# Need help to find Thevenin resistance and Thevenin voltage

Here is the question I am trying to solve(there are 2 things I need to find. These are Vth and Rth):

Here is my attempt: ( please check Vth if it is true )

I couldn't find Thevenin resistance at all

How can I solve my question above like this example?

There are 2 solutions to find Rth:

1.One of them is connect a and b and obtain Isc.

2.Other one is diactivate independent sources and connect a voltage source between a and b. I solved solution 1 before. But I couldn't obtain result with that. This is the second solution that I applied. Can someone check if it is correct ? (I added the example I was inspired by below) This is an example I imitated the solution

• Well, the setup for your (1) KCL equation looks right to me. I'd have included Isc so that I could set it to 0 or to 1. But other than that, looks okay. Why didn't you solve it for Va? The approach should be that you set up your KCL equation there, but include Isc in it, then solve for the function, Va(Isc). Once solved, you will have Isc present. Set Isc=0 and solve. That gives Vth=Va(0). Then set Isc=1 and solve for Va(1). The voltage difference will be Rth=(Va(1)-Va(0))/(1-0)=Va(1)-Va(0). Jun 14, 2023 at 0:24
• If I solve the first equation, Va is obtained as just a number not a function. Because the only thing that not known is Va. You can say "we don't know Ix as well" but actually no. We know Ix. It is (600-Va)/10 . If we put this expression into the equation we obtain Va just a number not a function. Jun 14, 2023 at 12:31
• I edited the question. Can you look at again ? Jun 14, 2023 at 13:38
• I think I now see a disconnect between our mutual ways of thinking about the problem. You are wondering about the idea of observing the open-circuit voltage and then shorting out a to b and observing the short circuit current. I was considering this differently. I looked at Isc as being an external current source. Set first to 0 and then set next to 1 A, forcibly injecting a current, and then observing the voltage at A. These two ideas get the same result. But they are different approaches. Jun 14, 2023 at 20:46
• hey can you look my other 2 questions? I need help. I have an exam tomorrow :( Jun 15, 2023 at 18:50

Looks to me like you did just fine finding that $$\V_{_\text{TH}}=300\:\text{V}\$$. That's also what I get. So I think your question is mostly about finding $$\R_{_\text{TH}}\$$.

There are lots of ways to go. It's not uncommon to see the suggestion of shorting the two terminals together and measuring the current. But this method really isn't universally applicable. For example, suppose the circuit is just a resistor going from terminal A to terminal B? You will measure $$\V_{_\text{OC}}=0\:\text{V}\$$ and $$\I_{_\text{SC}}=0\:\text{A}\$$. Which tells you nothing much about $$\R_{_\text{TH}}\$$.

A mathematically minded type would say that you need to inject an infinitesimally small current and see what infinitesimally small voltage change occurs as a result of it. This mathematical approach is excellent as it treats $$\R_{_\text{TH}}=\frac{\text{d}\,V}{\text{d}I}\$$, the linear slope at the operating point of the internal circuit, without changing anything about the large scale operating point of it. This method may be required in exceptional cases (or in certain numerical simulations.) But it's rarely needed.

The method I've settled on and is also widely used and well-documented is the idea of simply placing a current source at the two terminals, setting the current source to $$\0\:\text{A}\$$ and taking a voltage reading and then setting the current source to $$\1\:\text{A}\$$ and taking another voltage reading. This will work fine if the circuit is no more than a simple resistor, too. So it's my "go-to" approach.

Let's take the KCL result you provided above, which I agree with:

\begin{align*} \frac{V_a-0\:\text{V}}{R_2}+\frac{V_a-V_2}{R_4}&=\frac{V_1-V_a}{R_1}+I_a \\\\ \frac{V_a}{R_2}+\frac{V_a-10\cdot I_x}{R_4}&=\frac{V_1-V_a}{R_1}+I_a \end{align*}

I follow your work except that I've added an injectable current, which will be set first to zero and then to one, measuring the value of $$\V_a\$$ each time.

The solution is

\begin{align*} \frac{V_a}{R_2}+\frac{V_a-10\cdot I_x}{R_4}&=\frac{V_1-V_a}{R_1}+I_a \\\\ \frac{V_a}{R_2}+\frac{V_a-10\cdot \frac{V_1-V_a}{R_1}}{R_4}&=\frac{V_1-V_a}{R_1}+I_a \\\\ \frac{V_a}{R_1}+\frac{V_a}{R_2}+\frac{V_a}{R_4}+\frac{10\cdot V_a}{R_1\,R_4}&=\frac{V_1}{R_1}+\frac{10\cdot V_1}{R_1\,R_4}+I_a \\\\ V_a\cdot\left(\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_4}+\frac{10}{R_1\,R_4}\right)&=V_1\cdot\left(\frac{1}{R_1}+\frac{10}{R_1\,R_4}\right)+I_a \\\\ V_a\cdot 0.3\:\mho&=V_1\cdot 0.15\:\mho+I_a \\\\ V_a=V_1\cdot 0.5+I_a\cdot \frac1{0.3\,\mho}&=\frac12 V_1 + 3.\overline{3}\:\Omega\cdot I_a \\\\ \therefore\quad V_a\left(I_a\right)&=\frac12 V_1 + 3.\overline{3}\:\Omega\cdot I_a \end{align*}

It's pretty quick to see that $$\R_{_\text{TH}}=3.\overline{3}\:\Omega\$$. I didn't even need to inject any current. I could, of course. But there's no point because the Thevenin resistance just stands out in the open, now.

But if you are using a numerical solver for something like this, then you'd just plug in two different values for the current. (For simplicity, these two currents are usually taken to be zero and one amp.) So, just using a numerical solver, find:

$$R_{_\text{TH}}=\frac{V_a\left(I_a=1\:\text{A}\right)-V_a\left(I_a=0\:\text{A}\right)}{\left(I_a=1\:\text{A}\right)-\left(I_a=0\:\text{A}\right)}=V_a\left(I_a=1\:\text{A}\right)-V_a\left(I_a=0\:\text{A}\right)$$

• It seems I found true results. Thank you so much for your labor Jun 15, 2023 at 10:11