A series LCR circuit is connected to an A.C source of voltage V and angular frequency w.(omega.) When only the capacitor is removed, the current lags behind the voltage by a phase angle 'phi' and when only the inductor is removed , the current leads teh voltage by tha same phase angle. Find the current flowing and the average power dissipated in teh LCR circuit.? Please REply fast Please.

Given
Voltage in the circuit is $V=V_0 \sin \omega t$ (say)

When only capacitance is removed, the circuit will be LR, in which phase change θ between voltage and current is given by
$\tan \theta =\frac{\omega L}{R} --\left(1\right)$
When only inductance is removed, the circuit will be CR, in which phase change θ (since phase change is same) between voltage and current is given by
$\tan \theta =\frac{1}{\omega CR} --\left(2\right)$
From Eqs. (1) and (2) we get
$\frac{\omega L}{R}=\frac{1}{\omega CR}\Rightarrow \omega L=\frac{1}{\omega C}--\left(3\right)$
In LCR Circuit
Phase change between voltage and current is given by
$\tan \varphi =\frac{{\displaystyle \frac{1}{\omega C}}-\omega L}{R}=0$[using eq.(3)]  ---(4)
Current in the circuit is given by
$$\begin{gathered} i=\frac{V_0 \sin \left(\omega t+\varphi \right)}{\sqrt{R^2+{\left({\displaystyle \frac{1}{\omega C}}-\omega L\right)}^2}} \\ From eq. \left(3\right) and \left(4\right) we get \\ i=\frac{V_0 \sin \left(\omega t\right)}{R}=\frac{V}{R} Answer \end{gathered}$$

Average power dissipated in the circuit is given by
$$\begin{gathered} P_{avg}=\frac{1}{2}{V}_0{i}_o \cos \varphi \\ From eq \left(4\right) , we have \\ \tan \varphi =0\Rightarrow \cos \varphi =1 \\ \Rightarrow P_{avg}=\frac{1}{2}{V}_0{i}_o \cos \varphi =\frac{1}{2}{V}_0{i}_o =\frac{1}{2}\frac{V_0^2}{R} Answer \end{gathered}$$