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 \, \phi=\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
$ \, tan \, 30^{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 $ .
$\therefore P_{a v g}=V_{r m s}I_{r m s} \, cos \, \phi$
$\Rightarrow P_{a v g}=V_{r m s}I_{r m s}cos0^\circ $
$\Rightarrow P_{a v g}=V_{r m s}I_{r m s}$
$\Rightarrow P_{a v g}=\frac{V^{2}_{r m s}}{Z}\left[a t r e s o n a n c e Z = R\right]$
$=\frac{\left(220\right)^{2}}{200}=242W$