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Alternating Current question

2002 · Q136
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Alternating Current question

2002 · Q136

NEETPhysicsAlternating CurrentMCQ+4 / −1
For a series LCR circuit the power loss at resonance is
  1. A
    V2[ωL−1ωC]{{{V^2}} \over {\left[ {\omega L - {1 \over {\omega C}}} \right]}}[ωL−ωC1​]V2​
  2. B
    I2Lω{I^2}L\omegaI2Lω
  3. C
    I2R{I^2}RI2R
  4. D
    V2Cω{{{V^2}} \over {C\omega }}CωV2​
View written solutionFree

Correct answer: C

The impedance Z of a series LCR circuit is given by,

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

At resonance, XL = XC, hence Z = R.

Let, supply voltage = VR = V

∴\therefore∴ R.M.S. current, I = VR{V \over R}RV​

∴\therefore∴ Power loss = VI = I2R

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