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An electric bulb is rated 220V and 100W. When it is operated on 110V, the power consumed will be:
A). 100W
B). 75W
C). 50W
D). 25W

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Last updated date: 22nd Mar 2024
Total views: 315.9k
Views today: 6.15k
MVSAT 2024
Answer
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Hint – In this question use the concept that the rated voltage and rated power of an electric bulb are 220 V and 100 W. Use the concept that the power consumed is the ratio of the square of the voltage of the electric bulb to the resistance of the electric bulb that is $P = \dfrac{{{V^2}}}{R}$. This will help approaching the problem.

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

Complete step-by-step solution-
Given data:
Rated voltage and rated power of an electric bulb are 220 V and 100 W.
Now we have to calculate the power consumed by the electric bulb when it is operated on 110 V.
Now as we know that the power consumed is the ratio of the square of the voltage of the electric bulb to the resistance of the electric bulb.
$ \Rightarrow P = \dfrac{{{V^2}}}{R}$....................... (1)
Where, $P$ = power consumed by the electric bulb
              $V$= voltage on which the bulb is operated
              $R$ = resistance of the electric bulb.
Now when it is operated on 220 V it consumed 100 W so from the above equation we have,
$ \Rightarrow 100 = \dfrac{{{{\left( {220} \right)}^2}}}{R}$.......................... (2)
Now when it is operated on 110 volts, so let the power consumed on that voltage by the electric bulb be ${P_1}$ watts.
Now from equation (1) we have,
$ \Rightarrow {P_1} = \dfrac{{{{\left( {110} \right)}^2}}}{R}$...................... (3)
Now divide equation (3) by equation (2) we have,
$ \Rightarrow \dfrac{{{P_1}}}{{100}} = \dfrac{{\dfrac{{{{\left( {110} \right)}^2}}}{R}}}{{\dfrac{{{{\left( {220} \right)}^2}}}{R}}}$
Now simplify this equation we have,
$ \Rightarrow \dfrac{{{P_1}}}{{100}} = {\left( {\dfrac{{110}}{{220}}} \right)^2}$
$ \Rightarrow \dfrac{{{P_1}}}{{100}} = {\left( {\dfrac{{110}}{{220}}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^2} = \dfrac{1}{4}$
$ \Rightarrow {P_1} = \dfrac{{100}}{4} = 25$W.
So this is the required power consumed by the electric bulb when operated on 110 V.
So this is the required answer.
Hence option (D) is the correct answer.

Note – Power rating basically refers to the highest power that is allowed to flow through an electrical device. Corresponding power rating involves the rating for potential and power. It can also be considered as how much electricity the appliance eventually needs to work properly.
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