# Transient analysis with charged and uncharged inductor in series

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.

• 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. Sep 23 at 18:19
• Are the two inductor magnetically shielded? Sep 23 at 18:48
• @MarcusMüller The differential equations are unsolvable without knowing i(0+). Sep 23 at 18:58
• @Franc The question doesn't specify but let's assume they are coupled. What would i(0+) be? Sep 23 at 18:58
• as said, assume i(t) is differentiable. You know i(0⁻). Sep 23 at 19:43

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.