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PhysicsCurrent ElectricityElectrical PowerMedium2 minPYQ_2002

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An electric bulb marked $40 W$ and $200 V$ , is used in a circuit of supply voltage $100 V$ . Now its power is

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Solution:

Actual power of bulb $(P_{1}) = 40 W$ Actual voltage of bulb $(V_{1}) = 200 V$ and supply voltage $(V_{2}) = 100 V$ Power $(P) = \frac{V ^{2}}{R} \propto V^{2}$ . Therefore $\frac{P _{1}}{P _{2}} = \frac{V _{1}^{2}}{V _{2}^{2}}$ or, $\frac{40}{P _{2}} = \frac{( 200 ) ^{2}}{( 100 ) ^{2}} = 4$ or $P_{2} = \frac{40}{4} = 10 W$ (where $P_{2} =$ power when voltage is $100 V$ ).

Stream:AIIMSSubject:PhysicsTopic:Current ElectricitySubtopic:Electrical Power

⏱ 2mℹ️ Source: PYQ_2002

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