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QJEEM21Ph3d04: JEE Main 2021 Physics - Class 12 JEEM21Ph3d04 Solution: A sinusoidal voltage of peak value 250 V is applied to a series …

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

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At resonance power (P)

P=(Vrms)2R

∴ P=(250/2)28

⇒ P = 3906.25 w

⇒ P ≅ 4 Kw

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