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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACSeries AC CircuitsMedium2 minPYQ_2009

PhysicsMediumsingle choice

Power dissipated in an LCR series circuit connected to an a.c. source of emf $ε$ is :NEET 2009 (Mains)

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Answer:

Solution:

$P_{av} = E_{rms} \cdot I_{rms} cos ϕ = ε \cdot \frac{ε}{z} \cdot \frac{R}{Z} = \frac{ε ^{2} R}{z ^{2}} = \frac{ε ^{2} R}{R ^{2} + ( ω L - \frac{1}{ω C} ) ^{2}}$

Stream:NEETSubject:PhysicsTopic:EMI/ACSubtopic:Series AC Circuits

⏱ 2mℹ️ Source: PYQ_2009

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