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Suppose switch is closed at t = 0. Right after the switch is closed there will be the voltage rise from the ac source. As the voltage rises from the ac source there should be a corresponding rise in the current through an inductor during the first positive half cycle of an ac voltage, as there is also a change in voltage during the first positive half cycle of the voltage from the ac source.

Then why in the graph we see no change in current in the inductor during the first positive half cycle rise of ac voltage from the source?

enter image description here

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    \$\begingroup\$ The graph is only approximate. Try this circuit in a simulator and see what happens. \$\endgroup\$
    – The Photon
    Apr 15, 2020 at 1:39
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    \$\begingroup\$ There was a question about a similar situation recently. \$\endgroup\$
    – The Photon
    Apr 15, 2020 at 1:43
  • \$\begingroup\$ So the positive half cycle of the voltage is the time for the transient? \$\endgroup\$
    – Alex
    Apr 15, 2020 at 1:51
  • \$\begingroup\$ can you please explain that initial transient response intuitively without going through the diffrential equations? @the photon \$\endgroup\$
    – Alex
    Apr 15, 2020 at 2:07
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    \$\begingroup\$ Try it in a circuit simulator and see what happens. Also try adding a resistor in series with the inductor (maybe just a fraction of an ohm). \$\endgroup\$
    – The Photon
    Apr 15, 2020 at 2:50

2 Answers 2

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The graph as drawn is incorrect, or at least is only correct if we assume a switch in the circuit as well, being closed at wt=pi/2.

It's convenient when dealing with real world AC waveforms to split it into two parts, the initial transient, and the long term behaviour. The author is unfortunately ignoring the initial transient, and plotting the steady state current, which does not match the voltage waveform shown.

He could have saved you much confusion by simply starting his graph at the peak of the signal, V=cos(t). Then the initial transient would have been zero, and so could safely have been ignored.

Plotting the initial transient and the long term behaviour separately gives you this

enter image description here

The total voltage is in purple, sin(wt), starting from 0. This is what's shown in your diagram as the voltage.

We can split out the yellow curve, Vss for steady state, which starts at pi/2, when the voltage is at its peak, when we close our switch. You can see the green Iss current waveform rising quickly when the voltage is high, staying level when the voltage is zero, and falling when the voltage is negative. This is the current waveform drawn in your diagram. Crucially, at the next voltage peak at 5pi/2, the current has returned to zero. The voltage and current waveforms continue repeating every 2pi from here. They both have an average of zero.

What your diagram ignores is the initial voltage Vi in dark blue, the bit from wt=0 to pi/2. As it's positive, the initial current Ii, orange waveform, increases from t=0. At pi/2 when I throw my switch, Vi goes to zero, and Ii now continues indefinitely at the same value. Because there is no resistance in the circuit, there is no voltage drop across any resistance to reduce the current. Note that in a real circuit, there would be some resistance, and this initial transient current would die away to zero, with a time constant of L/R.

Finally we can add the partial solutions together to get the final result, Vtotal = Vi + Vss, and Itotal = Ii + Iss.

You'll note the total current is oscillatory, but has a a positive offset. In the real world, this offset would decay due to finite resistance. In the ideal world of your diagram, this initial transient persists indefinitely, as it would in a superconductor.

This diagram illustrates quite nicely why transformers and inductors have an 'inrush' current and need time delay fuses. The 'design' current of an inductor or transformer primary will be the green Iss curve. You'll notice that the peak of the light blue Itotal curve is twice the Iss peak, which usually exceeds the saturation current of the core, leading to a dramatic loss of inductance, and so a large further increase in current.

You may think that both the Iss and the Itotal curves continue indefinitely, and you'd be right. So why are we splitting out a transient and a steady state case? The answer is that in the real world, resistance causes the initial transient to decay, and so eventually at large t, regardless of the switch-on phase which determines the transient, the transient decays away and all current curves end up as the Iss curve.

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  • \$\begingroup\$ can you please explain that initial transient response intuitively without going through the diffrential equations? \$\endgroup\$
    – Alex
    Apr 15, 2020 at 4:25
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    \$\begingroup\$ @Alex Not really. Intuition only dawns after years of drawings graphs, going through differential equations, and getting a feel for how the world works. How about visually, which is what I've tried above? \$\endgroup\$
    – Neil_UK
    Apr 15, 2020 at 9:25
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    \$\begingroup\$ @Alex if you start from exactly V=0, with a perfect inductor, then yes. If you start from peak V, then as shown it will oscillate above and below zero current. The magnitude of the offset, whether positive, negative, or zero, depends on the voltage starting phase. \$\endgroup\$
    – Neil_UK
    Apr 18, 2020 at 15:45
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    \$\begingroup\$ @Alex Not sure what you're talking about here. The initial voltage in dark blue, I explcicitly show the current rising as a result of that. The current is the integral of the voltage, the general equation being I(t) = integral_from_0_to_t(v(t))dt+constant_of_integration. In this case with v(t) = sin(wt), I(t) will be an offset cosine(t), note the constant of integration offsets the light blue current trace, whereas it doesn't offset the green current trace, because of different starting boundary conditions. \$\endgroup\$
    – Neil_UK
    Apr 26, 2020 at 8:54
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    \$\begingroup\$ The rate of change of current is proportional to the voltage. In the first positive half cycle, the voltage goes from zero, to max, back to zero again, and the rate of change of current (not the current, but the rate of change of current) goes from zero (flat), to max (steeply rising), and back to zero again (flat at the top). \$\endgroup\$
    – Neil_UK
    Apr 26, 2020 at 11:19
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The waveform is wrong. You are correct, the current should rise during the first half cycle. When the power is switched on, there is an initial response that is different from the steady state response. That is not shown in this case.

To see what it looks like for switch-on at voltage zero crossing, look at the answer by @Andy aka here: What is the difference beteeen a solenoid's make current and inrush current?

Image from that answer:

enter image description here

The current starts to rise from zero as soon as the voltage begins to rise from zero. Since there is no reverse current at the start, the current rises above the normal steady state AC current. It keeps rising until the voltage reaches zero. Since the current rose higher than normal, the reverse voltage does not take the negative peak as low as it will at steady state. The current is a sine wave, but it has a DC offset that lasts for some period of time. The decay of the offset assumes that this is a real circuit with losses rather than and ideal circuit with no losses as shown in the diagram question.

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  • \$\begingroup\$ "When the power is switched on, there is an initial response that is different from the steady state response. That is not shown in this case." can you please explain that initial response intuitively without going through the diffrential equations? @charles cowie? \$\endgroup\$
    – Alex
    Apr 15, 2020 at 2:06
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    \$\begingroup\$ Revised answer re comment. \$\endgroup\$
    – user80875
    Apr 15, 2020 at 2:45
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    \$\begingroup\$ +1 but one thing to mention is that in the OP's circuit (with no lossy element), the transient will actually last forever, Output voltage will always be > 0, and will never decay to a sine without offset. \$\endgroup\$
    – The Photon
    Apr 15, 2020 at 16:26
  • \$\begingroup\$ Thank you @ThePhoton; edited \$\endgroup\$
    – user80875
    Apr 15, 2020 at 17:26
  • \$\begingroup\$ can you please give me the general equation of this initial current for this case .? . during the first positive half cycle ? \$\endgroup\$
    – Alex
    Apr 26, 2020 at 7:47

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