# I Inductive 'kick'

1. Feb 8, 2017

### mcairtime

Hello,

I am wondering why such a large voltage can be generated when the switch in a circuit containing an inductor is opened. The following is from a set of Electromagnetism course notes:

"The 1.5 V supply in the circuit below is insufficient to light the neon lamp. A neon lamp needs about 80 V across it before it will light.

http://file:///C:/Users/CPTEMP~1/AppData/Local/Temp/msohtmlclip1/01/clip_image002.jpg [Broken] The switch is closed and the current builds up to its maximum value. When the switch is opened, the current rapidly falls to zero. The magnetic field through the inductor collapses (changes) to zero producing a very large induced e.m.f. for a short time. The lamp will flash.
"

I have worked through the mathematics of calculating the emf generated both when the switch is first closed (equal to the emf of the supply) and when the switch is opened, when it can take more or less any value you want by tweaking the values of the resistance of the neon lamp etc. However, I feel like I still do not understand the situation in a fundamental way i.e. in terms of the rate of change of magnetic flux linkage.

If the emf generated in the inductor is 1.5 V when the switch is closed and over 80 V when it is opened then the rate of change of magnetic flux linkage in the inductor must be much greater when it is open. Why is this the case?

Hope it's clear what I'm after but feel free to seek clarification if needs be of course.

Last edited by a moderator: May 8, 2017
2. Feb 8, 2017

### cnh1995

When a current carrying inductor is open circuited, the current drops to zero in almost no time. There is a sharp decrease in its magnetic field. This rapid collapse of the field induces a voltage spike across the inductor, given by E=Ldi/dt.
An inductor tries to maintain the current through it (by inducing a back emf) and prevents it from changing instantaneously.

3. Feb 8, 2017

### Staff: Mentor

Are you familiar with the differential equation relating voltage and the change in current for an inductor?
$$v(t) = L\frac{di(t)}{dt}$$
That is the starting point for calculating the voltage across an inductor when there is a change in the current through it. In a real circuit, the peak voltage when you open a switch in series with an inductor is limited by the parasitic capacitance across the inductor. That forms an LC circuit that determines the peak transient voltage. Alternately, you can get a spark at the switch if that voltage is lower than the parasitic LC circuit's peak voltage.

Does that help?

Last edited by a moderator: May 8, 2017