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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityElectrical PowerEasy2 minPYQ_2020

PhysicsEasysingle choice

If $R_{1}$ and $R_{2}$ are the resistances of the filaments of $200 W$ and $100 W$ electric bulbs operated at $220 V$ , then $(\frac{R_{1}}{R_{2}})$ is

Options:

Answer:

Solution:

Power of bulb, $P = \frac{V^{2}}{R}$

$∴ \frac{P_{2}}{P_{1}} = \frac{R_{1}}{R_{2}}$

Or $\frac{R_{1}}{R_{2}} = \frac{100}{200} = 0.5$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Current ElectricitySubtopic:Electrical Power

⏱ 2mℹ️ Source: PYQ_2020

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