Suppose we have a circuit with a voltage source, a switch open and an inductor all in series. If we close the switch, the potential difference of the voltage source is instantaneously applied to the inductor. As the current starts to build up, induced voltage from the inductance opposes it. If the induced voltage (back-emf) is equal and opposite to the applied voltage, and the net voltage is zero, what drives the current then? 

  • Main question would be that if induced emf equals to applied voltage then how does current flow

P.S I have looked everywhere for this question but couldn't find any.

  • $\begingroup$ I've removed some comments which should have been combined into an answer. $\endgroup$
    – rob
    Feb 1, 2017 at 7:40

2 Answers 2


Lenz's law states that the induced current is in such a direction as to oppose the change producing it.

Back emf and a complete "conducting" circuit will result in an induced current.

The back emf can never exactly equal the applied voltage as then the current would be zero and not changing which would mean that there cannot be an back emf.

So you can think of it as follows.
As soon as a current starts to flow an emf is induced which produces an induced current which tries to oppose that change of current in the circuit produced by the applied voltage.
That induced current slows down the rate at which the current in the circuit increases.

So when deriving equations relating current to time in such a circuit it is convenient to say that at time = 0, when the switch is closed, the current is zero because the applied voltage and the back emf are equal in magnitude.

What you are usually not concerned with is a time scale equal to that whilst the switch is being closed.

Here is an attempt to illustrate the complexity of what happens even as the switch is being closed.
The switch is a capacitor and when open has charges stored on it.
As the switch is being closed the capacitance of the switch increases and so a very small current starts to flow around the circuit as the capacitor charges up.
The change of current is opposed by the induced current produced by the back emf.
As the distance between the contact of the switch continues to get less, the capacitance of the switch continues to increase and a changing current continues to flow.
When the contacts finally close there is already a very small but changing current flowing around the circuit which is being opposed by the induced current produced by the back emf with the back emf slightly less than the applied voltage.

What actually happens during the switch closing process is complicated by the fact that now you have an inductor, resistor and capacitor in the circuit and also that lumped circuit element analysis might be inappropriate to use over such a small time scale?.

  • $\begingroup$ But isnt back emf at every moment equal to voltage with which current is flowing so eventhough if current is increasing and trying to achieve max value with which it should flow by utilising maximum volatage of source then at every moment back emf should equal that value of source voltage and wont allow current to flow so what causing back emf value to not equal to applied at every instant.P.S would be lot of help if you think my concept about back emf isnt good so can you explain me about back emf or some refrence book or site to study it. $\endgroup$
    – Mahin
    Feb 1, 2017 at 7:44
  • 1
    $\begingroup$ Here is an argument to show that Lenz can never win. Lenz says that the induced current (produced by the back emf) opposes the change producing it. If Lenz actually stops the change then there is no back emf and no induced current and no opposition to the change so the change continues. $\endgroup$
    – Farcher
    Feb 1, 2017 at 7:55
  • $\begingroup$ did things i said in previous comment were correct anf if yes than what is paradox or how does this paradox come about $\endgroup$
    – Mahin
    Feb 1, 2017 at 7:58
  • 1
    $\begingroup$ Here is an argument to show that Lenz can never win. Lenz says that the induced current (produced by the back emf) opposes the change producing it. If Lenz actually stops the change then there is no back emf & no induced current & no opposition to the change so the change continues.As has been pointed out to you applied voltage minus back emf $L\frac {dI}{dt}$ equals voltage across the resistor $IR$.It is convenient to ignore everything that happens before the switch is closed & say that at that time the current is zero & the applied lovage & the back emf are equal. It is a good approximation $\endgroup$
    – Farcher
    Feb 1, 2017 at 8:01
  • 3
    $\begingroup$ The back emf reduces the rate of change of current it does not stop the rate of change. $\endgroup$
    – Farcher
    Feb 1, 2017 at 8:10

First of all, the condition that you are talking about, induced emf = voltage, is achieved at $t=0$. This is because at that instant no current is flowing in the wire and hence it follows from initial conditions and Kirchoff's law. (No current is flowing at $t=0$ since in $dt$ time from $t=0$ if $di$ has a finite value then $di/dt$ would give an impossibly large emf without anything to support this amount of energy.Update- Therefore di is infinitesimal, current increases infintesimally from 0 and is 0).

Now, the voltage after this just keeps on decreasing as $di/dt$ decreases. You get this from the equation, $$ L\ di/dt+iR=V,$$ which gives $i=I(1-e^{-(R/L)t})$ where $I=V/R$. (Derive this!) The derivative of this function is $di/dt$ and it approaches zero with time. Thus the back emf keeps on decreasing from initial conditions and there is no question of it being equal to voltage.

  • $\begingroup$ Can yoy please modify your answer first you are not going through all steps of equation you are deriving so its hard for me to keep up secondly you should give better reasoning for equation steps and also provide me with atleast the source or refrence book to study this following topic if you aren't able to know that how to code in mathematical equation in this forum $\endgroup$
    – Mahin
    Feb 1, 2017 at 7:12
  • $\begingroup$ Yeah, I actually don't know how to code in mathematical equations in this forum. You could simply google the derivation or go to this site -physics.info/circuits-rl. Now once you get the equation, its just basic maths to plot the graph. It approaches a finite value and hence the derivative approaches 0. Therefore, back emf decreases and becomes 0 after a long time. $\endgroup$
    – TheFool
    Feb 1, 2017 at 7:19
  • $\begingroup$ @TheFool Welcome to Physics! For guidance in typesetting mathematics, see this page in the help center. I've gone ahead and edited your post for you; have a look at the source or the edit history to see the changes. $\endgroup$
    – rob
    Feb 1, 2017 at 7:45
  • $\begingroup$ @TheFool I do not understand why $\frac{di}{dt}$ would be impossibly large? The way you have written it you have assumed $di$ to be finite but $dt$ to be infinitesimal. Why cannot they both be infinitesimal? $\endgroup$
    – Farcher
    Feb 1, 2017 at 8:04
  • $\begingroup$ No, that is actually my argument for di to be infinitesimal and hence current be 0. di can not have a finite value as that would mean that di/dt is impossibly large. $\endgroup$
    – TheFool
    Feb 1, 2017 at 9:05

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