Answer
Verified
315.9k+ views
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.
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE