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When an inductor is connected with a voltage source we get equal and opposite voltage on inductor against the source voltage. That equal and opposite voltage gradually decreases with time which allows the current(caused by the source voltage) to rise gradually.

My question is what makes the equal and opposite voltage in an inductor to fall gradually ?

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  • \$\begingroup\$ Reall good question and one that is troubling when you dig below the surface. I've never seen a good answer to this so if anyone has got one please comment me in! \$\endgroup\$ – Andy aka May 19 '18 at 20:00
  • \$\begingroup\$ @Andyaka i might have got the half answer related to this. you may find it interesting and may complete the answer. electronics.stackexchange.com/questions/293123/… \$\endgroup\$ – Alex May 19 '18 at 20:10
  • \$\begingroup\$ @Andyaka , Neil has a good answer :) \$\endgroup\$ – user103380 May 19 '18 at 22:23
  • \$\begingroup\$ @KingDuken think again what the problem is. \$\endgroup\$ – Andy aka May 19 '18 at 22:56
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When an inductor is connected with a voltage source we get equal and opposite voltage on inductor against the source voltage.

Yes

That equal and opposite voltage gradually decreases with time which allows the current(caused by the source voltage) to rise gradually.

No. If the voltage source is an ideal voltage source, the voltage across the inductor remains constant. The voltage across the inductor causes the current to rise. If the inductor is ideal (zero resistance), then the current will increase linearly with time.

My question is what makes the equal and opposite voltage in an inductor to fall gradually

It doesn't in the ideal case, see above.

In practice, the voltage source might have a significant output impedance, which will cause its voltage to drop as the current increases. But note it's the increase in current that causes the voltage drop, not the other way around.

The inductor (unless it's superconducting) will also have some resistance. There will be a voltage drop across this resistance. Even with an ideal voltage source, the voltage across the inductive part of the inductor will fall, and so the rate of current rise will slow. Eventually, the current will rise to the point that all of the source voltage is being dropped across the resistance, and the current stops increasing.

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  • \$\begingroup\$ In a perfect inductor with a constant applied voltage, the back emf equals that applied voltage so, how can current flow? Maybe it's a physics question really? \$\endgroup\$ – Andy aka May 20 '18 at 9:40
  • \$\begingroup\$ @Andyaka This old question had some activity on it, which caused me to revisit. Are you still asking how current flows? See my other answer. \$\endgroup\$ – Neil_UK Apr 18 '20 at 13:45
  • \$\begingroup\$ I asked that years ago LOL. \$\endgroup\$ – Andy aka Apr 18 '20 at 16:38
  • \$\begingroup\$ @Andyaka I know you asked it years ago, I can see the date on the question, that's why I asked if you were still asking. \$\endgroup\$ – Neil_UK Apr 18 '20 at 16:58
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Whether the voltage drops on the source would depend in practice on how fast the source can follow a changing load. It is the current that is changing during the initial "energizing" time period for the inductor. This might appear as a voltage change at the source if the source cannot follow the changing load fast enough.

The "energizing" of the inductor means energy is input into the process of creating magnetic fields around the inductor, which takes some time. The energy storage and release mechanism of an inductor is essentially the creation and collapsing of the surrounding magnetic fields, respectively. These processes take some time which corresponds to the "gradual decrease" being observed.

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schematic

simulate this circuit – Schematic created using CircuitLab

R represents the sum of the voltage source internal resistance and inductor wire resistance.

If R is 0 ohms then voltage across the inductor is constant. Let i be the current in the inductor, and suppose i=0 at t=0.

\$ V=L\frac{di}{dt} \rightarrow i = \frac{V}{L}t\$

So, current rises at a constant rate to infinity, and voltage across the inductor is always constant.

In a real circuit, R is not zero, so:

\$ V=L\frac{di}{dt} + Ri \rightarrow i = \frac{V}{R}( 1 - e^{-\frac{R}{L} t} ) \$

After the current settles to V/R, voltage across a real inductor having internal resistance, will thus not fall to zero. If the inductor is modeled with a perfect inductor (having no resistance) and a resistor in series, then voltage across the perfect model inductor will fall to zero as current settles, and voltage will appear across the resistor.

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When an inductor is connected with a voltage source we get equal and opposite voltage on inductor against the source voltage.

Yes.

That equal and opposite voltage gradually decreases with time ...

No.

...which allows the current(caused by the source voltage) to rise gradually.

No. The current started rising gradually the moment the voltage was applied. The back EMF is given by \$V=L\frac{dI}{dt}\$. For there to be a back EMF, there has to be a changing current.

My question is what makes the equal and opposite voltage in an inductor to fall gradually ?

I wonder what makes you think the back EMF from the inductor falls below the applied voltage?

If the inductor has resistance, then the voltage drop caused by current through the resistance reduces the voltage seen by the inductance, and with that, the rate of change of current drops. A reduced rate of change of current means a reduced back EMF. You'll notice that the back EMF was always equal to the voltage seen by the inductance all along.

In the circuit theory model of inductors, the applied voltage and change of current are always related by \$V=L\frac{dI}{dt}\$. One does not cause the other, they both occur at the same time.

There is a deeper model for an inductor, that of a shorted transmission line. Let's put some figures on it for the sake of illustration. Say we have a 100 ohm transmission line with an electrical length of 10 uS, shorted at the far end, and apply a step 10 V to it.

At t=0, a current of 100 mA will begin to flow, while a +10 V + 100mA step is launched into the line. After 10 uS the step hits the far end and, finding a short circuit, which allows any current but zero volts, reflects a -10 V + 100mA step. The current in the s/c at the end of the line is now 200 mA. After another 10 uS, the step reaches our launch point, where it finds a short circuit, the zero output impedance of our voltage source. It therefore reflects a +10 V +100 mA step into the line. The current supplied by the source now rises to 300mA. Rinse and repeat, indefinitely.

As the wave travels the length of the line and back, you can see that the current in the line is rising by 100 mA every 10 uS. The shorted line is behaving as an inductor, with a rate of current rise proportional to the applied voltage. The inductance exhibited is proportional to the line impedance, and to the length of the line. We can increase the line impedance and reduce the length while keeping their product constant to get the 'steppy' current increase as smooth as we like.

There is a model deeper than transmission lines, but it's quantum mechanics, with photons being emitted and absorbed to transmit the electrical effects, and I'm frankly not up to that.

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