In a series RLC circuit (AC):L = 10 , mHC = 0 | vedclass

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(C) At resonance,the impedance of the circuit is purely resistive,so $Z = R = 50 \, \Omega$.The peak voltage is $V_0 = 10 \, V$.The root mean square v

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(C) At resonance,the impedance of the circuit is purely resistive,so $Z = R = 50 \, \Omega$.The peak voltage is $V_0 = 10 \, V$.The root mean square voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \, V$.The root mean square current at resonance is $I_{rms} = \frac{V_{rms}}{Z} = \frac{10 / \sqrt{2}}{50} \, A$.The power dissipated in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi$.At resonance,the phase angle $\phi = 0^{\circ}$,so $\cos \phi = 1$.Substituting the values: $P = \left( \frac{10}{\sqrt{2}} ight) 	imes \left( \frac{10 / \sqrt{2}}{50} ight) 	imes 1$.$P = \frac{100}{2 	imes 50} = \frac{100}{100} = 1 \, W$.

In a series $RLC$ circuit $(AC)$:
$L = 10 \, mH$
$C = 0.01 \, \mu F$
$R = 50 \, \Omega$
If the supply voltage is $V = 10 \sin \omega t$,find the power dissipated at resonance in $W$.

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In a series $RLC$ circuit $(AC)$:$L = 10 \, mH$$C = 0.01 \, \mu F$$R = 50 \, \Omega$If the supply voltage is $V = 10 \sin \omega t$,find the power dissipated at resonance in $W$.