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

For a series LCR circuit the power loss at resonance is:
  1. $\frac{V^2}{\left(\omega L-\frac{1}{\omega C}\right)}$
  2. $r L \omega$
  3. $I^2 R$
  4. $\frac{V^2}{C \omega}$

Solution

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

Asked in: NEET 2002

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