# Why isnt the angle between EMF and LC impedance 90 degrees in AC circuit?

I have a generic LC circuit where the inductor and capacitor are in series and we have alternating EMF. I'm trying to find the the impedance of the circuit with phasors.

A phasor diagram shows inductive reactance $$\frac{\pi}{2}$$ anticlockwise from the EMF phasor (taken as reference with it pointing along "+x axis") and capacitive reactance $$\frac{\pi}{2}$$ clockwise from EMF phasor. I subtract the 2 reactances since they are parallel (**I mean the phasors are anti-parallel in the phasor diagram) and so I believe that whichever reactance is more dominant, the phase difference the impedance makes with the EMF phasor is the same as that for the more dominant reactance, i.e. either $$+\frac{\pi}{2}$$ or $$-\frac{\pi}{2}$$. But I came to find out that it is actually:$$\theta=\tan^{-1}(|\omega l-\frac{1}{\omega C}|)$$

I cannot reason why. Why is the phase difference as such?

• Your argument is correct, the angle between voltage on the system and current in it is plus or minus 90 degrees ($\pm \pi/2$). Where did you find out that formula for phase shift? It seems wrong for different reason as well: generally, argument of $\tan^{-1}$ should be dimensionless quantity, not quantity having dimension of Ohm (which $\omega L$ has). May 4, 2022 at 13:52
• Some videos online. I'll double check other sources. Thanks for confirmation May 4, 2022 at 13:58