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In a parallel LC circuit, does an applied amplitude of voltage change anything when it comes to choosing the capacitor and inductor (except the voltage rating of course)?

I know that bigger L causes bigger Q, but besides from that, will the same current flow if I use a really small capacitor (speaking about ideal elements now) and a really big inductor at resonant frequency, or are we searching for specific capacitors when it comes to reaching the highest current at given supply voltage (and frequency ofcourse, but we choose L and C appropriately to achieve resonance every time)?

What are the differences when we change the values of capacitance and inductance while still maintaining the same resonant frequency (besides Q factor)?

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    \$\begingroup\$ besides Q factor also impedance (at \$\omega \neq \omega_0\$) changes \$\endgroup\$ – Curd Aug 21 '17 at 10:06
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For a given applied voltage (\$V_A\$/\$X_L\$) across a parallel resonant LC circuit that is tuned to resonate, the current that flows in the inductor is \$V_A\$/\$X_L\$ and the current that flows in the capacitor is \$V_A\$/\$X_C\$ AND, importantly these two currents are identical in magnitude at resonance.

However they are opposite in sign hence the net current taken from \$V_A\$ is zero amps (after the initial transient has settled down). But, there are still currents circulating between L and C of equal magnitude.

I know that bigger L causes bigger Q, but besides from that, will the same current flow if I use a really small capacitor (speaking about ideal elements now) and a really big inductor at resonant frequency

No, the value of current (although not taken from the supply at resonance) is determined by that supply and either \$X_L\$ or \$X_C\$.

What are the differences when we change the values of capacitance and inductance while still maintaining the same resonant frequency (besides Q factor)?

If you double L and halve C to achieve the same resonant frequency, then the circulating current halves. If you halve L and double C then the circulating current doubles.

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  • \$\begingroup\$ Thank you!! :) So we choose L in a way to make sure that C fills up to the same peak voltage as voltage across L does in any cycle. If we increase C, we have to reduce L to allow more current to flow so the C still fills up to the same voltage? If this is true, doesn't then make sense to choose even smaller L than that to fill the capacitor to even higher voltage which would further increase the current? \$\endgroup\$ – MaDrung Aug 21 '17 at 10:14
  • \$\begingroup\$ @MaDrung I'm unsure what you mean by "fill up". Generally every application is different - if you have freedom to choose L and C then both components take the same magnitude of current at resonance and at resonance, the applied voltage has the minimum current taken from it. Maybe you need to specify a particular application? \$\endgroup\$ – Andy aka Aug 21 '17 at 10:22
  • \$\begingroup\$ If I keep the same voltage amplitude and C and just reduce the L, the resonant frequency increases. But what happens to resonating current? Does it increase when I'm at the new resonance compared to when I was at the old resonance with previous system, or does it stay the same? \$\endgroup\$ – MaDrung Aug 24 '17 at 12:09
  • \$\begingroup\$ With L taken to a lower value and the applied frequency moved to the new higher resonant frequency you still have an AC voltage across an inductor and that inductor will still take current. Because L is smaller (say half) then the resonant frequency will have increased by \$\sqrt2\$. Are you with this so far.... part 2.... coming.... \$\endgroup\$ – Andy aka Aug 24 '17 at 13:16
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    \$\begingroup\$ Current at the previous frequency is twice if inductance halves. And current at new resonant frequency is 1.4142 times. \$\endgroup\$ – Andy aka Aug 24 '17 at 14:56
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A useful concept is the impedance, given by \$\sqrt{\frac{L}{C}}\$, though most oscillator designers will talk about the 'L to C ratio' of their resonator.

If you scale the loss resistance, whether external source or load, or internal to the inductor, by the impedance, the Q will stay the same.

The peak capacitor voltage is the impedance * the peak inductor current.

Generally, our external loads are fixed, and we find that we get best Q from pushing the impedance in one direction or the other.

If we are going for a high power application, and find we run out of drive current before voltage handling, then we'd increase the impedance, and vice versa.

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