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In an AC circuit, the reactance of capacitive and inductive elements are determined by the frequency of the AC signal. When we try to emulate DC analysis by putting $\omega$ = 0 , the capacitive reactance becomes infinite, and the inductive reactance goes to zero. We see that the capacitor fully impedes flow of current in the circuit, while the inductor essentially becomes short. This makes it seem that the capacitor instantaneously opens the circuit, while inductors seem to immediately make themselves short.

However, in an ideal DC circuit, we know that the capacitor never completely 'opens' the circuit, and the inductor never completely 'allows' current to flow. There is an exponential function obtained; we actually get the answers we obtained from AC formulae only at infinite time. Why does this happen? Why is the time delay thing missing in AC analysis? Why does AC not carry this information?

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When we try to emulate DC analysis by putting $\omega$ = 0 , the capacitive reactance becomes infinite, and the inductive reactance goes to zero.

That is correct, but it only applies to steady state conditions, that is, where $\omega$ = 0 for a long time. If the frequency is zero for a long time, it is the equivalent of saying you constant voltages and currents in the circuit. And under those conditions the ideal capacitor looks like an open circuit and the ideal inductor looks like a short circuit.

This makes it seem that the capacitor instantaneously opens the circuit, while inductors seem to immediately make themselves short.

That don't do so instantaneously, that is, at the instant the frequency is zero. As I said above, it only applies after a long time when transient voltages and currents no longer exist.

However, in an ideal DC circuit, we know that the capacitor never completely 'opens' the circuit, and the inductor never completely 'allows' current to flow. There is an exponential function obtained;

The exponential conditions only exist in a DC circuit typically immediately after a switching event. Not under steady state conditions.

For circuits involving DC voltage and or DC current sources, there basically three stages (1) steady state, (2) conditions immediately after switching, and (3) transient conditions between switching and steady state.

  1. The circuit conditions have existed for a long time so that there are no longer any transient (changing in time) currents and voltages. Under these conditions, the current through and ideal capacitor is zero and the voltage across an ideal inductor is zero.

  2. A switching event occurs. The event occurs at $t=0$. At the instant after switching, whatever the voltages across ideal capacitors or currents through ideal inductors were before the switching event will be the same the instant after the event. You can't change the voltage across an ideal capacitor instantaneously and you can't change the current through an ideal inductor instantaneously. These rules come directly from the basic current voltage relationships for the capacitors and inductors.

  3. Following the switching event, currents can start to change in inductors and voltages can start to change across capacitors. It is during this "transient" period that exponential relationships between voltage and current can exist.

  4. After a long time following (3), that is when t goes to infinitely, the exponentials disappear. Voltages and current once again become constant and the steady state rules in (1) above return.

What I am trying to ask is why the transient state thing seems to be missing in AC analysis.

As @The Photon and @Samuel Wier have pointed out, in ac circuits one is usually only interested in steady state conditions, that is, any switches in the circuit are closed a long time. But if there were switching in the circuit, there would be transients. The following link gives an example of a series RL circuit switched onto an ac source:

http://www.robots.ox.ac.uk/~gari/teaching/b18/background_lectures/1P2-Circuit-analysis-II-Notes-Moore.pdf

The circuit response in the example is the solution to a first order linear differential equation. The solution is composed of the transient response, which vanishes after a long time, and the steady state response.

Hope this helps.

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  • $\begingroup$ I know the Physics behind this, having studied these in somewhat detail in my 12th, so I am not unaware of transient and steady state conditions. What I am trying to ask is why the transient state thing seems to be missing in AC analysis. Is it that the double differential equation (in AC circuits) that forms gives i(t) = v(t)/z only under steady state conditions? (z here denoting impedance) $\endgroup$
    – Aabesh Ghosh
    Commented Sep 10, 2019 at 17:07
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    $\begingroup$ @AabeshGhosh - The AC analysis assumes that enough time has gone by that any transients due to the exact starting conditions have died out. So to make a proper comparison to DC conditions, you should have done a comparison to a DC situation in which transients have also long died out. But you didn't. Instead, you tried to compare it with a condition shortly after the DC voltage changed (presumably from 0 voltage) and time-dependent transients still existed. This is what Bob D was pointing out to you. $\endgroup$
    – user93237
    Commented Sep 10, 2019 at 19:15
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    $\begingroup$ @AabeshGhosh - Also, as for your question why consideration of transients is missing in the AC analysis, oftentimes the details of the transients during AC startup (or DC startup) are unimportant because one is usually interested in times long after the startup. $\endgroup$
    – user93237
    Commented Sep 10, 2019 at 19:18
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    $\begingroup$ @AabeshGhosh, RE "why the transient state thing seems to be missing in AC analysis.", it's because AC analysis is a steady state analysis. It's useful despite neglecting transient effects because it's often simpler to calculate and comprehend than a transient analysis. If you haven't already, you'll probably soon learn Laplace analysis, which bridges the gap between the two (but which I have never actually used in practice in a 20 year engineering career). $\endgroup$
    – The Photon
    Commented Sep 10, 2019 at 21:14
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    $\begingroup$ @AabeshGhosh In response to your comment to me, I have revised my answer to include a link which provides an example of the transient response in an ac circuit. Hope it helps. $\endgroup$
    – Bob D
    Commented Sep 11, 2019 at 15:21

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