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

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Question

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

The value of x to the nearest integer is ______.

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Solution

A sinusoidal voltage of peak value 250 V is applied to a series LCR circuit, in which R = 8 Ω, L = 24 mH, and C = 60 μF. The value of power dissipated at resonant conditions is 'x' kW. The value of x to the nearest integer is 4.

Explanation:

Peak voltage [V0] = 250 V

So, VRMS = `"V"_0/sqrt2 = 250/sqrt2"V"`

Resistance [R] = 8 Ω

Inductor [L] = 24 mH

Capacitor [C] = 60 µF

The dissipated power at the resonant condition

P = `(["V"_"RMS"]^2)/"R" ⇒ (250/sqrt2)^2/8`

 ⇒ 3906.25 W

Approx. P = 4 kW

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