Suppose we have an L circuit with DC power supply of emf \$\mathscr{E}\$


simulate this circuit – Schematic created using CircuitLab

This suggests that \$\frac{dI}{dt}=\frac{\mathscr{E}}{L}\$
As at \$t=0\$, \$I=0\$
So, current increases at a constant rate from \$0\$.

But, I am not able to understand why this is so physically.

My interpretation
The above situation is analogous to the situation in which we have a coil placed between the Helmholtz coil, of which we are increasing the current at constant rate. Thus magnetic field increases at constant rate between the Helmholtz coil so the magnetic flux increases at a constant rate in the first coil and a constant emf will be induced in the first coil by Faraday's law. But in this case, "we" are increasing the current in Helmholtz coil at a constant rate. But how this is achieved in the original circuit?
KVL forces that voltage across the inductor has to be \$\mathscr{E}\$, and then by Faraday's law thus \$\frac{dI}{dt}\$ has to be constant.
But I am not able to understand how this has been achieved in a circuit when we close the switch.
When we close the switch, suddenly the charge carriers gets accelerated and thus \$\frac{dI}{dt}\neq 0\$, but the inductor does not like that. So it will induce emf instantly and when \$V_L=\mathscr{E}\$, \$I=0\$. As that corresponds to opposing batteries of same emf. As long as \$V_L=\mathscr{E}\$, no current in the circuit, thus no change in current and thus \$\frac{dI}{dt}=0\$. So, as a result the inductor's induced voltage will go to \$0\$. When it starts decreasing, \$V_L-\mathscr{E}\neq 0\$, thus \$I\$ starts increasing and \$\frac{dI}{dt}\neq 0\$ but inductor will again increase the induce emf to oppose this \$\frac{dI}{dt}\$. As soon as \$V_L=\mathscr{E}\$, \$I=0\$ again. So, this suggests that \$V_L=\mathscr{E}\$ always, which is also given by KVL.
But this corresponds to \$I=0\$ at all the time. But this leads to the problem that as current is not changing then \$V_L\$ should be \$0\$. But I think that one can argue that as \$V_L\$ starts to decrease from \$\mathscr{E}\$, net voltage in the circuit won't be \$0\$ and this leads to the change in the current and thus to oppose this change in current inductor has to maintain the \$V_L=\mathscr{E}\$. So, the KVL and Faraday's law still holds in the sense to oppose the "tendency" of changing current.
Same is the case of the limiting friction, it won't be 0, till the object has tendency to have a relative motion with the surface. But as the guiding force exceeds the limiting friction the object moves. In that case, there is an upper limit to the value of limiting friction (depends on the type of surface) which when the guiding force exceeds that value, the object starts moving. But here the guiding force is constant (i.e, constant voltage source), the opposing force has no upper limit (i.e, \$V_L\alpha\frac{dI}{dt}\$ which has no upper bound).
But the point is that the above analysis suggests that we have a constant current in the circuit and as at \$t=0\$, \$I=0\$, thus at all other times \$I\$ should be \$0\$. One can give another argument that for \$\frac{dI}{dt}\$ to be non-zero, the charge carriers have to be accelerated, but the net voltage in the circuit is \$0\$. So, there is nothing that accelerates the charge carriers. Thus they move with constant speed, thus passes with same amount through a given area at all the times, thus constitute constant current.

My question is how \$\frac{dI}{dt}\neq 0\$ as suggested by KVL? What actually cause this?
Definitely one can't say that as \$V_L=\mathscr{E}\$, thus \$\frac{dI}{dt}\neq 0\$. This is kind of a circular argument. I have seen several posts and physics books which bypasses the question by making the above argument.
Can somebody please help me in understanding this. I am very confused.

  • \$\begingroup\$ Take a step back: phi = L I. And dphi/dt = v. If v is constant, it means that the rate of change of phi is constant. So, what does that say about phi, and its proportional quantity I ? \$\endgroup\$ Mar 18 at 6:52
  • \$\begingroup\$ Iti, at deeper levels the question gets far more interesting and Faraday looks like the incredible genius he was, while Kirchhoff only states the mundane obvious about a situation, by comparison. But at the electronics view, it's simple. An increasing current (constant rate of change) is required in order to sustain the necessary back emf. Or read through the 2nd volume (start at the beginning, not just chapters 16 and 17!) of Feynman's Lecture series. \$\endgroup\$ Mar 18 at 7:00
  • \$\begingroup\$ It is interesting that people do not have a problem with a capacitor being charged by a constant current source. And they feel intuitive (bracket water analogy) that the voltage across the capacitor will then ramp up in a straight line. Because the capacitor stores energy in terms of voltage (in an electric field). But in the inductor, we have an opposite situation. The energy is stored in terms of a current (in the magnetic field), thus, the constant voltage will "charge" the inductor and the current must ramp up in a straight line. As in a capacitor charged from a constant current source. \$\endgroup\$
    – G36
    Mar 18 at 7:27
  • \$\begingroup\$ @G36 Yes, volt-seconds (Webers) are a kind of dual for charge. Some of the difficulty, perhaps, comes from the fact that capacitor charge is "countable", like coins or tokens, independent from time, while inductive charge 'Webers' aren't countable without also keeping track of time. \$\endgroup\$ Mar 18 at 7:51
  • \$\begingroup\$ The back emf is a red herring; it doesn't somehow stop current flowing because it equals the applied voltage; the effect of the back emf is has nothing to do with the current flow. Why you bring Helmholtz coils into your question is beyond me? When we apply a voltage to a resistor, we seen an opposing voltage generated by the resistor but we don't get in a muddle about that do we? \$\endgroup\$
    – Andy aka
    Mar 18 at 13:20

