Finding Instantaneous UL and iL After Commutation for Transients (LC) Physics

[gneill]

But there are two loops that interact. You only need to solve for the current in the second loop, but the first loop still plays a part!

 

[builder_user]

But there are two loops that interact. You only need to solve for the current in the second loop, but the first loop still plays a part!

So,I do not need first loop as equatation to solve?

 

[gneill]

You need the first loop equation in order to eliminate its current (say, i1) from the second loop's equation. But you don't have to solve for i1.

You have solved circuits with two loops before. You know that the loop currents in both loops affect each other in the solution. So you have to write the equations for both loop currents (i1, i2), and eliminate i1 from the the loop 2 equation. Solve for i2.

 

[builder_user]

You need the first loop equation in order to eliminate its current (say, i1) from the second loop's equation. But you don't have to solve for i1.

You have solved circuits with two loops before. You know that the loop currents in both loops affect each other in the solution. So you have to write the equations for both loop currents (i1, i2), and eliminate i1 from the the loop 2 equation. Solve for i2.

But I know J - so it's i1?

 

[gneill]

But I know J - so it's i1?

I don't believe that your circuit with J in series with R2 is correct. The original circuit (with the switch closed) has J in parallel with R2 and in parallel with C1. I don't see how you could go from there to having J in series with R2.

If J were a voltage source, then yes, I would agree. Then you would have the Thevenin equivalent of J and R2.

 

[builder_user]

Correct result is sth like this.(I've only changed direction of Uc. It was mistake but it wasn't a big problem)
i(t)=7.3*e^(-2863t)-3.89*e^2.73t*cos⁡(133t)-0.4175*e^2.73t*sin⁡(133t)