Hi, first post, glad to be here!

Have a hard time understanding Thévenin equivalent and this problem is not a exception. The question is "The inductor L is initially uncharged. Set up the expression for the current iL (t) through the inductor"

I understand the fact that Rth = R1 // R2+R3.
But when transformering the current source to voltage source, V=I*R we only calculate with the R2 resistor. From my perspective the current can flow in both branches and the voltage source should be V=I * R2 // R1+R3.

And I don´t understand the calculations of the voltage divider to get Vth. Would be thankful if someone could explain.

Best regards

 

The conversion of the current source and 4 Ω resistor to an equivalent voltage source is just a source-transformation technique used for circuit analysis. You could certainly analyze the circuit without doing so. You would discover the 4/9 of the current goes through the 3 Ω resistor, resulting in 32 V appearing across it.

Vth is the voltage appearing between the terminals where the inductor was, which is the voltage on the left side minus the voltage on the right.

Vth = Vleft - Vright

Using the bottom node as the reference node

Vleft = Vsource - V_R2
Vright = V_R3

Where the voltage across the resistors are taken with the current flowing clockwise in the circuit and the positive terminal being the terminal that the current enters.

So you have

Vth = Vsource - V_R2 - V_R3

The reason that their Vth calculation doesn't make sense is that they were extremely sloppy and made five different mistakes in their work.



If you have the option, run away from whomever wrote that.