2 Answers 2


You are over thinkling it. The defining relationship says it all. $$v_{L}=L\frac{di_{L}}{dt}$$

  1. If a voltage is applied across an inductor the current will change at a rate defined by the inductance. The higher the inductance, the slower the rate of change. So the inductance opposes the rate of change but cannot stop it.
  2. If the current is changing in the inductor (for whatever reason) then there will be a voltage across it. The faster the change and the greater the inductance, the higher the voltage that is created. The inductor opposes the rate of change with a voltage, but cannot stop it.

The inductor has no likes or dislikes. Its behaviour is described completely by the defining relation.

Addition to speak to the Helmholtz part of the question.

The above situation is analogous to the situation in which we have a coil placed between the Helmholtz coil, of which we are increasing the current at constant rate

Not so! If there is an inductor with nothing connected and aligned with the magnetic field of the Helmoltz coil, then a voltage will appear scross the inductor by Faraday's Law.$$v=-N\frac{d\phi}{dt}$$

There is no current in the inductor. The inductor is not required for the voltage field to exist. It is alreaady there. Placing the several series connected loops of wire so that they align with the voltage field allows the voltage to be displayed by the wire.

If the inductor is shorted from one terminal to the other a current will appear in the inductor driven by the "induced" voltage.

But how this is achieved in the original circuit?

In the original circuit, an external voltage is applied across the inductor. By Faraday's Law, the rate of change of flux is constrained to a fixed value different from zero.$$\frac{d\phi}{dt}=-\frac{1}{N}V$$

This is like accelerating a mass. The actual velocity can be zero but the acceleration can be very large at the same time.

The applied voltage also places an electric field along the wire accelerating the electrons.

But Farady's Law does not include inductance or current. So let's take a closer look by introducing current into the law by writing:$$v=-N\frac{d\phi}{di}\frac{di}{dt}$$

Inductance can easily be recognized$$L=-N\frac{d\phi}{di}$$as the ability for current to establish a magnetic flux in a region of space. It depends on the medium and the number of turns.

it is not evident what is the cause of constant rate of change of current which leads to a constant emf across the inductor.

So the applied voltage does double duty.

  1. The current is produced by the electric field created by the voltage.
  2. The rate of change of current is constrained by the voltage and the inductance to a fixed value.

The constant emf is because the applied voltage is making it so. The constant applied voltage leads to a constant rate of change of current. Not the other way around.

  • \$\begingroup\$ thanks for the reply. If we have a situation that we have an inductor between the Helmholtz coils, then it is evident that as we increase the current at constant rate through Helmholtz coil, magnetic field between it increases and thus the magnetic flux through the inductor and this induce a constant emf. But in the case of the circuit shown in the question, it is not evident what is the cause of constant rate of change of current which leads to a constant emf across the inductor. \$\endgroup\$
    – Iti
    Mar 18 at 8:22

$$V_L = L\times\frac{\mathrm{d}I }{\mathrm{d}t}$$ This equation indicates that inductance voltage depends not on the 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 is.

At the beginning (at time 0+) we apply a voltage across the inductor by doing this we are attempting to cause a sudden change in the current. The induced voltage now steps in and tries to keep the current down to its initial value (0A) (Lenz law) and this induced voltage must be exactly equal to the applied voltage. And the current starts to rise up. But why the current start to rise up in the first place? Well, according to Faraday, the current takes time to build up in an inductor only because of "induced voltage".

Remember this equation \$V_L = L\times \frac{ΔI}{Δt}\$ and if \$ \frac{ΔI}{Δt} =\$ 0A. However, that cannot be, because a zero rate of change in current implies no induced voltage. In other words, the very existence of "induced voltage" depends on the fact that the current is changing, and it must change. We have a negative feedback mechanism here. If the current changes to fast (fast rate of change) the induced voltage will be larger than the applied voltage and this will reduce the rate of change so that this is true \$ΔI = \frac{V_L}{L}\times Δt\$. Or if the rate of change is too small then the induced voltage will be lower than the applied voltage and the current will increase the rate of change to sustain the rate of change of current at a constant value.

In short. By definition, the voltage measured across an inductor at any moment is the "induced voltage". And to get this "induced voltage" across the inductor the current must change as it was described by Faraday.


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