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Question

Physics Question on Electric Power

Power dissipated in an LCRLCR series circuit connected to an a.c. source of emf ε\varepsilon is :

A

ε2R/R2+(Lω1Cω)2\varepsilon^{2} R / \sqrt{ R ^{2}+\left( L \omega-\frac{1}{ C \omega}\right)^{2}}

B

ε2R/[R2+(Lω1Cω)2]\varepsilon^{2} R /\left[ R ^{2}+\left( L \omega-\frac{1}{ C \omega}\right)^{2}\right]

C

ε2[R2+(Lω1Cω)2]/R\varepsilon^{2} \sqrt{\left[R^{2}+\left(L \omega-\frac{1}{C \omega}\right)^{2}\right]} / R

D

ε2[R2+(Lω1Cω)2]R\frac{\varepsilon^{2}\left[ R ^{2}+\left(L \omega-\frac{1}{ C \omega}\right)^{2}\right]}{ R }

Answer

ε2R/[R2+(Lω1Cω)2]\varepsilon^{2} R /\left[ R ^{2}+\left( L \omega-\frac{1}{ C \omega}\right)^{2}\right]

Explanation

Solution

Pav=ErmsIrmscosϕP _{ av } = E _{ rms } \cdot I _{ rms } \cos \phi
=εεzRz=ε2Rz2=\varepsilon \cdot \frac{\varepsilon}{ z } \cdot \frac{ R }{ z }=\frac{\varepsilon^{2} R }{ z ^{2}}
=ε2RR2+(ωL1ωC)2=\frac{\varepsilon^{2} R }{ R ^{2}+\left(\omega L -\frac{1}{\omega C }\right)^{2}}