Q78 / 90 · JEE Main 2010 · Physics

In a series LCR circuit $R=200 \Omega$ and the voltage and the frequency of the main supply is $220 \mathrm{~V}$ and $50 \mathrm{~Hz}$ respectively. On taking out the capacitance from the circuit the current lags behind the voltage by $30^{\circ}$. On taking out the inductor from the circuit the current leads the voltage by $30^{\circ}$. The power dissipated in the LCR circuit is

  • A
    $305 \mathrm{~W}$
  • B
    $210 \mathrm{~W}$
  • C
    Zero W
  • Correct answer
    D
    $242 \mathrm{~W}$

Solution

The given circuit is under resonance as $X_L=X_C$ Hence power dissipated in the circuit is $$ \mathrm{P}=\frac{\mathrm{V}^2}{\mathrm{R}}=242 \mathrm{~W} $$

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

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