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PhysicsCurrent ElectricityElectrical PowerMedium2 minPYQ_2024

PhysicsMediumsingle choice

An electric bulb rated $50 W - 200 V$ is connected across a $100 V$ supply. The power dissipation of the bulb is:

Options:

Answer:

Solution:

Rated power \& voltage gives resistance $R = \frac{V ^{2}}{P} = \frac{( 200 ) ^{2}}{50} = \frac{40000}{50} R = 800 P = \frac{( V _{applied} ) ^{2}}{R} = \frac{( 100 ) ^{2}}{800} P = 12.5 watt$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Electrical Power

⏱ 2mℹ️ Source: PYQ_2024

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