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If we connect an AC source with a pure inductor when AC source is going through 0 volts we get the following graph of the voltage and current.

There are few things in the graph about which i am confused.

In the first graph/diagram "A" of the picture which shows the voltage when it starts to flow and the corresponding current, the current doesnt seem to be lagging behind the voltage by 90 degrees. What is the reason behind that? Is the statement "current lags behind the voltage in an inductor BY 90 degree" is valid just for the steady state voltage shown in graph "B"?

Second confusion is that for the first half cycle of the voltage in graph "A", current is rising both for increasing and decreasing voltage. How is this possible that current is rising in the inductor for both increasing and decreasing voltage?

enter image description here

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    \$\begingroup\$ Can you plot a voltage across a capacitor when we connect an empty capacitor across an AC current source \$I= I_O \: sin \: \omega t\$? One cycle will be enough. \$\endgroup\$
    – G36
    Apr 26, 2020 at 20:23
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    \$\begingroup\$ We have a DC offset because of the current is starting from 0A and the voltage too at power-up (initial condition). \$\endgroup\$
    – G36
    Apr 26, 2020 at 20:36
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    \$\begingroup\$ But can you tell me fist do you understand what in this circuit i.stack.imgur.com/IWAIg.png the voltage across the capacitor keeps rising despite the fact that the current decreases for t > 10ms? \$\endgroup\$
    – G36
    Apr 26, 2020 at 20:53
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    \$\begingroup\$ Yes, good answer. And exactly the same thing happening in the inductor during positive -cycle. This is why the current in the inductor is rising all the time during a positive-cycle. And remember that the inductor is the opposite of a capacitor. And stores the energy in the form of a magnetic field. And because of this an inductors are not so intuitive to understand as capacitors is. \$\endgroup\$
    – G36
    Apr 26, 2020 at 21:08
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    \$\begingroup\$ Yes, this is exactly what I meant. See with the Capacitor version (capacitor driven by current source ) i.stack.imgur.com/cFxy8.png \$\endgroup\$
    – G36
    Apr 26, 2020 at 21:23

2 Answers 2

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Actually, the current is lagging the voltage by 90 degrees throughout the entire graph. What is confusing you is that the current has a DC offset. This DC offset is actually the 'transient' solution, and it does not die off because there is no resistance (dissipation) in this circuit.

Second confusion is that for the first half cycle of the voltage in graph "A", current is rising both for increasing and decreasing voltage. How is this possible that current is rising in the inductor for both increasing and decreasing voltage ?

This is the behavior you expect because of the relationship between voltage and current:

$$V = L\frac{dI}{dt}$$

So if V is positive, \$dI/dt\$ is positive, and therefore the current I is rising. I is increasing any time V is positive, not when \$dV/dt\$ is positive.

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  • \$\begingroup\$ thank you. can you please give me the intuition that how will this transient die off when resistance is added in this circuit ? \$\endgroup\$
    – Alex
    Apr 26, 2020 at 20:06
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    \$\begingroup\$ @Alex There would be many ways of describing intuitively how a resistor would cause the dc current to die off. One way to think of it is that the voltage source is pushing current around symmetrically -- positive and negative. If the current has a dc offset, then a resistor pushes 'back' on the current asymmetrically, since a resistor always pushes 'against' the current. Eventually the dc goes away until the resistor is pushing symmetrically too, since the current going through it is symmetric. Not a great intuitive description, but maybe that will help. \$\endgroup\$
    – rpm2718
    Apr 26, 2020 at 20:22
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    \$\begingroup\$ @Alex If it had a resistance, you would see the current waveform start off with an offset, but then that offset would decay until it is gone....i.e. you reach the 'steady state' solution. \$\endgroup\$
    – rpm2718
    Apr 26, 2020 at 20:28
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    \$\begingroup\$ @Alex The offset is there because of the initial conditions (V=0, I=0). If you were to start it off with just the right conditions, such as V=0, I=-\$1/\omega L\$, there would be no offset (transient solution). \$\endgroup\$
    – rpm2718
    Apr 26, 2020 at 20:32
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    \$\begingroup\$ @Alex \$\omega\$ is frequency of your voltage sine wave, L is the value of your inductor. \$\endgroup\$
    – rpm2718
    Apr 26, 2020 at 20:37
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Yes, the inductor current lagging the voltage by 90 degreees in an AC circuit is valid for steady-state. Any transient solution has to die out before the steady-state solution dominates.

An inductor follows the relationship: $$\ V=L*di/dt $$ Therefore the current is proportional to the integral of the voltage across it. Increasing or decreasing, as long as the polarity is the same the inductor will "accumulate" current.

When the polarity of the applied voltage flips, you can see that the rate of change of current is zero at the zero crossing of the voltage.

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  • \$\begingroup\$ "Any transient solution has to die out before the steady-state solution dominates." how will transient die in this case. normally resistance do the job. but can you please give me an intuition that how resistor will finish transient in this case ? \$\endgroup\$
    – Alex
    Apr 26, 2020 at 19:04
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    \$\begingroup\$ Well in the case of a pure inductance and ideal source you're right, the inductor will "remember" the transient and it could be an offset to the steady-state solution. Steady-state by definition means "after any transient solution has died out". If the transient solution persists due to zero damping (no resistance) then you will never reach steady-state. (See the third graph from the top in your question.) \$\endgroup\$
    – John D
    Apr 26, 2020 at 19:17
  • \$\begingroup\$ can you give me an intuition that how damping will happen in this case ? i am not getting it. \$\endgroup\$
    – Alex
    Apr 26, 2020 at 19:34

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