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Now this seems a bit silly question, but I'm confused about it. The picture shows an inductor which is connected with a battery whose value is 5V. It means we are creating a constant 5 volts current push through the inductor at ANY given time.now suppose in some moment at $t$ (which is ofcourse at the very beginning) the inductor is producing a back-emf of value 3volts, that means, in effect A value of 2 volts (applied voltage-back emf) should appear across the inductor, (at that moment)which contradicts KVL. What is wrong here? I know i'm ignoring the resistance of the inductor, maybe that could be the reason, but can somebody please explain this in a systematic manner?

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  • \$\begingroup\$ There is no constant 5V current pushing anywhere. There is constant 5V voltage over the inductor terminals. Always. What happens internally in the inductor like back-EMF is internal to the inductor. \$\endgroup\$
    – Justme
    Commented Nov 25, 2020 at 10:54
  • \$\begingroup\$ If the switch is closed then The EMF across the inductor is always 5 volts. Always. It's not possible for the back-EMF to be 3 volts. It's always 5 volts. Back-EMF and EMF are the same thing. \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Nov 25, 2020 at 17:00
  • \$\begingroup\$ We know the formula for back emf V=L. di/dt, so isnt it possible in some moment t, L. di/dt=3V? Also L. di/dt is a function of t, then why this will not change when t changes? and the changing procedure permanently stops once back emf V=0 that is , di/dt is zero. \$\endgroup\$
    – Sayan
    Commented Nov 25, 2020 at 17:37

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The back emf from a pure inductor with no series resistance when connected to a 5 volt source is 5 volts. It's as simple as that. If the inductor is producing a back emf of 3 volts then it's not a pure inductor and will have series resistance that is dropping 2 volts.

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It means we are creating a constant 5 volts current push through the inductor at ANY given time.

First thing, the volt is a unit of potential difference or emf, not a unit of current. The phrase "5 volts current" makes no more sense than "5 kilograms tall" or "7 meters of weight".

now suppose in some moment at \$t\$ (which is ofcourse at the very beginning) the inductor is producing a back-emf of value 3volts...

There's no reason you can assume a priori that the inductor has a certain back-emf. The back emf will be whatever the rest of the circuit determines it to be, not some arbitrary value.

You must solve the circuit to find out what the back-emf is, not assume a value and apply that to the circuit equations.

hat means, in effect A value of 2 volts (applied voltage-back emf) should appear across the inductor, (at that moment)which contradicts KVL. What is wrong here?

What's wrong is, with no justification whatsoever, you have assumed you knew the inductor's back-emf before solving the circuit. You should have solved the circuit to find out the back-emf instead.

If your instructor has posed this as a problem and asserted that the back-emf is 3 V, then it is just one of those problems (which bears little relation to anything you will ever encounter working as an engineer) where they arbitrarily reveal certain circuit variables and expect you to determine the others. In this case you should probably assume the remaining potential in the KVL loop is taken up by the parasitic resistance of the inductor. But this is inconsistent with your saying that the time $\t$ being analyzed is "at the very beginning", because the inductor current in the instant after the switch is closed will remain at 0 A, and thus the resistive voltage drop will be 0 V.

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  • \$\begingroup\$ I didn't say the inductor has a back emf of 3 V, what i understand is, back emf is depends on the rate at which the current ischanging, like, from very beginning its 5, then 4.9, then 4.8 and so forth upto zero. i just said the back emf decreases from 5V to 0 volts, so there can be a moment t for which L. di/dt=3, of course this varies with time and finally settles to 0 \$\endgroup\$
    – Sayan
    Commented Nov 25, 2020 at 17:44
  • \$\begingroup\$ In the circuit as drawn, the back emf is always 5 V for t>0, so the current increases indefinitely. There is no moment when the back emf is 3 V. \$\endgroup\$
    – The Photon
    Commented Nov 25, 2020 at 18:09
  • \$\begingroup\$ @Sayan, you said, "suppose in some moment at $t$ (which is ofcourse at the very beginning) the inductor is producing a back-emf of value 3volts". I think my interpretation of "producing a back-emf of 3 volts" and "at the very beginning" are the ones that most people would make. These assumptions are not consistent with the circuit model you drew. Remember that I can't read your mind, and I only know as much about what you are thinking as include in your question. \$\endgroup\$
    – The Photon
    Commented Nov 25, 2020 at 18:15

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