Timeline for Parallel resonant circuit depending on voltage
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 24, 2017 at 14:56 | comment | added | Andy aka | Current at the previous frequency is twice if inductance halves. And current at new resonant frequency is 1.4142 times. | |
| Aug 24, 2017 at 13:59 | comment | added | MaDrung | So, when I reduce the L, the current both at previous frequency and the new higher one is now higher? Bassicaly L acts just the opposite of C in a way. | |
| Aug 24, 2017 at 13:19 | comment | added | Andy aka | ... part 2 here. Frequency has risen by 1.4142 but the inductor impedance has fallen by 2 at the original frequency but will have fallen by 1.4142 at the new frequency so the current taken by the inductor is 1.4142 times higher. Does this clear things up? | |
| Aug 24, 2017 at 13:16 | comment | added | Andy aka | 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.... | |
| Aug 24, 2017 at 12:09 | comment | added | MaDrung | 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? | |
| Aug 21, 2017 at 10:22 | comment | added | Andy aka | @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? | |
| Aug 21, 2017 at 10:14 | comment | added | MaDrung | 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? | |
| Aug 21, 2017 at 9:59 | vote | accept | MaDrung | ||
| Aug 21, 2017 at 9:58 | history | answered | Andy aka | CC BY-SA 3.0 |