Having a hard time working out Thevenin equivalents when I see circuits like this:

Vs - 30V, R1 - 100 ohms, R2 - 1000 ohms, R3 - 1500 ohms, R4 - 900 ohms


So, right off that bat, I'm not sure I understand which resistors are in series or parallel. I thought that R3 and R4 are in series, and (R3 + R4) is in parallel with R2, and all of that is in series with R1. It doesn't seem correct, as I end up with the wrong answer.

If I was right, that would make it look like a voltage divider.

So, I come up with:

$$(R3 + R4) || R2 = \frac{1}{\frac{1}{1500\Omega + 900\Omega} + \frac{1}{1000\Omega}} = 705.88 \Omega $$

And then to figure out the Vth, I do:

$$V_{th} = 30V * \frac{705.88}{705.88 + 100} = 26.28V$$

I know that's wrong (I have the solution to this problem), but how do I really figure it out?

For Rth, I shorted out Vs, and figured out that ((R1 || R2) + R3) || R4.

That gives me...

$$((R1 || R2) + R3) || R4) = \frac{1}{\frac{1}{((\frac{1}{\frac{1}{100\Omega} + \frac{1}{1000\Omega}}) + 1500\Omega)} + \frac{1}{900\Omega}} = 574.8 \Omega $$

which actually comes out correctly!

Thanks in advance

  • 2
    \$\begingroup\$ The 'Vth' you've worked out is the voltage across R2. Now apply a voltage divider with R3 and R4 to get Vth=Vab \$\endgroup\$
    – Chu
    Commented Jun 24, 2017 at 23:26
  • \$\begingroup\$ Do you understand what @Chu wrote? You are working through things pretty well, but you just need to apply what he wrote here. You are almost there. Does it make sense? \$\endgroup\$
    – jonk
    Commented Jun 25, 2017 at 0:26
  • \$\begingroup\$ The voltage I calculated above was all the resistors, I think the voltage across R2 is 27.27. Can I just treat this as 2 voltage dividers? If R2 is 27.27 and then calculating the votage across R4 I get 10.22, but the answer in the solution I have is 9.85V. \$\endgroup\$
    – stbenjam
    Commented Jun 25, 2017 at 13:37
  • \$\begingroup\$ But your question says you've calculated 26.28V, not 27.27V, and \$\frac{900}{900+1500}\times 26.28 = 9.86\: V\$, as required! \$\endgroup\$
    – Chu
    Commented Jun 25, 2017 at 13:53
  • \$\begingroup\$ ... 27.27V is wrong \$\endgroup\$
    – Chu
    Commented Jun 25, 2017 at 14:00

4 Answers 4


I would just like to give another way of calculating the desired voltage by finding a Norton equivalent, which in result simplifies the circuit and should give you no trouble.

Find the current that goes through ab by short-circuiting the R4 resistor (remove it from the circuit and just place a wire instead of it). You have already found the thevenin resistance so you just have to find the current through ab (which has no R4 resistor) and multiply the two. $$I_{ab}R_e=V_T$$ I find this the easiest way of solving this.


Take a fresh look at the network. The voltage across a-b is the voltage across across R4. While you have calculated the total network resistance, that does not directly relate to the voltage across R4.

I am stopping short of giving you the answer. If you need further assistance, indicate so in the comments to this question and I will elaborate further.

  • \$\begingroup\$ I'm still not sure, can you help me understand the effect of the resistors on the voltage here? I'm just doing problems from tuttle.merc.iastate.edu/ee201/practice/thevenin/thevenin.php, and there's a ton of them that are like this. I thought originally, I could treat this as 2 voltage dividers, but I end up with 10.22V across R4 instead of the solution of 9.85V. It seems close, but I'm missing something in my calculation there. \$\endgroup\$
    – stbenjam
    Commented Jun 25, 2017 at 13:43

The resistance you have calculated just gives the voltage remaining after drop at resistor R1. But the Rth is not that one because there is still one resistor in series (i.e R3) within the resistance you have calculated. So you have to subtract the effect of R3 which will give you the voltage across R4 which is also the required Rth. Hope this one helps.....


I think the easiest way to intuitively solve this, is by determining the Thevenin equivalent in two steps, like so:

First you solve the Thevenin equivalent for Vs, R1 and R2 only


simulate this circuit – Schematic created using CircuitLab

In the first stage, you find that

$$V_{th1} = \frac{R_2}{R_1+R_2}V_s$$ $$R_{th1} = R1||R2$$

In the second stage, you find the actual Thevenin equivalent from the intermediate one.

$$V_{th} = \frac{R4}{R_{th1} + R_3 + R_4}V_{th1}$$ $$R_{th} = (R_{th1} + R_3) || R_4$$

The second way of solving this, is what I consider the "general" way of solving it: by using KCL or KVL laws to find the Thevenin voltage. This should always work, but it is also the most tedious. Using KCL laws, calling the middle voltage \$v_1\$, you can find that

$$\begin{align} &in\ v_1)\quad& \frac{v_1 - V_s}{R_1} + \frac{v_1}{R_2} + \frac{v_1 - V_{th}}{R_3} &= 0 \\ &in\ V_{th})\quad& \frac{V_{th} - v_1}{R_3} + \frac{V_{th}}{R_4} &= 0 \end{align}$$

A third way of solving this, is by using the EET (Extra Element Theorem). In which case you can leave out one element to find the output voltage almost immediately. For example, using R2 as the extra element, you can find \$R_d\$ and \$R_n\$ by leaving it out, making the schematic much easier.

$$\begin{align} \frac{V_{th}}{V_s} &= H_\infty\frac{1 + \frac{R_n}{R}}{1 + \frac{R_d}{R}} \\ &= \frac{R_4}{R_1 + R_3 + R_4}\cdot \frac{1 + \frac{0}{R_2}}{1 + \frac{R_1 || (R_3 + R_4)}{R_2}} \\ &\Downarrow \\ V_{th} &= \frac{R_4}{R_1 + R_3 + R_4}\frac{1}{1 + \frac{R_1 || (R_3 + R_4)}{R_2}} V_s \end{align}$$

I have to admit that it takes a little while to get used to this method, but it can certainly pay off in some cases.


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