Hi,

I have trouble finding the equivalent Thevenin resistance for this network:

I managed to do most of the other exercises and looked on the internet for quiet a while, but really can't find it.

I've found E(th) though. I redrew the circuit with the 300 ohm resistance in series with a parallel combination of 400 ohm and 2k on the one side and 1k on the other side. R(tot) then is 1005,8 ohm, I(tot) = 14,9 mA. Voltage over the 300 ohm resistor is 4,47V, over the 400 ohm resistor 1,755V, substracted from the 15V voltage source gives 8,775V for E(th).

But I just can't find the equivalent Thevenin resistance. I redrew it but don't know if it's correct...

I thought, looking "in" to the circuit, the first thing you the load "sees" is a parallel combination of the 400 and 2k resistances, solve for them, put them in parallel with 1k and the whole in series with 300 gives R(th), but that's incorrect. Any clues?

Thanks!

 

Hmm, calculate Rth through this diagram

PS And there is a error in answer
Rth is equal =479.532Ω



So for sure Rht isn't equal (2KΩ+400Ω)||1KΩ||300Ω

 

so you moved the voltage source, I can tell that and... nothing more

I realise my equivalent circuit is not correct, but with yours I still don't know what to do with the diagonal resistor. 2000 ohms in parallel with 400 + 1000 in series and that as a whole in parallel with 300 doesn't work. Without the 1000 it would be 2000 // 400 + 300, but now, I really can't figure it out

 

OK,
I drew this diagram slightly different and I drew the current path to.
Maybe this will help.

 

Got it!

I guess

2k//(400+(1k//300)) ?

 

Yes, you got it.

 

Yes, you got it.

Thanks for your help!