Q125 / 200 · NEET 2002 · Physics

For a series LCR circuit the power loss at resonance is:

  • A
    $\frac{V^2}{\left(\omega L-\frac{1}{\omega C}\right)}$
  • B
    $r L \omega$
  • Correct answer
    C
    $I^2 R$
  • D
    $\frac{V^2}{C \omega}$

Solution

Power loss is given by $$ P=V I \cos \phi $$ At resonance current and voltage are in the same phase, that is $\phi=0$. $$ \therefore P=V I \cos 0=V I=I^2 R $$

Hence, the correct answer is option: 'C'.

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