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As a student who is now studying resonance circuits, I wondered why at the resonance frequency, in a series connection, when the capacitor's reactance and the inductor's reactance are equal to each other, a minimum impedance is obtained, but in a parallel connection, when the reactance of the capacitor and the inductor are equal to each other, a maximum impedance is obtained?

What confuses me here is that the same condition exists but different phenomena occur.

One of the conclusions I reached is that this happens because there is a minimal impedance between the capacitor and the inductor (because they have the same reactance but at opposite directions), which will cause most of the current to flow in these branches, so the charge will continue to move between the inductor and the capacitor and will not return to the voltage source, which indicates maximum resistance. I don't know if my conclusion is correct.

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    \$\begingroup\$ Your idea has validity. You might consider exciting a parallel LC with a current source instead of a voltage source, and find the resulting voltage between the two nodes. Do be careful about simulating perfect L and perfect C - voltage may tend ever larger cycle after cycle. In real life, there are always losses that require adding at least one resistor somewhere. \$\endgroup\$
    – glen_geek
    Commented Aug 24 at 0:41

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Very good observation!

In the parallel resonant circuit, the capacitor and inductor do "exchange" energy between themselves, and because their currents cancel out at resonance, the source does not need to supply much current, which translates into a maximum impedance. Remember that there's a 180 degree difference between their phases. On the other hand, in a series circuit, their impedances cancel each other out directly, leading to minimal impedance and maximum current through the circuit because resonance becomes purely resistive.

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  • \$\begingroup\$ Thank you very much, Really appreciate the help. \$\endgroup\$
    – Nir Moyal
    Commented Aug 24 at 12:32

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