# Regarding the inductance model

1. Aug 18, 2013

### MechatronO

The inductor below is conducting an arbitrary current, with the switch on.

Suddenly the switch is set off and and voltage over the inductor raises to the negative infinity according to

Vl = L*di/dt

However, voltage is defined as energy per charge.

V=J/C

The energy stored "in" the inductor is finit and

El = L*i^2/2

Combining these equations, would the voltage instead then raise to

Vl = L*i^2/(2*C) ?

Where C is the sum of all free charges or something, relevant for the equation?

I guess what I've wrote above is far to simplified to describe what would happen, but let's consider this thought experiment also.

Imagine that there are only two relevant charges in the whole system. When the current is switched off, this would get an infinite potential energy distributed on a finite number of charge.
This would thus have inifinite energy created out of a finite amount of energy, thus the Vl=Ldi/dt model must begin to show serious flaws at some point?

EDIT: And no we don't consider things such as parasitic capacitance, eddy current or anything like that here.

2. Aug 19, 2013

### Staff: Mentor

Should be: V = -L*di/dt

Should be: Energy = L*i^2/2

EI does not have units of energy.

Last edited by a moderator: May 6, 2017
3. Aug 19, 2013

### vanhees71

The problem with this problem is that it is not well defined or very complicated to solve. Just opening the switch gives you a situation of an open circuit, and you don't know what happens. You can treat the opened switch as a capacitor with some assumed capacitance to make the problem well defined. It could of course get worse and you have a spark across the opened switch, because in the very first moment a very large EMF is induced in the circuit.

Another possibility is to switch such that the battery is substituted by a shorc-circuit at the moment you switch and then kept. Imho that would make the most sensible question and a good exercise in integrating quasistationary initial-value problems in circuit theory.

4. Aug 19, 2013

### MechatronO

NascentOxygen: Ei sure does have the dimension energy.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/indeng.html

vanhees71: Let's assume no capacitance at all and that the voltage could reach to infinity therefore.

Why would changing the battey to a short circuit at the moment of switching off change the situation?

The point of all this is not really to study a practical circuit, but just a curiosity to see wether the Ldi/dt model has a shortcoming or non defined result, as ohm's model (law) has when R reaches zero in

I = U/R

5. Aug 19, 2013

### Staff: Mentor

If E is potential and I is current, EI does not have units of energy.

6. Aug 20, 2013

### MechatronO

Aha, you see i in El i just the letter E with a badly implemented subscript L, representing energy in the inductor L. EL would be better, as I even myself confused i with L in the last post.