AC Circuits

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Question

A sinusoidal voltage of peak value 250V is applied to a series LCR circuit, in which $R = 8 Ω$ , 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……………...

Moderate

Solution

At resonance , $X_{L} {=X}_{C}$
Hence, impedance, Z = R
$\begin{array}{l} {Power = V}_{rms} i_{rms} = \frac{{V_{r m s}}^{2}}{R} = \frac{250}{\sqrt{2}} \frac{250}{\sqrt{2}} \times \frac{1}{8} = 3.906 k W \end{array}$

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