In this circuit all the components that have values are known, values of components that aren't shown are considered unknown. If the switch is off, then current I1' is known and it's 100mA. Now, i'm not really sure how to find the value of the current I3 after the switch is closed. I tried to solve it and i did it this way:

1. While switch was off, I found Thevenin generator for the part of the circuit right from the R3, Rt was 50 Ohms and i could't find the value of Et
2. Now, Et and E3 were in series, so were the resistors, so i summed up values of resistances Rt and R3 and it was 200 Ohms, and placed one E instead of Et and E3 (considering E=E3+Et)
3. Since i have the value of Ig1 and I1' i found the current through the E and R3 resistor and it was 75mA
4. After that, i found the voltage R1*I1'-E1 because it was equal to I3*200-E1 so i could find E and it was 23V
5. When the switch turned on i wasn't considering original circuit, i used circuit with E=23V and R=200Ohms right from the branch with switch.
6. Using node voltage method i found out that I3=29mA (i considered I3 is current through R and E, but i'm not sure if that's correct because circuit looked different at the begging so i'm not sure is that the I3 i need to find)

 

There are four unknown components and you are only given one known current to compensate. My initial reaction is that the problem isn't adequately constrained. But we aren't trying to solve for all of the voltages and currents, just one of them (I3 is a slightly different circuit), so perhaps we are okay and the approach you are taking is reasonable. Notice that if you find the Thevenin equivalent for everything to the right of the branch with the switch (so V3 on), you end up with a known resistance and unknown voltage. But at that point you have a circuit with one unknown component and one known current, so you can find the unknown Thevenin voltage very easily (almost by inspection). With that you have everything you need to find I3 when the switch is closed.

I just did it very quickly (meaning that I might have made a mistake and I haven't checked my result) and I get a value of I3 after the switch is closed that is between 50mA and 70mA.

 

There are four unknown components and you are only given one known current to compensate. My initial reaction is that the problem isn't adequately constrained. But we aren't trying to solve for all of the voltages and currents, just one of them (I3 is a slightly different circuit), so perhaps we are okay and the approach you are taking is reasonable. Notice that if you find the Thevenin equivalent for everything to the right of the branch with the switch (so V3 on), you end up with a known resistance and unknown voltage. But at that point you have a circuit with one unknown component and one known current, so you can find the unknown Thevenin voltage very easily (almost by inspection). With that you have everything you need to find I3 when the switch is closed.

I just did it very quickly (meaning that I might have made a mistake and I haven't checked my result) and I get a value of I3 after the switch is closed that is between 50mA and 70mA.

So basically, what i did is correct, and the value i found is actually the same as the I3 in the original circuit? I mean that's what i really need to know. Thanks!

 

So basically, what i did is correct, and the value i found is actually the same as the I3 in the original circuit? I mean that's what i really need to know. Thanks!

What you did is basically correct, but I never said that the value of I3 that you got was correct.

Have you checked your work? Always check your work (and note that I pointed out that I hadn't had time to check mine previously).

So let's check my work. I said that I got an answer between 50mA and 70mA. Since the value I got for I3 when the switch is open is 75mA that means that I am claiming that closing the switch will make I3 go down. Does this make sense? Well, with 100 mA of current in the left branch the voltage on the top node is 8V. When we close the switch current will flow downward in R2 which will drop the voltage on the top node. With a lower voltage on the top node and a Thevenin voltage on the far right of 23V, that means that I3 must increase. Hence, my previous answer MUST be wrong.

That also means that your answer of 29mA MUST be wrong.

You can also check this by plugging the answer into the circuit and see if you get a circuit that works out. The most obvious way, to me, to do that would be to keep the Thevenin equivalent for the entire right half in place (23V in series with 200Ω) and treat the 25mA current source as an unknown. With 29mA flowing in the Thevenin part of the circuit, you get a voltage on the top node of 23V - 200Ω·29mA = 17.2V. With the top node voltage of 17.2V you get a current downward in R2 of 13.2V/300Ω = 44mA. With 17.2V on the top node, you get a current in R1 of 29.2V/200Ω = 146mA. So what must the current source be to satisfy KCL on the top node? Does it match 25mA upward? If not, something is wrong.

Having failed my sanity check, I didn't even need to do that much. I just redid the problem (since my earlier work is at the office) and I did overlook a component. The correct answer is something a bit more than 75 mA and not only does it satisfy the sanity check that I3 had to increase, but it also passes the detailed check described above.

The nice thing about most engineering problems, even in the real world, is that you can almost always verify the validity of an answer from the answer itself. You may not have solved the right problem, but you should almost always know that you correctly solved the problem you did solve.

 

Thanks a lot for the explanation, it really makes sense, but that's my main problem, i have idea every time on how to solve something, but i'm rarely sure if it's correct or not. Guess i have to practice more, anyway, thanks again.

 

Yep, lots and lots of practice.

A very good way to get that practice AND boost your grades is to always, always, ALWAYS check your work. If nothing else, solve the problem a second time using a different method. That will not only get you good practice with different techniques, it will also allow you to start internalizing when different techniques are better suited than others.