(A) The resonance angular frequency $\omega$ is given by the formula: $\omega = \frac{1}{\sqrt{LC}}$.Given $L = 8 \, mH = 8 \times 10^{-3} \, H$ and $

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$2500 \, rad/s$,$5\sqrt{2} \, A$

(A) The resonance angular frequency $\omega$ is given by the formula: $\omega = \frac{1}{\sqrt{LC}}$.Given $L = 8 \, mH = 8 \times 10^{-3} \, H$ and $C = 20 \, \mu F = 20 \times 10^{-6} \, F$.Substituting these values: $\omega = \frac{1}{\sqrt{8 \times 10^{-3} \times 20 \times 10^{-6}}} = \frac{1}{\sqrt{160 \times 10^{-9}}} = \frac{1}{\sqrt{16 \times 10^{-8}}} = \frac{1}{4 \times 10^{-4}} = 2500 \, rad/s$.At resonance,the impedance $Z = R = 44 \, \Omega$.The $RMS$ current $I_{rms} = \frac{V_{rms}}{R} = \frac{220}{44} = 5 \, A$.The amplitude of the current (peak current) $I_0 = I_{rms} \sqrt{2} = 5\sqrt{2} \, A$.

Class 12 Physics Alternating Current Half Power Frequency , Quality Factor ,Resonance in AC Circuit

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For the series $LCR$ circuit shown in the figure,what is the resonance angular frequency and the amplitude of the current at the resonating frequency? The source voltage is given as $220 \, V$ ($RMS$ value).

A
$2500 \, rad/s$,$5\sqrt{2} \, A$
B
$2500 \, rad/s$,$5 \, A$
C
$2500 \, rad/s$,$\frac{5}{\sqrt{2}} \, A$
D
$25 \, rad/s$,$5\sqrt{2} \, A$

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Half Power Frequency , Quality Factor ,Resonance in AC Circuit

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For the series $LCR$ circuit shown in the figure,what is the resonance angular frequency and the amplitude of the current at the resonating frequency? The source voltage is given as $220 \, V$ ($RMS$ value).

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$A$ series $LCR$ circuit is shown in the figure. Where the inductance of $10 \ H$,capacitance $40 \ \mu F$,and resistance $60 \ \Omega$ are connected to a variable frequency $240 \ V$ source. The current at resonating frequency is (in $A$)

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