# Using Thevenin's theorem, how do you find the Thevenin voltage?

The above image shows a circuit. It is required that we simplify the circuit using Thevenin's theorem to find the Thevenin voltage, current and resistance.

I've done a countless number of problems but only succeed in finding the Thevenin resistance. Having difficulties with finding voltage, in turn, unable to find the Thevenin current.

• Please show the methods you used for solving, and the steps you took, and we will be able to help you with where you went wrong
– MCG
Apr 23, 2019 at 8:19
• Define the Thevenin voltage. The problem becomes the usual circuit analysis. Apr 23, 2019 at 8:47
• @LongPham Are you able to use nodal analysis? Or mesh? Your problem will present a Thevenin voltage of 0 V. Shorting that will produce a current of 0 A, making such an approach to finding the Thevenin resistance difficult. You can instead inject a current of 1 A at node A(+) and solve for the Thevenin resistance (which is $\frac{36}{37}\:\Omega$. Regardless, you need to say something more about the methods you are familiar with and permitted to use. (Clearly, you or someone simulated the circuit to get those numbers.)
– jonk
Apr 23, 2019 at 8:56
• @jonk looks like the teacher or professor put together package of questions in a "quiz" perhaps to help with revision - looks like a neat job actually. Apr 23, 2019 at 9:27
• @jonk My apologies. I find the Thevenin resistance first by open circuit the current source and short the voltage source and calculate the equivalent resistance, so $R_{th} = 36//1 = \frac{36}{37} \Omega$. Next I find $V_{th}$ by using nodal analysis and by hand. $V_{th} = 0V$. Apr 23, 2019 at 10:24

I'll save myself a little time and avoid re-writing your schematic. You have labeled the parts and just need to label the nodes. The bottom node I'll consider "ground" or $$\0\:\text{V}\$$, labeled on your schematic as B(-). I'll number the remaining nodes from left to right, as $$\V_1\$$, $$\V_2\$$, and $$\V_3\$$ (which is also labeled on your schematic as A(+).)

Four equations in four unknowns (to read how I do nodal, see this section called nodal, done my way):

\begin{align*} \frac{V_1}{R_1}+\frac{V_1}{R_2}&=I_{V_1}+\frac{V_2}{R_1}+\frac{0\:\text{V}}{R_2}\\\\ \frac{V_2}{R_1}+\frac{V_2}{R_3}&=9\:\text{A}+\frac{V_1}{R_1}+\frac{V_3}{R_3}\\\\ \frac{V_3}{R_3}+\frac{V_3}{R_4}+I_{V_1}&=\frac{V_2}{R_3}+\frac{0\:\text{V}}{R_4}\\\\ V_1&=9\:\text{V}+V_3 \end{align*}

This solves out as $$\V_3=0\:\text{V}\$$. So this means that $$\V_\text{TH}=0\:\text{V}\$$. (The other values are $$\V_1=9\:\text{V}\$$, $$\V_2=63\:\text{V}\$$, and $$\I_{V_1}=4.5\:\text{A}\$$.)

If you don't have it already installed, I recommend getting sympy and sage. They are free and work well. For example, here's the above solved using sympy:

var('v1 v2 v3 r1 r2 r3 r4 iv1')
eq1=Eq( v1/r2+v1/r1, iv1+v2/r1+0/r2 )
eq2=Eq( v2/r1+v2/r3, 9+v1/r1+v3/r3 )
eq3=Eq( v3/r3+v3/r4+iv1, v2/r3+0/r4 )
eq4=Eq( v1, 9+v3 )
ans=solve( [eq1,eq2,eq3,eq4], [v1,v2,v3,iv1] )
ans[v3].subs({ r1:12, r2:1, r3:14, r4:36 })
0


I recommend sympy as it makes this all a piece of cake and its free to use.

Next, inject $$\1\:\text{A}\$$ by adding a current source straight into node $$\V_3\$$ (or A(+) if you like) in order to inject that current and see what happens to that node voltage.

