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May28-05, 02:41 PM
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I told Andrew I would look at some equations and then come back. The equation I’m concerned about is the differential equation for the RL-circuit:
[itex]u = R \cdot i + L \cdot \frac{di}{dt}[/itex]
This equation says that the inductor will hinder the effective voltage [itex]R \cdot i[/itex] to build up. But to me this equation doesn’t make sense, although I would agree it is correct. Let me explain this in the case of a positive DC voltage.

First of all, from the physical (not mathematical!) viewpoint, why wouldn’t the term [itex]L \cdot \frac{di}{dt}[/itex] exceed the voltage? Because then we would have a negative voltage over the inductor causing the current to flow backwards (or at least decrease). But this is an absurdity since the current has to always increase to keep the term positive. I have no complaints this far.
But you have, nevertheless, some misconceptions. First, understand what the different terms mean.

u is the applied voltage (it is a number fixed by the power supply/battery) not some "effective voltage"; iR is the voltage drop across the resistor; and Ldi/dt is the voltage drop across the ends of the inductor.

The fact that the electric field generated by the battery is a conservative field (ie : the work done by the field on a charge is path independent, or the work done over a closed loop is zero) requires that the voltage gain provided by the battery equal the total voltage drop across various elements in the circuit. This concept is contained within Kirchoff's Voltage Law.

At t=0, there is no current and hence no drop across the resistor. So, the entire potential drop is across the inductor, or Ldi/dt = u. Since u > 0, di/dt > 0. This tells you that i must increase from its initial value of 0, causing a potential drop to appear across the resistor. For the equation to work (which it must, for the reason stated in the previous paragraph), as iR increases, Ldi/dt must decrease, so as the keep the sum constant. Since Ldi/dt started out equal to u and has henceforth only decreased, it can not exceed u (in theory).

The second question is: why would the current increase at all. If you solve the equation you will (of course) find an increasing current. The funny thing is that, according to me, the equation doesn’t give a physical reason why the current would increase. Looking at a certain time the effective voltage is [itex]R \cdot i[/itex], which is enough to drive a current [itex]i[/itex]. It is not enough to drive a current [itex]i+di[/itex]!
The equation never intends to give a physical reason why the current increases. The equation (as many others) merely tells you how to find the current at a certain time. The reason why the current increases is that the net force on the electrons is non-decreasing over timescales that are large compared to electron-phonon scattering times. In the absence of the inductor, when u=0, i=0; and as u instantaneously changes to some non-zero number (by throwing a switch), the current too reaches a non-zero value "instantaneously" and stays steady at that value. So, the current increases simply because of the appearnce of a driving force - the electric field generated by the battery. Thus, it is seen that the damping force from the resistor has an extremely short time (virtually zero) constant. This is indeed true and this time constant would be of the order of the relaxation time between scattering off the atoms in the resistor. This number is of the order of nanoseconds, and hence may be neglected as far as the big picture is concerned. This, however, is what you are talking about when you speak of the effective mass of the electron having a role in the damping (not the Ldi/dt term). Now you know why this contribution is not important to the behavior over longer time-scales.

When an inductor is included in the circuit, there is an addition retarding force coming from the induced magnetic field whose value is proportional to the rate of change of the field/current (and hence disappears in the steady state). This slows the current growth over time scales that are much bigger than the relaxation (or scattering) time.

Now my idea is that in real life the term [itex]L \cdot \frac{di}{dt}[/itex] would be somewhat smaller than [itex]u - R \cdot i[/itex]. Then that would allow for current to build up, since you then have a “net voltage”.
From how you are using this term, it would be unphysical to have a "net voltage". This would require that the work done by the electric field over a closed loop be non-zero, resulting in a violation of energy conservation.