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When an AC input is applied, the inductor in initial condition (uncharged) does not act as an open circuit, even though the magnitude of instantaneous AC voltage constantly varies.

Why is its behavior different from the switching condition in a DC input circuit even though the AC voltage can be considered as a constantly switching input?

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    \$\begingroup\$ Your assertion is wrong as far as I can tell. An inductor always behaves following a simple well-known formula from Faraday. \$\endgroup\$
    – Andy aka
    Jul 17, 2021 at 9:32
  • \$\begingroup\$ Should that not be DOSE act as an open circuit. Of a step change in voltage is applied to an inductor then initially zero current flows. The worst case in usual circuits is to apply an AC voltage at its' zero crossing - when the inductor will draw maximum current as the voltage increases. \$\endgroup\$
    – Russell McMahon
    Jul 17, 2021 at 12:53

3 Answers 3

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When an AC input is applied, the inductor in initial condition (uncharged) does not act as an open circuit

That's a capacitor you're talking about. An inductor is a short circuit.

even though the magnitude of instantaneous AC voltage constantly varies.

Why is its behavior different from the switching condition in a DC input circuit even though the AC voltage can be considered as a constantly switching input?

An .AC analysis is not done with a time domain waveform that alternates in time, instead a complex signal is used which is flat all across the frequency, but varies in the circuit according to its topology (unless you have pure resistors or similar). So there's no varying signal.

Therefore an inductor's initial conditions have no meaning in an .AC analysis, while in a .DC or .TRAN they do, because those IC influence the initial operating point of the time domain response.

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  • \$\begingroup\$ An inductor is often treated as an open circuit when considering the initial (\$t=0^+\$) response to transient stimulus. And a capacitor is treated as a short circuit for the same conditions. \$\endgroup\$
    – The Photon
    Jul 17, 2021 at 14:13
  • \$\begingroup\$ Also, there's no evidence OP was asking about how SPICE does different analyses. \$\endgroup\$
    – The Photon
    Jul 17, 2021 at 14:15
  • \$\begingroup\$ @ThePhoton OP's title and text with initial conditions lead me to believe it's about some sort of simulation, and since it seems to insist on the AC part, I thought it may be about SPICE (and why I thought there is some confusion about .AC/.TRAN). If it isn't, I'll gladly remove the misleading answer. \$\endgroup\$ Jul 17, 2021 at 14:28
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When an AC input is applied, the inductor in initial condition (uncharged) does not act as an open circuit, even though the magnitude of instantaneous AC voltage constantly varies.

Who says it doesn't?

Here's a simple LR circuit: enter image description here

And here's the response (voltage stimulus in blue, inductor current in green): enter image description here

You can see that even though the voltage jumps to 1 V, the inductor current does not change in the first milliseconds afterwards. That is, the inductor does not respond, on a short time scale, to changes in the voltage applied to it. This is exactly what we mean when we say it "behaves like an open circuit" on short time scales or in the time immediately after a change in the stimulus.

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The time scales for alternating current are different for say a grid supplied AC motor and a 10 kHz pulse with a 1us rise time. The frequency content of the voltage supply depends on this rate of dV/dt which affects the current and inductor impedance as the inductor current will not change instantly with voltage, rather it accumulates or integrates this change slowly according to a time constant of L/R.

The series inductor raises the impedance with frequency content of a step pulse but a low frequency grid voltage has more time to flow.

Thus, a rapid pulse may instantly appear across the inductor but the load R current only begins to change at a ramp rate of dI/dt= V/L for that step voltage. But again, the time scales may differ.

$$V_L=L ~dI/dt$$

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