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Why are the curves for growth and decay of current in inductors of exponential nature i.e $$ 1-e^{-x} $$ and $$ e^{-x} $$ respectively. I understand the mathematical derivation for the current i(t). But I want to know the factors like change in flux, application of voltage across the inductor, etc cause the growth and decay of the current to be of exponential nature.

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    \$\begingroup\$ It's only approximately exponential because the inductor is imperfect, otherwise current increase with constant applied voltage would be perfectly linear with respect to time. \$\endgroup\$ – Spehro Pefhany Nov 21 '18 at 20:41
  • \$\begingroup\$ Since you understand the mathematical nature of di/dt functions, the question actually is: explain the nature of inductor impedance as being iomegaL. \$\endgroup\$ – Ale..chenski Nov 21 '18 at 21:08
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    \$\begingroup\$ You could start at this chapter from Feyman's Lectures series: laws of induction or go a little earlier in his lectures to look at the A-field (which is to me more important than B-) by going to magnetic field in various situations. I'd also recommend another book I think provides a much more accessible and intuitive approach, called "Matter & Interactions," 3rd edition, by Chabay and Sherwood. \$\endgroup\$ – jonk Nov 21 '18 at 21:20
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    \$\begingroup\$ \$1-e^{-x}\$... \$\endgroup\$ – Chu Nov 22 '18 at 0:40
  • \$\begingroup\$ If you go halfway, then the forcing voltage has dropped to 50%. Once you reach 90%, then only 10% voltage is left to force the behavior. Once you reach 99%, then only 1% voltage is left to force the behavior. All these points are self-similar, and the differential equations are not hard to write; the e-to-the-X solution comes out because the function is also its own derivative, within a constant. \$\endgroup\$ – analogsystemsrf Nov 22 '18 at 3:38
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This is quite the question.

Let me start by saying that the factor \$e\$ has been invented to make our life a bit easier. Any power can be written as another power with a different base as follows:

$$ a^b = \left(e^{\ln(a)}\right)^b = e^{\ln(a)\cdot b} $$

The usual way how this relationship occurs is when a quantity changes linearly according to its own value in some way. Ie. if the quantity increases, then the rate-of-change of that quantity will change proportionally.

Your question could be considered invalid though, because there are a lot of circuits with inductors and capacitors that don't change exponentially. For example:

The current through the capacitor is constant, and the voltage changes linearly. There is no exponential involved.

An example of where it does seem to be the case would be:

(V1 is a step function)

The current through the capacitor depends linearly on its own voltage, because of the added resistor. The behavior of a capacitor is such that \$i = C\frac{dv}{dt}\$ so this means a derivative will ultimately be linearly dependent on its own value. This circuit will result in a solution with an exponential.

Rather than talking about flux, voltage, current, etc. you should consider that having a solution that is exponentially changing is something much more basic.

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