i need your help to solve this simple circuit using Thevenin, i used Ohm to solve but using thevenin the values was diferent, so i need your help, the question is the voltage at R4 and the current at R6, thanks a lot.

Thevenin Circuit

I do this: enter image description here

  • 1
    \$\begingroup\$ Show your work and we might point out what you messed up. Without knowing what you did, though, we can hardly figure out what you did wrong. \$\endgroup\$
    – The Photon
    Commented Mar 2, 2017 at 4:41
  • \$\begingroup\$ What you showed is correct so far. That should give you the correct result for voltage across R4. For current through R6, if required to use thevenin, you'll have to go back and find a new equivalent source connected at the terminals of R6. \$\endgroup\$
    – The Photon
    Commented Mar 2, 2017 at 5:27
  • 1
    \$\begingroup\$ @ThePhoton Except that it looks like the OP computed \$R_{TH}=\left(40\:\Omega\vert\vert 120\:\Omega\right)+80\:\Omega=320\:\Omega\$ instead of \$\left(40\:\Omega\vert\vert 120\:\Omega\right)+80\:\Omega=110\:\Omega\$ for that step. \$\endgroup\$
    – jonk
    Commented Mar 2, 2017 at 5:35
  • \$\begingroup\$ Can you do \$I_6\$? \$\endgroup\$
    – skvery
    Commented Mar 2, 2017 at 8:02
  • \$\begingroup\$ @jonk yeah my mistake, to find the Voltage at R6 now i get the resistence 110 and voltage 7.5v can i use Ohms law? \$\endgroup\$ Commented Mar 2, 2017 at 13:38

2 Answers 2


Douglas, the question regarding the voltage across \$R_4\$ should be easily solved using your own approach. CASE 1 below shows what I believe you arrived at, though as I said I think you miscalculated the resulting \$R_{TH}\$. I've shown what I think the equivalent should be on the right side of it. You should be able to easily apply Ohm's law now to work out the voltage across \$R_4\$. Compute the total current and apply that to \$R_4\$.


simulate this circuit – Schematic created using CircuitLab

I've also shown above your \$R_6\$ question. I hope I followed the approach you were being taught by breaking the connections around that resistor. From there, it's very simple to work out the value of \$R_{\vert\vert}\$, replacing four resistors with just one. The rest should be again quite easy to do, as you have a simple circuit and it should be easy to calculate the total current in it. That will also, of course, be the current in \$R_6\$.

  • \$\begingroup\$ R4 perfect , i forgotten put back R4....My mistake, but R6 i still dont understand, what you do OK, no problem, but after that i use the voltage of R3 to find the current? \$\endgroup\$ Commented Mar 2, 2017 at 18:32
  • \$\begingroup\$ @DouglasNickson Just sum up the series resistance of all three resistors and use that to compute the loop current. Since all of the loop current must also flow through \$R_6\$, you have your answer for the current in \$R_6\$. \$\endgroup\$
    – jonk
    Commented Mar 2, 2017 at 18:35
  • \$\begingroup\$ The sum of 60 + 20 + 20? This is not ohms law? It's not necessary find RTH and VTH? \$\endgroup\$ Commented Mar 2, 2017 at 18:41
  • \$\begingroup\$ @DouglasNickson No, it's not necessary. But if you did, \$R_{TH}\$ would just be the sum of \$R_1\$ and \$R_{\vert\vert}\$. \$\endgroup\$
    – jonk
    Commented Mar 2, 2017 at 19:01
  • \$\begingroup\$ hum, but in this part of the circuit when i open the terminals at R6 the voltage will not be 0? So How can i find VTh? \$\endgroup\$ Commented Mar 2, 2017 at 20:53

Simple.... Your Rth is wrong... Rth = ((20+20)//120) +40+40 Then solving this we get it as =>(40//120)+80 =>(40*120/(120+40)) + 80 => (4800/160)+80 => 30+80 => 110 Rth=110 ohms...... Now Vth as u already calculated is 7.5 Voltage across r4 is Vth * r4/(r4+Rth) 7.5*40/(40+110).........=> 7.5*4/15 ===2 v

  • \$\begingroup\$ It's possible use Ohms law to find the voltage at R4? Because using Ohms law the voltage that a get was 2.72v \$\endgroup\$ Commented Mar 2, 2017 at 13:33
  • \$\begingroup\$ Show how you solved it using ohm's law.... May be there is some mistake \$\endgroup\$ Commented Mar 3, 2017 at 3:33
  • \$\begingroup\$ yeah, i did some things wrong, but now using Ohms is correct, i find 2v, but, can't find the current at R6 using Thevenin, can you help me please? \$\endgroup\$ Commented Mar 3, 2017 at 3:37

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