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Below circuit is at resonance. Shouldn't the circuit offer infinite impedance to the voltage source ? I thought, then the current through the source would be 0A. But the simulation shows that the circuit draws a constant current of -1A. Why is this so ?

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

EDIT : As suggested, with 0.1 ohm resistor in series with the inductor, the current slowly dies out

enter image description here

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    \$\begingroup\$ Two main things. (1) Wait longer for the simulation to settle (milliseconds.) (2) Set the frequency to 159.155kHz. And if you want, give the inductor a .IC command perhaps? \$\endgroup\$
    – jonk
    Jun 9, 2018 at 7:23
  • \$\begingroup\$ @jonk that was my first thought too and I kept a delay of 100u. If I increase the delay to ms, simulation might take longer. I'll give it a try.. ty :) \$\endgroup\$
    – AgentS
    Jun 9, 2018 at 7:26
  • \$\begingroup\$ Pardon my ignorance byt may I ask whats .IC command for the inductor ? \$\endgroup\$
    – AgentS
    Jun 9, 2018 at 7:27
  • \$\begingroup\$ @jonk I've delayed the simulation upto 10ms. The circuit is still drawing the same -1A. Perhaps the simulation tool itself has an issue ? \$\endgroup\$
    – AgentS
    Jun 9, 2018 at 7:29
  • \$\begingroup\$ Sorry you meant initial conditions, got you :) I don't know how to set it in circuit lab, I'll google bit \$\endgroup\$
    – AgentS
    Jun 9, 2018 at 7:35

3 Answers 3

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Why is this so?

This is because at the start of the simulation (take a look during the first 50 \$\mu\$s for instance) the sinewave is positive for 2 quarts of a cycle. During this period the coil is 'charged' to 2 A. If you'd have applied a cosine (I think you can't in this CircuitLab), you would of course have seen an infinite current spike in the capacitor but also that in the first quarter cycle the coil's current would go to 1 A, then in the next back to zero, then go to -1 A, then 0 etc. etc. In other words: your expected behavior.

This is all a matter of bad timing, the initial build up of flux in the coil is not compensated with a braking down of it because of the two positive quart cycles at the start of the simulation.

The same 'problem' appears when you connect a transformer to the grid. Done at the 'wrong' time, i.e. at zero voltage crossing, the inductor will be 'charged' to the normal operating amplitude, then however 'overcharged' to double the peak flux during normal operation. Mostly that's a region of heavy saturation, associated with a huge inrush current which can reach factors more than the mere double of the current amplitude during normal operation.

In your case however there's no resistance in the circuit, so this 'charge' will never be consumed and from the source will be drawn a constant 1 A, instead of average 0 A. Add a small resistor of 0.01 \$\Omega\$ in series of L and then see what happens.

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  • \$\begingroup\$ Awesome! With that series resistor the current is slowly averaging to 0A (pls see edit in the question for a snapshot). So if I understand your explanation correctly, in the first half of the cycle, the inductor is getting charged way more than it normally should and with out a resistor it maintains that constant additional energy flowing in the circuit. I feel I get this. Thank you so much :) \$\endgroup\$
    – AgentS
    Jun 9, 2018 at 13:51
  • \$\begingroup\$ Exactly, it will get charged to double the amplitude of the 'regular' flux. In a normal transformer (with resistance) that's connected at the time that the voltages 'goes through' 0, the same happens, but the DC component (transient) will be consumed after a few cycles. (If you accept this as the answer you could kindly flag it as such. ;) \$\endgroup\$
    – HarryH
    Jun 10, 2018 at 12:05
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C1 takes zero current, as you'd expect.

L1 takes an offset current, because of the phase the voltage happened to start.

The DC current through V1 is correctly shown as the sum of these two components.

If you want this DC current to 'settle down' to zero as time goes on, then you need to add a small resistance in series with the L. This will generate a voltage drop due to the current flow, this voltage will steer the DC current down to zero, with a time constant of L/R. At the moment, the L is ideal, and this persistent DC current is simply ideal behaviour.

The classical mathematical solution to this sort of problem contains both steady state and transient terms. The oscillatory behaviour is steady state. The 1A DC is a transient. Normally the transient would die out with an L/R time constant. In your ideal case with R=0, L/R is infinity, and the transient does not die out.

Delaying the start of the simulation won't necessarily change the start phase of the sinewave from V1. Depending on what simulator you're using, you should be able to change this in the V1 source parameters directly. If you have a cosine option, this should give you zero offset current. Bonus marks for why sine that starts at zero gives you an offset current, and cosine that starts at maximum voltage will give you zero offset current.

Even if you apply a cosine voltage waveform, the transient behaviour still last indefinitely, it's just that the transient has value zero.

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  • \$\begingroup\$ \$V = \sin(t) \$ gives \$i = 1-\cos(t)\$. Clearly delaying the start will not help. The constant term doesn't go away with out energy dissipating element in the circuit. If we add a resistor, that constant term changes to a decaying exponential. Awesome! I should probably keep more faith in math haha.. Thank you for opening my eyes :) \$\endgroup\$
    – AgentS
    Jun 9, 2018 at 13:59
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For an inductor: -

$$V = L\dfrac{di}{dt}$$

And, it follows from this that if you apply a voltage sinewave that is rising through 0 volts at the instant you apply it, the rate of change of inductor current must be zero.

This means that the current begins at the positive peak of a sinewave. But, because that current cannot rise instantly in an inductor, the waveform of current has to be a wholly positively biased AC sinusoidally shaped waveform starting at zero amps (t = 0) and rising to twice the peak current compared to the peak current attained if the voltage had been applied at a peak.

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  • \$\begingroup\$ I see now how the current through inductor depends on the phase with which the voltage is applied. Kinda amazing how precisely the inductor honors its math equations! \$\endgroup\$
    – AgentS
    Jun 9, 2018 at 13:34
  • \$\begingroup\$ @rsadhvika that formula is the bottom line for inductors. Everything else about inductors is based on this relationship. \$\endgroup\$
    – Andy aka
    Jun 9, 2018 at 14:38

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