Question

Calculate the power dissipated in the LCR circuit with $R=200\Omega$ and on taking out the capacitance from the circuit, the current lags behind the voltages by $30^{\circ }$ while it leads by the same when the inductor is taken out of the circuit. The rms valur and frequency of voltage supply is $220V$ and $50Hz$ respectively.

• A. $242W$
• B. $305W$
• C. $210W$
• D. Zero

Answer 1

Answer: (A)

The phase angle is given by $tan\varphi =\frac{X_L-X_c}{R}$ .
When only the capacitance is removed, the phase difference between the current and voltage is
$tan30^{\circ }=\frac{X_L}{R}$
$\Rightarrow X_L=\frac{R}{\sqrt{3}}$
When only the inductance is removed, the phase difference between current and voltage is
$tan30^o=\frac{X_c}{R}\Rightarrow X_c=\frac{R}{\sqrt{3}}$
Since, $X_L=X_c$ , the circuit is in resonance. So the phase angle between voltage and current is $0^{\circ }$ .
$\therefore P_{avg}=V_{rms}{I}_{rms}cos\varphi$
$\Rightarrow P_{avg}=V_{rms}{I}_{rms}cos{0}^{\circ }$
$\Rightarrow P_{avg}=V_{rms}{I}_{rms}$
$\Rightarrow P_{avg}=\frac{V_{rms}^2}{Z}\left[atresonanceZ=R\right]$
$=\frac{{\left(220\right)}^2}{200}=242W$