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Alternating Current question

2009 · Q141
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Alternating Current question

2009 · Q141

NEETPhysicsAlternating CurrentMCQ+4 / −1
Power dissipated in an LCR series circuit connected to an A.C. source of emf ε\varepsilonε is
  1. A
    ε2R2+(Lω−1ω)2R{{{\varepsilon ^2}\sqrt {{R^2} + {{\left( {L\omega - {1 \over {\omega }}} \right)}^2}} } \over R}Rε2R2+(Lω−ω1​)2​​
  2. B
    ε2R2+(Lω−1Cω)2R{{{\varepsilon ^2}\sqrt {{R^2} + {{\left( {L\omega - {1 \over {C\omega }}} \right)}^2}} } \over R}Rε2R2+(Lω−Cω1​)2​​
  3. C
    ε2RR2+(L−1Cω)2{{{\varepsilon ^2}R} \over {\sqrt {{R^2} + {{\left( {L - {1 \over {C\omega }}} \right)}^2}} }}R2+(L−Cω1​)2​ε2R​
  4. D
    ε2R[R2+(Lω−1Cω)2]{{{\varepsilon ^2}R} \over {\left[ {{R^2} + {{\left( {L\omega - {1 \over {C\omega }}} \right)}^2}} \right]}}[R2+(Lω−Cω1​)2]ε2R​
View written solutionFree

Correct answer: D

Power dissipated, P = ErmsIrms cosϕ\phi ϕ

cosϕ\phi ϕ = RZ{R \over Z}ZR​

But Irms = ErmsZ{{{E_{rms}}} \over Z}ZErms​​

∴\therefore∴ P = Erms2.RZ2E_{rms}^2.{R \over {{Z^2}}}Erms2​.Z2R​

Also we know, Z = R2+(XL−XC)2\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} R2+(XL​−XC​)2​

So, P = Erms2.RR2+(XL−XC)2E_{rms}^2.{R \over {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}}}Erms2​.R2+(XL​−XC​)2R​

= ε2R[R2+(Lω−1Cω)2]{{{\varepsilon ^2}R} \over {\left[ {{R^2} + {{\left( {L\omega - {1 \over {C\omega }}} \right)}^2}} \right]}}[R2+(Lω−Cω1​)2]ε2R​

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