# How to calculate current in this circuit?

update:

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

• Insert initial conditions using .ic as a spice directive and try it. Jan 3, 2021 at 17:29
• If the two inductors are in series in this manner, what interpretation can you ascribe to L1(0)=2 A,L2(0)=0? Can you reconcile it with Kirchoff's Current Law? Jan 3, 2021 at 17:36
• assuming the problem is valid and does not use ideal elements, assume L1 is also a resistor of R1 ohms and L2 of R2. then assume that the connection L1-L2 is 0 ohms (because why not?)...
– Abel
Jan 3, 2021 at 19:17

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}$$

• can we analyse the circuit by not using laplace? i am studying circuit analysis in RLC chapter.I have not studied laplace transform yet. Jan 4, 2021 at 4:33
• Yes, you write the differential equations (e.g. $v=L\frac{di}{dt}$) for the inductor voltages and solve the circuit with KVL. Then solve for desired quantities using initial conditions. Laplace just greatly simplifies this by making it an algebra problem. Jan 4, 2021 at 12:45
• When L1 discharge some current pass through R and charge L2 in the same time.L1 discharge current also depend on L2 and R.L1 cannot be replaced with constant current source. Can you write differential equation for this example? Jan 4, 2021 at 15:42
• I added a picture to show the differential voltages across the inductors (which are also impressed across their parallel resistors). From these you can write KVL and KCL equations and then solve them using the initial conditions. Surely your teacher is explaining this. Jan 4, 2021 at 16:36
• now i can solve 2 DE form kvl and kcl.i get the same answer as yours. Jan 5, 2021 at 16:32

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.

• What is wrong with it? At time t=0 The circuit is re-configured with switch action to produce the circuit you see for t>= 0. Jan 3, 2021 at 18:12
• @relayman357 There is no switch in the OP's circuit. However, if there were the circuit would be invalid when the switch closed. For inductors, $v_L = L\frac{di}{dt}$, so when the switch closes one or both of the inductor currents must change instantaneously, which requires an infinite voltage, which leads to numerically undefined results. Jan 3, 2021 at 18:28
• Yes, good point @ElliotAlderson. Agree. Jan 3, 2021 at 18:29
• if i add a resister parallel to l1 and l2 how to calculate? Jan 3, 2021 at 19:01
• I would use Laplace transform methods. Jan 3, 2021 at 19:02

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.