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

Given:

$$R=200\,\Omega,\qquad V=220\,V,\qquad f=50\,Hz$$

Let the applied voltage be

$$e=E_0\sin\omega t$$

For an RL circuit,

$$i=\frac{E_0}{\sqrt{R^2+X_L^2}}\sin(\omega t-\phi)$$

and

$$\tan\phi=\frac{X_L}{R}$$

Since the current lags by $$30^\circ$$,

$$X_L=R\tan30^\circ$$

Similarly, for the RC circuit,

$$i=\frac{E_0}{\sqrt{R^2+X_C^2}}\sin(\omega t+\phi)$$

Since the current leads by $$30^\circ$$,

$$X_C=R\tan30^\circ$$

Therefore,

$$X_L=X_C$$

So, in the original LCR circuit, the two reactances cancel:

$$Z=\sqrt{R^2+(X_L-X_C)^2}=R$$

Hence,

$$I_0=\frac{E_0}{\sqrt{R^2+(X_L-X_C)^2}}=\frac{E_0}{R}$$

Since $$V=\frac{E_0}{\sqrt2}$$,

$$I=\frac{V}{R}=\frac{220}{200}=1.1\,A$$

The circuit is effectively purely resistive, so

$$P=I^2R$$

$$P=(1.1)^2(200)=242\,W$$

Hence, the answer is $$242\,W$$.