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(D) The average power dissipated in an $LCR$ series circuit is given by $P = E_{rms} I_{rms} \cos \phi$.The impedance of the circuit is $Z = \sqrt{R^2

Vedclass · Accepted Answer

$\frac{\varepsilon^2 R}{R^2 + (\omega L - \frac{1}{\omega C})^2}$

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(D) The average power dissipated in an $LCR$ series circuit is given by $P = E_{rms} I_{rms} \cos \phi$.The impedance of the circuit is $Z = \sqrt{R^2 + (X_L - X_C)^2}$,where $X_L = \omega L$ and $X_C = \frac{1}{\omega C}$.The power factor is $\cos \phi = \frac{R}{Z}$.The root-mean-square current is $I_{rms} = \frac{E_{rms}}{Z}$.Substituting these into the power formula: $P = E_{rms} \cdot \frac{E_{rms}}{Z} \cdot \frac{R}{Z} = \frac{E_{rms}^2 R}{Z^2}$.Substituting $Z^2 = R^2 + (\omega L - \frac{1}{\omega C})^2$ and $E_{rms} = \varepsilon$,we get $P = \frac{\varepsilon^2 R}{R^2 + (\omega L - \frac{1}{\omega C})^2}$.

Power dissipated in an $LCR$ series circuit connected to an $A.C.$ source of $e.m.f.$ $\varepsilon$ is

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