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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?

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  • $\begingroup$ 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). $\endgroup$ May 4, 2022 at 13:52
  • $\begingroup$ Some videos online. I'll double check other sources. Thanks for confirmation $\endgroup$
    – Doobius
    May 4, 2022 at 13:58

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It's not clear whether they are in series or parallel, because the first paragraph says they are series and the second paragraphs says they are in parallel. If they are in series, add the reactances to get the impedance. If they are in parallel, take the reciprocal of the sum of the reciprocals (i.e., the sum of the susceptances). Either way, you will get an impedance that is either 90 or -90 degrees.

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  • $\begingroup$ I meant to say that the Phasors representing inductive and capacitive reactance are parallel (specifically anti-parallel) in the phasor diagram. The actual inductor and capacitor are in series in the circuit. Edited my q. But yeah, thanks for the confirmation. $\endgroup$
    – Doobius
    May 4, 2022 at 13:18

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