using phasor diagram for a series LCR circuit connected to an ac source of voltage V=V`Sinomegat.derive the realtion for the current flowing in the circuit and the phase angle between the voltage across the resistor and the net voltage in the circuit.

Dear Student,

Please find below the solution to the asked query:

When we connect a Resistor, a Capacitor and an Inductor in series with an AC source of emf $V=V_0 \sin \omega t$,

Then as we know the for the three omponents, the voltages and the current phasor diagrams will be as shown in the figure below.



By using phasor diagram, there will not be any derivation for current. We can write by using the phasor diagram shown in the above figure.

Now,  we can write the impedance of the circuit as,

$$\begin{gathered} V=\sqrt{V_R^2+{\left(V_L-V_C\right)}^2} \\ If we consider the net impedence of the circuit is Z, then \\ IZ=\sqrt{I^2{R}^2+I^2{\left(X_L-X_C\right)}^2}\Rightarrow Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2} \\ Where X_L = \omega L is called inductive reac\tan ce and \\ {X}_C=\frac{1}{\omega C} is called capacitive reac\tan ce. \\ Now, we can write the equation for the current as, \\ I=\frac{V}{Z}=\frac{V_0}{Z}\sin \omega t\Rightarrow I=I_0 \sin \omega t \\ Now, if we consider the angle between V_R \& V as \varphi , then from the phasor diagram we can write, \\ \tan \varphi =\frac{X_L-X_C}{R} \end{gathered}$$



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