Hi i got the trouble and i miss the concept about v out in thevenin i have the circuit like this enter image description here so i used voltage divider for finding the voltage at point 1, but how to finding voltage at Vo , is that same the result of voltage at 2 , and voltage between R2 and R3 , i thought at R2 and R3 we need voltage divider calculation also , from voltage at point 2, is that true? . if The calculation like that there is a voltage difference between two Vout


First iam finding the voltage at point 1 enter image description here

i got v1= 12V then vo use voltage divider like this enter image description here

so Vo is Vth? i got missconception about this

2.Finding for Rth RTh= (R3||R2)||(R1+R4)+R5+R6


  • \$\begingroup\$ Please, expand your anwer with the work you have done. You can use LaTeX markup for equations. \$\endgroup\$ Commented Oct 29, 2015 at 1:45
  • 2
    \$\begingroup\$ Possible duplicate of Multiple voltage dividers \$\endgroup\$
    – The Photon
    Commented Oct 29, 2015 at 16:36
  • \$\begingroup\$ You are making the same error as in this previous question. You need to consider that R2+R3 have an effect on the first divider (the one you've caluculated using just R1, R4, R5, R6). \$\endgroup\$
    – The Photon
    Commented Oct 29, 2015 at 16:37
  • \$\begingroup\$ Where did you choose your ground reference? \$\endgroup\$
    – jippie
    Commented Nov 1, 2015 at 10:22

1 Answer 1


I assume the question is asking for the Thévenin equivalent of this circuit as seen from the port indicated by Vo (across R2). Start by applying simple rules like combining series resistors. So let's say R56 = R5 + R6 = 2K. Likewise R14 = R1+R4= 2K.

No transform E1 and R56 to a [Norton] current source with I1 = 6mA. Since this is a current source, its resistor (call it R56' if you want), is now parallel with I1.

So now combine R56' with R14 (they are in parallel) to get R1456 = 1K. Transform back I1 and R1456 to a voltage source with E1456 = 6V.

Now this source sees a (series) voltage divider made by R1456', R2, and R3. And you want the voltage drop across R2, which is 1/3 of E1456 since R1456' = R2 = R3 = 1K, so Vo = 2V.

Also from these transformation, Rth = R2||(R1456+R3) = 2/3K.

If none of what I said makes any sense to you, you should probably watch some lectures like https://www.youtube.com/watch?v=UyPMMtW-2dc and https://www.youtube.com/watch?v=S-3NWNrAQRQ.


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