The four equations in four unknowns now is:

\begin{align*} \frac{V_1}{R_1}+\frac{V_1}{R_2}&=I_{V_1}+\frac{V_2}{R_1}+\frac{0\:\text{V}}{R_2}\\\\ \frac{V_2}{R_1}+\frac{V_2}{R_3}&=9\:\text{A}+\frac{V_1}{R_1}+\frac{V_3}{R_3}\\\\ \frac{V_3}{R_3}+\frac{V_3}{R_4}+I_{V_1}&=1\:\text{A}+\frac{V_2}{R_3}+\frac{0\:\text{V}}{R_4}\\\\ V_1&=9\:\text{V}+V_3 \end{align*}

Note that all I did was add $$\1\:\text{A}\$$ on the right side of the $$\3^\text{rd}\$$ equation. This now solves out as $$\V_3=\frac{36}{37}\:\text{V}\$$. From this, you can infer that $$\R_\text{TH}=\frac{36}{37}\:\Omega\$$.

Here's sympy, again:

var('v1 v2 v3 r1 r2 r3 r4 iv1')
eq1=Eq( v1/r2+v1/r1, iv1+v2/r1+0/r2 )
eq2=Eq( v2/r1+v2/r3, 9+v1/r1+v3/r3 )
eq3=Eq( v3/r3+v3/r4+iv1, 1+v2/r3+0/r4 )
eq4=Eq( v1, 9+v3 )
ans=solve( [eq1,eq2,eq3,eq4], [v1,v2,v3,iv1] )
ans[v3].subs({ r1:12, r2:1, r3:14, r4:36 })
36/37


There are many other methods you can apply. But this is pretty bullet-proof and I can write out nodal equations all day long almost as fast as I can type. So that's the method of my choice, anyway.

• I understand your working out, however, the answer given is 1.728347x10^(-15) V. Just still a little confusing why. Apr 24, 2019 at 8:48
• @LongTran The answer given clearly wasn't generated by nodal analysis, but instead was generated by placing the circuit into a Spice simulator. Those are numerical solver errors. This tells me you are being taught by "cheaters" who can't be bothered to actually solve these using nodal. They want you to solve them one way, while they use Spice to do their work for them.
– jonk
Apr 24, 2019 at 11:58
• Interesting method :) Apr 24, 2019 at 16:37
• @LongPham The method of outflowing (on left) vs inflowing (on right) currents being set up? (Which I first observed when reading Spice code.) Or the method of using sympy as a solver? Or?
– jonk
Apr 24, 2019 at 17:38
• @LongPham It really helps to keep the signs straight because it is easy to see things as "inward" and "outward." I made far more mistakes when I kept trying to remember which to subtract from which. Then I read Spice code. And in trying to follow its reasoning, I got a sudden moment of vision and saw. This works well with inductors and capacitors as well, by the way.
– jonk
Apr 24, 2019 at 20:05

I've done a countless number of problems but only succeed in finding the Thevenin resistance. Having difficulties with finding voltage, in turn, unable to find the Thevenin current.

The I-V characteristic of a linear 1-port network is a straight line. The Thevenin voltage is where this line crosses the x-axis (the voltage with a 0 A load). The Thevenin resistance is the slope of the line ($$\\frac{dV}{dI}\$$).

You can use any method you like to find the equation of this straight line. Normally, you will find two points on the line by considering what will be the output of the circuit with two different test loads. For some circuits it might be useful to consider loading the network with a voltage source, for some with a current source, for some with a resistor or two different resistors. It's up to you. Now use the two points you found to get the equation of the line.

There is no "Thevenin current" so there's no point trying to find it. It might be useful to try to find the Norton equivalent source current. That's just the place where the I-V curve crosses the y-axis (the current with a 0 V load).

The Thevenin voltage is equal to the open circuit voltage of the terminals in question.

If you apply superposition to your circuit, you will find the solution for $$\V_{a,b}\$$ to be fairly trivial.