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What I understand(FACTS) :

In for example a LC resonance circuit the output can many times greater than the input voltage. At resonance the impedance of the inductor is the same as the impedance of the capacitor.

At the resonance frequence Xl=Xc

What I fail to understand

If the impedance's are equal, how can the output voltage be larger than the input voltage? Although not correct, you would almost start to reason that the output voltage is half the input voltage(voltage divider). Where the output voltage is the voltage across the capacitor.

It seems so counter intuitive.

Why is the voltage higher at the output? What is a intuitive way of explaining why the output voltage is higher than the input voltage. Or is it one of those circuit where it only starts to make sense when you deriver the transfer function. source elprocus.com

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  • \$\begingroup\$ first, define where in your circuit your "output voltage" is being measured. Then mathematically (with an equation) define what you meant when you said "the impedances must be equal". Being exact here makes a difference! \$\endgroup\$ – Marcus Müller Oct 30 '18 at 20:51
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    \$\begingroup\$ The impedances aren't equal they are complex conjugates at resonance. Big difference \$\endgroup\$ – sstobbe Oct 30 '18 at 20:59
  • \$\begingroup\$ In the circuit you show, the voltage across C at resonance is not just large, it is infinite. Or equivalently: without any R, the peak voltage across C will be higher on each cycle, ramping upwards without limit. \$\endgroup\$ – wbeaty Oct 31 '18 at 0:59
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Why is the voltage higher at the output? What is a intuitive way of explaining why the output voltage is higher than the input voltage. Or is it one of those circuit where it only starts to make sense when you drive the transfer function.

Inductors and capacitors store energy over time, this means that the voltage or current can get higher than the initial voltage or current:

enter image description here Source: libretext: 14.5: Oscillations in an LC Circuit

This only happens at the resonant point formed by the capacitor and inductor, the circuit wants to oscillate at a certain frequency (the Q point), and will do so. The energy in the system is preserved but you can get higher voltage.

This property of inductors is how a DC to DC converter works, we charge inductors to create a magnetic field, then switch them off which momentarily creates a higher voltage.

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  • \$\begingroup\$ your answer shows the oscillation between the energy in the inductor and capacitor but does not give any insight of why the output voltage is higher than input voltage. The fact that energy is stored in the capacitor over time is a bit confusing w.r.t. the picture as it shows a oscillating energy. \$\endgroup\$ – Navaro Oct 30 '18 at 22:51
  • \$\begingroup\$ @Navaro, calculate the output voltage at resonance using the voltage divider rule. What do you get? \$\endgroup\$ – The Photon Oct 31 '18 at 5:55

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