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Physics - EMI/AC Question with Solution | TestHub

PhysicsEMI/ACSeries AC CircuitsMedium2 minQB

PhysicsMediumnumerical

A sinusoidal voltage of peak value $250 V$ is applied to a series $L C R$ circuit, in which $R = 8Ω, L = 24 mH$ and $C = 60 μ F$ . The value of power dissipated at resonant condition is $x$ $kW$ . The value of $x$ to the nearest integer is _______.

Answer:

4.00

Solution:

At resonance power, $P = \frac{( V _{r m s} ) ^{2}}{R}$

$P = \frac{( 250/ 2 ) ^{2}}{8} = 3906.25 W$

$\approx 4 kW$

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

⏱ 2mℹ️ Source: QB

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