Self induction: Why is induced voltage smaller than the applied voltage?

Question

Imagine a circuit with a voltage source, a switch and an inductivity all connected in series.

First, the switch is open and there's no current and no magnetic field around. If we close the switch, the potential difference of the voltage source is instantaneously applied to the inductivity. Lenz's Law tells us that the induced voltage from the inductivity will always be such that the change in the magnetic flux is reduced. And we have Faraday's Law which tells us that the induced voltage is equal to the change in the magnetic flux.

What is the theoretical reason why the change in magnetic flux must be smaller than the applied voltage? Why does for t -> infinity always run a current trough the inductivity as it would be just a resistor? Is it just because the wires have a resistance? What would then happen when taking superconductors instead of wires? Is it the inner resistivity of the voltage source then?

Answer 1

Well there are many questions in your question and I'll try to address them one by one.

As you may know,

$L* di/dt = V (t)$

one important thing to note about inductors is that unless there is an impulsive excitation(that is, the input is bounded), it's current can not make instantaneous jumps.

Therefore, when you apply the voltage, it will still have a 0 current, it will have a high slope, but 0 current nonetheless at $0^+$ where $0^+$ denotes the instant right after 0.

Ok, so using $L* di/dt = V (t)$ and basic first order circuit analysis coupled with a bit of differential equations, you get the current equation:

$i(t) = V/R *(1 - e^ {-tR/L})$

R is the total thevenin resistance in the circuit. This includes the internal resistance of the inductor. This is the Zero state response of the inductor and it does not take into account any form of initial condition, however it is not hard to implement that into the formula. As you can see, the current will only reach its final value as t -> infinity.

If you assume the inductor circuit has zero resistance, it would be a short circuit between the terminals of the voltage source. If you assume there is an output resistance of voltage source, you can separate it from the voltage source and use in the calculation as R.

Change in the magnetic flux is basically the self induced EMF. The differential equation basically points out to the fact that most things in the inductor occur in an exponential decay kind-of fashion. I hope this answers at least some of your questions. Let me know in the comments if anything is not clear.

Edit: Just to be clear * is multiplication in my answer, not convolution or something like that.