of the appliance. It is the maximum power that can be dissipated. And it is the voltage supplied by the current draw. In order to calculate the power rating, we should know about the resistance and the voltage requirements of the appliance. The power rating can be given in units of watts or kilowatts.
The expression for the power rated in the bulb is given as,
$\Rightarrow P = \dfrac{{{V^2}}}{R}$
Where, $V$ is the voltage and $R$ is the resistance and $P$ is the power.
From the above expression, the resistance of the bulb is given as,
$\Rightarrow R = \dfrac{{{V^2}}}{P}$
Substituting the values in the above expression,
$\Rightarrow R = \dfrac{{{{\left( {220\;{\text{V}}} \right)}^2}}}{{100\;{\text{W}}}} $
$\Rightarrow 484\;\Omega $
Thus, the resistance of the bulb is $484\;\Omega $.
The power consumed when the bulb is operated in $110$ volts is given as,
$\Rightarrow P = \dfrac{{{{\left( {110\;{\text{V}}} \right)}^2}}}{{484\;\Omega }} $
$\Rightarrow P = 25\;{\text{W}} $
The power consumed is $25\;{\text{W}}$.
The answer is option (D).

Note
We have to note that if the resistance of the appliance is greater, then the power consumed will be smaller. Thus if the operating voltage is higher, then more power will be consumed.