In circuit shown in the picture below all impedances are known, when switch is off, current I1 is also known I1=20mA. The task is to find the same current when the switch is on. Impedances:
Z1=(20+j40) Ohms, Z2=(120-j80)Ohms, Z3=(80+j120)Ohms, Z4=(140-j180)Ohms, Z5=(100+j100)Ohms.


I've tried to solve it this way:

Since i know all impedances and i don't know the voltage on E1 and E4 and current of Ig5 i thought i should find Thevenin generator for the part of the circuit left from the Z2 (including the branch with Z2) when the switch is off(since then i know the current I1). Impedance of Thevenin generator is Zt=40(2-j)V now i ave circuit with Z1 and Zt in series and two voltage generators which can be replaced with one whose voltage is sum of their voltages. With known current it's easy to determine E(E=E1+Et) and i got E=2V. When the switch is on i have only Zt impedance left, and E which is now known value, so I1'=E/Zt=(100+j50)mA.

So, is this correct and is this valid approach, i mean am i missing something or making mistake somewhere? Any suggestion, advice or correction is welcome.

 

The problem with your approach is that I1 is the current flowing in Z1, but it is NOT the current flowing in Z2, nor is it the current flowing out of your Thevenin equivalent for the left hand side.

I don't follow where you get that Z1 and Zt would be in series or where your two voltage sources could be combined.

It would help a lot if you draw sketches of the transformations you are making instead of just trying to describe them in words.

 

The problem with your approach is that I1 is the current flowing in Z1, but it is NOT the current flowing in Z2, nor is it the current flowing out of your Thevenin equivalent for the left hand side.

I don't follow where you get that Z1 and Zt would be in series or where your two voltage sources could be combined.

It would help a lot if you draw sketches of the transformations you are making instead of just trying to describe them in words.

Ok, this is what i meant:

So thevenin generator includes all elements of the circuit except E1 and Z1 which means that circuit after implementing thevenin's theorem looks like this:

 

That helps a lot. From your verbal description it sounded like you were finding the Thevenin circuit for E1, Z1, and Z2 since you said that it was for the part of the circuit LEFT of (and including) Z2.

I'll try to look at it later. Got stuff I've got to do right now.

 

Cdummie, your approach seems valid, but I haven't checked the math on your actual answer.
Regards

 

Cdummie, your approach seems valid, but I haven't checked the math on your actual answer.
Regards

Thank you, that's what i needed to know, even if i made some math mistakes they can be fixed, more important thing is that approach is good, thanks again!