enter image description here

For the above question I drew the corresponding Laplace transform diagram, as follows (didn't draw the switch since it basically open circuit after \$t=0\$):


simulate this circuit – Schematic created using CircuitLab

For the inductor on the upper right, note that I plugged in the value of \$i(0^{-})\$ that is, \$3A\$, as that was the current that was flowing through it when the switch was closed for a long time (as \$\frac{12 V}{4\Omega}=3A\$).

The loop equation thus turns out to be:

$$\frac{12}{s}-4I(s)-2sI(s)+6-sI(s)-4I(s)=0$$ $$\implies I(s)=\frac{12+6s}{8s+3s^2}$$

Which on Inverse Laplace transform gives me the actual loop current in time domain as \$i(t)=\frac{3}{2}+\frac{1}{2}e^{-8t/3}\$.

Clearly, \$i(0^{+}) = \lim_{t\to 0^{+}}i(t)=\frac{3}{2}+\frac{1}{2}=2\$. Thus, \$i(0^{+})\$ is quite different from \$i(0^{-})\$, which is \$3\$ (in amperes).

Why is there a sudden jump in current between \$t=0^{-}\$ and \$t=0^{+}\$ ?

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    \$\begingroup\$ @TonyStewartEEsince1975 Which assumptions are you talking about? Also what do you mean by "there would be a singularity in voltage in 0 time"? \$\endgroup\$ – user186505 Apr 22 '18 at 14:21
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    \$\begingroup\$ V=LdI/dt as dt goes to 0, V goes to 1/0 \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 22 '18 at 14:27
  • \$\begingroup\$ @TonyStewartEEsince1975 Okay, so what would be the correct way to approach the question? \$\endgroup\$ – user186505 Apr 22 '18 at 15:07
  • \$\begingroup\$ @TonyStewartEEsince1975 What is HV? \$\endgroup\$ – user186505 Apr 22 '18 at 15:24
  • \$\begingroup\$ Let us continue this discussion in chat. \$\endgroup\$ – user186505 Apr 22 '18 at 15:24

It may help to recognize that this is analogous to connecting capacitors charged to different voltages in parallel. In the case of inductors it's conservation of flux rather than charge that is at play. So, by inspection, the current just after the switch opens will be 2/3 * I0.

As with the capacitor dual, you will also find that the energy is (apparently) not conserved because energy is lost in the switch opening. The energy in the inductors before the switch opening is 0.5*2*3^2 = 9J. The energy in the inductors after the switch opens is 0.5*2*2^2 + 0.5* 1* 2^2 = 8J.

  • \$\begingroup\$ Could you please say why you mention the word "apparently" there. In reality is energy conserved or lost as heat, in that case? \$\endgroup\$ – user186505 Apr 22 '18 at 16:37
  • \$\begingroup\$ It's lost as heat and EM radiation, yes. Of course energy is conserved. \$\endgroup\$ – Spehro Pefhany Apr 22 '18 at 16:51

Just like Charge should be conserved when two capacitors are brought parallel,

Flux should be conserved when two inductors are brought in series.

Equate the flux before and after switch was opened:

$$L_1I_{0^-} = (L_1+L_2)I_{0^+}$$

where \$I_{0^-} = 3A\$ , \$L1 = 2H \$ and \$L2 = 1H\$.

Means \$I_{0^+} = \frac{2}{3}I_{0^-} = 2A\$

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    \$\begingroup\$ Thank you for the answer. Someone already answered the same thing sometime back ^. But I appreciate the help. :) It sucks that I can green tick only one answer though. \$\endgroup\$ – user186505 Apr 22 '18 at 17:13

at t=0 loop current stays at 3A and decays to steady state 1.5A

Simulation in Browser


simulate this circuit – Schematic created using CircuitLab enter image description here


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