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If the voltage across a bulb rated 220 volts, 100 watt drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is:
A. 5%
B. 10%
C. 20%
D. 2.5%

Answer
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Hint: We know the formula for the power, that is power is equal to the square of the voltage divided by the resistance. From this relation we can find the required quantity to proceed for the solution of the problem. We can get the resistance from the given values of the rated power and voltage.

Complete step by step answer:
So the given quantities are,
Rated voltage= V=220volt,
Rated power =P=100watt
And the percentage of drop of the voltage =2.5%
As we know that, P=V2R
So, we can also write as,
R=V2P
To get the resistance of the bulb we put the values in the above formula, as follows
R=2202100
 R=48Ω
So as per the given condition (the voltage drop by 2.5% of its rated value) then the new voltage will be:
Vnew=2202.5×220100
Vnew=2205.5
Vnew=214.5V
Since now the voltage has been changed so the power will also be changed accordingly,
So new power will be;
Pnew=Vnew2R
Pnew=214.52484
Pnew=95.06W
Now we will calculate the percentage change (decrease) in the power as follows,
 So, the percentage change in the power= PinitialPnewPinitial×100
=10095.06100×100
=4.94
This is approximately 5%.
So, the percentage change in the power as per the change in the voltage is 5%.
Hence the option (A) is the correct answer.

Note:
Here we see that the power formula is in the terms of the voltage and the resistance. We have some other formulae of the power in the terms of the voltage and current and in the terms of the current and resistance. The power formula in the terms of the voltage and the current is power is equals to the voltage multiplied by the current.
Hence, P=VI .