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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\$ – KingDuken 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
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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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