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I have to do transient analysis in the following circuit.

When t<0.

L2 inductor is effectively bypassed and 10A flows through inductor L1.

At t = 0+ L1 behaves as a current source outputting 10A and L2 behaves as an open circuit.

I need the value for i(0+) to further solve the question. If I had to guess, I'd go with 0A because I feel like an open circuit is supposed to take priority over a current source.

enter image description here

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  • \$\begingroup\$ Hm, if "L_2 behaves as an open circuit" and "L_1 behaves as a current source", you would be getting infinite voltage over L_2. And while that is a valid analytical result for an inifinitely short duration, it makes it impossible to say what happens in the time after t=0. So, you need to actually write down your differential equations and solve this, analytically, instead of just going by rules of thumbs. \$\endgroup\$
    – Marcus Müller
    Sep 23 at 18:19
  • \$\begingroup\$ Are the two inductor magnetically shielded? \$\endgroup\$
    – Franc
    Sep 23 at 18:48
  • \$\begingroup\$ @MarcusMüller The differential equations are unsolvable without knowing i(0+). \$\endgroup\$
    – rjpj1998
    Sep 23 at 18:58
  • \$\begingroup\$ @Franc The question doesn't specify but let's assume they are coupled. What would i(0+) be? \$\endgroup\$
    – rjpj1998
    Sep 23 at 18:58
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    \$\begingroup\$ as said, assume i(t) is differentiable. You know i(0⁻). \$\endgroup\$
    – Marcus Müller
    Sep 23 at 19:43

1 Answer 1

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The initial current seems distracting, The focus should be on the voltage aross \$L_2\$.

Place a resistor \$R_x\$ in parallel with \$L_2\$.

Pretend that \$L_1\$ is a current source for now.

The voltage across \$L_2\$ is \$V_{L2}=R_x I(0^+)\$

The current in \$R_x\$ decreases while the current in \$L_2\$ increases.

All of the current in \$R_x\$ transfers to \$L_2\$ in \$5\frac{L_2}{R_x}\$ time constants as \$V_{L2}(0^+)\rightarrow 0\$

Set \$R_x=\text{TeraOhms}\$ to approximate removing it from the circuit. Then the duration of \$t=0^+\$ is pico-seconds. Take the \$\underset{R_{x}\rightarrow\infty}{\lim}\$.

So now you should be able to determine \$I(0^+)\$ and \$V_{L2}(0^+)\$.

Now also \$L_2\$ stops being a current source a becomes an inductor again.

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