How to calculate current in this circuit?

Question

update:

How do I calculate current in this circuit if switch close at t=0 and the circuit initial conditions are \$IL1(0)=2A\$?

Answer 1

To answer the question you pose in your comment (changeing the circuit to include parallel resistors) - I would find whatever quantities I was interested in with Laplace. This will give you closed form solutions. If you don't care about the current in the resistors you can obviously lump them together to simplify your circuit.

Here is the new circuit in Laplace domain. Solve it using usual circuit methods (the voltage source is \$L_{1}I_{0}\$ where \$I_0\$ is the initial current through inductor \$L_1\$. Once you find an equation for the quantity of interest, transform it back to time-domain. Here is a great table of Laplace transform pairs.

I simulated this with \$L_1 = 2mH\$, \$L_2 = 3mH\$, \$R_{EQ}=500\Omega\$, and \$I_0=2A\$ and the resulting current through the equivalent resistor is plotted below.

p.s. If you apply this same method to your original circuit and solve for the voltage across either inductor - you will see that the inverse Laplace will produce a dirac delta, \$\delta(t)\$. This would be the infinite voltage that Elliot Alderson explains in his comment. Also, here and here are a couple of other examples where I work out a Laplace solution.

UPDATE: Per your request, I added the differential equations (inductor voltages) which you can use in writing KVL and KCL equations, and then solve with your initial conditions.

UPDATE 2: Per your latest request, I show the network below that you would solve for the desired quantity (using differential equations and your initial conditions).

I solved for \$i_{L1}\$ using Laplace and show it below - your work should result in the same: $$i_{L1}(t)=0.80 + 1.20e^{\frac{-10}{3}t} \text{ A}$$

Answer 2

This is an invalid circuit, if you expect to use the conventional circuit analysis techniques and assume that the elements are ideal. So you can't do any meaningful analysis.

Your initial conditions specify that the current through the two inductors is different, but they are connected in series. Our definition of series is that the two elements must have the same current because of how they are connected together.

Answer 3

This impossible initial condition, would create an impossible infinite voltage at t>0 and lose energy in the arc. V=LdI/dt.

If you had a transfer switch of 0 Ohms you could compute the total energy in the inductors and assume the final energy is the same with the added inductance L1+L2 to compute the new current and ignore reality. $$E=1/2 (L_1I_1^2+L_2I_2^2)=1/2(L_1+L_2)I_{new}^2$$ and solve for I-new.