An electric bulb is rated as 220 V and 80 W. When it is operated on 11

An electric bulb is rated as 220 V and 80 W. When it is operated on 110 V, the power rating would be :

[amp_mcq option1=”80 W” option2=”60 W” option3=”40 W” option4=”20 W” correct=”option4″]

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UPSC NDA-1 – 2023

The power rating of an electric bulb is given by the formula P = V^2 / R, where P is power, V is voltage, and R is resistance. The resistance of the bulb remains constant (assuming its temperature change is negligible for this calculation). First, calculate the resistance using the rated values: R = V_rated^2 / P_rated = (220 V)^2 / 80 W = 48400 / 80 = 605 Ω. When operated on 110 V, the power consumed is P_new = V_new^2 / R = (110 V)^2 / 605 Ω = 12100 / 605 = 20 W.

– The resistance of the bulb is a fixed property.
– Power is proportional to the square of the voltage when resistance is constant (P ∝ V^2).
– Since the voltage is halved (110V is half of 220V), the power will be reduced by a factor of (1/2)^2 = 1/4.
– New Power = Rated Power * (New Voltage / Rated Voltage)^2 = 80 W * (110 V / 220 V)^2 = 80 W * (1/2)^2 = 80 W * (1/4) = 20 W.

This calculation assumes the resistance R is constant, which is a reasonable approximation for this type of problem, although in reality, the resistance of a filament bulb changes with temperature.

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