0
\$\begingroup\$

schematic

simulate this circuit – Schematic created using CircuitLab

Assuming SW1 is closed for along time, and current in L2 to be zero initially.

What sort of current waveform can I expect in L2 when SW1 is opened ?

\$\endgroup\$
4
  • \$\begingroup\$ In the topic you write current source, but here you have a voltage source. Which is it? \$\endgroup\$
    – PlasmaHH
    Dec 24, 2017 at 17:38
  • \$\begingroup\$ L1 behaves as current source, when SW1 is opened \$\endgroup\$
    – spaul
    Dec 24, 2017 at 17:38
  • 5
    \$\begingroup\$ I'm voting to close this question as off-topic because it's a homework question without a solution. \$\endgroup\$
    – pipe
    Dec 24, 2017 at 17:40
  • \$\begingroup\$ @pipe this is not a homework, you don't have to answer if you don't want, but don't pretend to a scientist. \$\endgroup\$
    – spaul
    Dec 24, 2017 at 18:46

2 Answers 2

1
\$\begingroup\$

Sticking to ideal components flux conservation applies :

$$\sum L_k\,i_k =\mathrm{const}$$ In a mesh made of inductors only.

a Dirac delta of voltage across inductors will step current so that total magnetic flux before and after discontinuity is the same. So in this very case will step from 1A to 0.5A

$$L_1\,i(0^-)=(L_1+L_2)i(0^+)$$ $$i(0^+)=\frac{1\mu\mathrm{H}\times 1\,\mathrm{A}}{1\mu\mathrm{H} + 1\mu\mathrm{H}}=0.5\,\mathrm{A}$$

and will keep flowing "forever".

The dual problem is much better known: one 1V charged capacitor is connected to another one, same value. Charge conservation applies:

$$\sum C_k\,v_k=\mathrm{const}$$ on a node of capacitors only

one Dirac delta of current steps voltage to 0.5V and stay so "forever". Total charge is the same

\$\endgroup\$
4
  • \$\begingroup\$ this idea seems reasonable, by magnetic flux do you mean total "flux linkage" ? \$\endgroup\$
    – spaul
    Dec 25, 2017 at 7:07
  • \$\begingroup\$ @user408669 yes have a look to my edits. It more than idea though, it's a basic physics law. Try prove it on your circuit, I'd like you to do it as an exercise. Hint: those discontinuities always push to strane things to happen,temporary remove it with one resistor across second inductor....and see what it comes out this way \$\endgroup\$
    – carloc
    Dec 26, 2017 at 9:27
  • \$\begingroup\$ I already did that exercise. One this I was wondering, you may notice that total energy stored is changing. Where is that energy going? \$\endgroup\$
    – spaul
    Dec 26, 2017 at 9:35
  • 1
    \$\begingroup\$ You are right, you will have some energy loss unaccounted by that model. The same applies when studying two capacitors. This kinda things happen when pushing models beyond what they can do. Simply this model is not good enough \$\endgroup\$
    – carloc
    Dec 26, 2017 at 11:00
2
\$\begingroup\$

In the instant after the switch is opened, L1 acts as a 1 A current source.

At the same time, L2 acts as a 0 A current source.

As in other cases where you put ideal current sources in conflict, what that means is that you have an incomplete model.

You will need to consider parasitics in the circuit, particularly the inter-winding capacitance of the two inductors, to make a reasonable model to predict the circuit behavior.

\$\endgroup\$
2
  • \$\begingroup\$ Thank you, I've encountered such situations before, like in derivation of input output relation in a boost converter the parasitic resistance of inductor has to be considered. So can you please provide an improved model for this question. \$\endgroup\$
    – spaul
    Dec 24, 2017 at 18:42
  • \$\begingroup\$ @user408669: I already told you in my answer that the inter-winding capacitance of the inductor is most important. \$\endgroup\$
    – The Photon
    Dec 24, 2017 at 23:47

Not the answer you're looking for? Browse other questions tagged or ask your own question.