Question

How does the sign of the phase angle $\varphi$, by which the supply voltage leads the current in an LCR series circuit, change as the supply frequency is gradually increased from very low to very high values?

Answer 1

Formula used:

$\tan \varphi =\frac{X_L-X_C}{R}$

We know that the phase angle$\left(\varphi \right)$ by which voltage leads the current in LCR series circuit is given by,

$\tan \varphi =\frac{X_L-X_C}{R}=\frac{\omega L-\frac{1}{\omega C}}{R}$

At small frequencies, $X_L<X_c\tan$ is negative, i.e.,$tan\varphi <0(forv<v_0)$


If frequency is large, $X_L<X_c$


So, $X_L-X_C$ is positive, i.e.,

$\tan \varphi >0(forv>v_0)$


For $X_L=X_C$, i.e, at resonant frequency

$\tan \varphi =0(forv=v_0)=\frac{1}{2\pi \sqrt{2C}}$

So, phase angle in series LCR AC circuit will change from a negative to zero at resonance frequency and then from zero to a positive value.