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In a purely Inductive AC Circuit, the voltage/emf Leads the current in phase by 90. The graph between Voltage and Current is given below:

enter image description here

Now can anyone here please tell me in a more intuitive way how the current is varies sinosuidally with emf in every quarter of cycle? Also how Inductor stores Energy in its Magnetic Field?

I don't need any Mathematical Explanation. Thanks...

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  • \$\begingroup\$ You might want to search the Physics SE site and see if this question has already been asked. \$\endgroup\$ – The Photon Feb 23 '17 at 16:54
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The voltage across an ideal inductor is \$V_L=L\dfrac{di}{dt}\$ or we can sometimes use this approximation \$ V_L = L*\frac{ΔI}{Δt}\$
This equation indicates that inductance voltage depends not on current which actually flows through the inductance, but on its rate of change.This means that to produce the voltage across an inductance, the applied current must change. If the current is kept constant, no voltage will be induced, no matter how large the current. Conversely, if it is found that the voltage across an inductance is zero this means that the current must be constant but not necessary zero. enter image description here

In summary: When the current is increasing di/dt > 0 The voltage across the coil VL must be positive because L times a positive number yields a positive voltage.

When the current is decreasing di/dt < 0, so V must be negative because L times a negative number yields a negative voltage.

When we have no change in current over time then we cant have any voltage V = L*di/dt = L * 0 = 0.

Knowing all of this, you can now use this V = L * dI/dt equation and this picture to better understanding Inductor AC behavior.

enter image description here

Or maybe this will help you

Why is there any voltage even present across the inductor? We always accept a voltage across a resistor without argument because we know Ohm’s law (V = I × R) all too well. But an inductor has (almost) no resistance it is basically just a length of solid conducting copper wire (wound on a certain core). So how does it manage to “hold-off” any voltage across it? In fact, we are comfortable about the fact that a capacitor can hold voltage across it. But for the inductor, we are not very clear! A mysterious electric field somewhere inside the inductor! Where did that come from? It turns out, that according to Lenz and/or Faraday, the current takes time to build up in an inductor only because of ‘induced voltage.’ This voltage, by definition, opposes any external effort to change the existing flux (or current) in an inductor. So if the current is fixed, yes, there is no voltage present across the inductor, it then behaves just as a piece of conducting wire. But the moment we try to change the current, we get an induced voltage across it. By definition, the voltage measured across an inductor at any moment (whether the switch is open or closed) is the ‘induced voltage.’

http://booksite.elsevier.com/samplechapters/9780750679701/9780750679701.PDF (from page 22 Understanding the Inductor)

How does an inductor store energy?

EDIT

To know at which "phase" the inductor is we must look at the current. What the current is doing at a given moment. Inductor stores energy in form of magnetic field. And the inductor is fully charged when IL=I_max and VL = 0V. Discharging phase ends when IL = 0A and VL=V_max.

enter image description here

So, from 90 to 180 degrees the inductor current is rising and ends at IL_max. This must be the Charging Phase.

From 180 to 270 degrees we have Discharging Phase.

From 270 to 360 degrees we have a Charging Phase but in the opposit direction.

0 to 90 degrees we have a Discharging Phase.

We can also look at instantaneous power, the product of the instantaneous voltage and the instantaneous current.

enter image description here

The positive power means that we are "absorbing" power from the source(circuit), the charging phase.

Negative power means that the inductor is releasing power back to the source(circuit), discharging phase.

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  • \$\begingroup\$ sorry for the late reply.@G36 The way you showed the Charging and Discharging Phase of Capacitor in my previous question, can you do the same for the inductor? \$\endgroup\$ – Perspicacious Feb 25 '17 at 9:07
  • \$\begingroup\$ @MritunJay I update my answer. \$\endgroup\$ – G36 Feb 25 '17 at 12:29
  • \$\begingroup\$ how can an inductor discharge its current at the very beginning of the cycle? \$\endgroup\$ – Perspicacious Feb 26 '17 at 4:41
  • \$\begingroup\$ @MritunJay Do not forget that all this figures and description show steady-state situation. They assume that the inductor was Turn-ON for a very long time before we start our analysis. Do you understand this ? \$\endgroup\$ – G36 Feb 26 '17 at 10:30
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I don't need any Mathematical Explanation. Thanks...

A pure inductor does not dissipate power so if you apply a sinewave voltage to its terminals, current has to flow sinusoidally but in such a way that the average power is zero. So the options are: -

  • no current flows therefore power is zero
  • the voltage is shorted out completely so power has to be zero
  • current is at 90 degrees to voltage hence average power is zero

Which one makes most sense and which one shows that energy is both stored and returned to the sinewave supply in successive half cycles of the waveform?

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