Suppose we have an L circuit with DC power supply of emf \$\mathscr{E}\$
simulate this circuit – Schematic created using CircuitLab
By KVL,
\$\mathscr{E}=L\frac{dI}{dt}\$.
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.