In a series $$LCR$$  circuit $$R = 200\Omega $$   and the voltage and the frequency of the main supply is $$220V$$  and $$50 Hz$$  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

Solution :
When capacitance is taken out, the circuit is $$LR.$$
$$\eqalign{ & \therefore \tan \phi = \frac{{\omega L}}{R} \cr & \Rightarrow \omega L = R\tan \phi = 200 \times \frac{1}{{\sqrt 3 }} = \frac{{200}}{{\sqrt 3 }} \cr} $$
Again , when inductor is taken out, the circuit is $$CR.$$
$$\eqalign{ & \therefore \tan \phi = \frac{1}{{\omega CR}} \cr & \frac{1}{{\omega c}} = R\tan \phi = 200 \times \frac{1}{{\sqrt 3 }} = \frac{{200}}{{\sqrt 3 }} \cr & {\text{Now, }}Z = \sqrt {{R^2} + {{\left( {\frac{1}{{\omega C}} - \omega L} \right)}^2}} \cr & = \sqrt {{{(200)}^2} + {{\left( {\frac{{200}}{{\sqrt 3 }} - \frac{{200}}{{\sqrt 3 }}} \right)}^2}} = 200\Omega \cr & {\text{Power dissipated }} = {V_{rms}}{I_{rms}}\cos \phi \cr & = {V_{rms}}\frac{{{V_{rms}}}}{Z}.\frac{R}{Z}\left( {\because \cos \phi = \frac{R}{Z}} \right) \cr & = \frac{{{V^2}_{rms}R}}{{{Z^2}}} = \frac{{{{\left( {220} \right)}^2} \times 200}}{{{{\left( {200} \right)}^2}}} = \frac{{220 \times 220}}{{200}} = 242W \cr} $$