Question

An electric bulb is rated $220V$ and $100W$. When it is operated on $110V$, what will be the power consumed?
$\begin{aligned} & A.100W \\\ & B.75W \\\ & C.50W \\\ & D.25W \\\ \end{aligned}$

Answer 1

The power is given by taking the ratio of the square of the voltage rated to the resistance offered. From this the value of the resistance is to be found first. Find out the current flow in the circuit. Then calculate the voltage drop across the bulb and finally estimate the power consumed by the bulb. This information will help you in solving this question.

Complete answer:
first of all, the power of the circuit is given by the equation,
$P=\dfrac{{{V}^{2}}}{R}$
Where $V$ be the electric potential and $R$ be the resistance.
Rearranging this equation in terms of the resistance will give,
$R=\dfrac{{{V}^{2}}}{P}$
The value of the voltage is given as,
$V=220V$
Power is mentioned as,
$P=100W$
Substituting the values in it,
$R=\dfrac{220\times 220}{100}=484\Omega$
Now let us calculate the current flow in the circuit. This can be found using the equation,
$i=\dfrac{{{V}'}}{R}$
Where ${V}'$ be the supplied voltage in the circuit. It is mentioned in the question as,
${V}'=110V$
Substituting the values in it will be shown as,
$i=\dfrac{110}{484}=0.2272A$
Therefore the current through the circuit is obtained. This will be the voltage consumed by the bulb. Therefore we can write that,
${{V}_{c}}=110V$
Now we have to calculate the power consumed by the bulb. This can be found by the equation,
$P=\dfrac{{{V}_{c}}^{2}}{R}$
Substituting the values in it will give,
${{P}_{b}}=\dfrac{110\times 110}{484}=25W$
Therefore the correct answer is obtained as ${{P}_{b}}=25W$.

So, the correct answer is “Option D”.

Note:
The power rating shown generally on the bulb is an indication about their consumption. It represents how much electricity the bulb will be transforming into light and heat over a specific amount of time. If it is kept on for one hour, then a $25W$ bulb will be consuming $25Watt-hours$ of energy.