Solveeit Logo
Login

Question

Question: Power dissipated in an LCR series circuit connected to an ac source of emf \(\varepsilon\)is...

Power dissipated in an LCR series circuit connected to an ac source of emf ε\varepsilonis

A

ε2R2+(ωL−1ωC)R\frac{\varepsilon^{2}\sqrt{R^{2} + \left( \omega L - \frac{1}{\omega C} \right)}}{R}

B

ε2[R2+(ωL−1ωC)2]R\frac{\varepsilon^{2}\left\lbrack R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2} \right\rbrack}{R}

C

ε2RR2+(ωL−1ωC)\frac{\varepsilon^{2}R}{\sqrt{R^{2} + \left( \omega L - \frac{1}{\omega C} \right)}}

D

ε2R[R2+(ωL−1ωC)2]\frac{\varepsilon^{2}R}{\left\lbrack \sqrt{R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2}} \right\rbrack}

Answer

ε2R[R2+(ωL−1ωC)2]\frac{\varepsilon^{2}R}{\left\lbrack \sqrt{R^{2} + \left( \omega L - \frac{1}{\omega C} \right)^{2}} \right\rbrack}

Explanation

Solution

: Average power, ,P=VrmsIrmscos⁡φP = V_{rms}I_{rms}\cos\varphi

Here, Z=R2+(XL−XC)2,cos⁡φ=RZZ = \sqrt{R^{2} + (X_{L} - X_{C})^{2}},\cos\varphi = \frac{R}{Z}

But Irms=VrmsZ∴P=Vrms2RZ2I_{rms} = \frac{V_{rms}}{Z}\therefore P = V_{rms}^{2}\frac{R}{Z^{2}}

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