I had asked this question before and thought I had found the help I needed, but unfortunately I found myself unable to solve the problem fully when I attempted it. The problem:


For the Thevenin resistance, I believe I'm on the right track but am not confident in my answer: R5+(R4 ||(R2+R3)), excluding R1 because I believe it will become irrelevant if I short the voltage source?

As for the Thevenin voltage, I wasn't sure how I was supposed to go about solving it. Any insight would be appreciated.

  • \$\begingroup\$ For Thevenin voltage you need to find the voltage on R4. Pretty easy. As you said, R1 is out as it is parallel to the source. R5 is out as well, as there is no current in it. \$\endgroup\$
    – Eugene Sh.
    Commented Sep 28, 2018 at 18:32
  • \$\begingroup\$ Just look at the circuit and see where the current flows and where it does NOT flow. For example, no current can flow through \$R_5\$. It's got nowhere to go. So if you can work out the current in \$R_4\$, then you can figure out the voltage across it. And if you know the voltage across it and if there is zero current in \$R_5\$ (no voltage drop, therefore), then you have your answer. \$\endgroup\$
    – jonk
    Commented Sep 28, 2018 at 18:40
  • \$\begingroup\$ Thank you for the inputs, I was able to see what I wasn't seeing before. \$\endgroup\$
    – shinryu333
    Commented Sep 28, 2018 at 19:03

2 Answers 2


so R5 is out, which means the thevenin voltage is the voltage Across R4. Using KVl in the loop , you obtain Vthev=V*R4/ (R4+R2+R3)

  • \$\begingroup\$ What? How is \$R_1\$ out? Finding \$V_{OC}\$ doesn't involve shorting \$R_1\$ out. -1 \$\endgroup\$
    – Shamtam
    Commented Sep 28, 2018 at 19:34
  • \$\begingroup\$ @Shamtam The existence of R1 does not change the result. Not because of shorting it, but because it is paralle to the voltage source and does not influence the R4 current. \$\endgroup\$
    – Eugene Sh.
    Commented Sep 28, 2018 at 19:42
  • 1
    \$\begingroup\$ yeah okay fair point as it might confused , i changed the answer ,that R1 is out is just to solve for Rth. \$\endgroup\$
    – Navaro
    Commented Sep 28, 2018 at 19:50

Your approach to solving for \$R_{th}\$ is correct.

The quick answer is to note that no current flows through \$R_5\$, which means there's no voltage drop across it. Thus, all you need to do is solve for the voltage across \$R_4\$ in the simple two-loop circuit that remains.

I'll leave the rest up to you.


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