1
\$\begingroup\$

enter image description here

I understand from here I'm supposed to solve the current flowing through \$R_1\$. So I have to remove the \$R_1\$ and add terminals A and B and solve the resistance between them? Would it be easier to solve something like this by first redrawing the diagram?

I just tried to do it so, that I first removed the \$R_1\$ resistor (which now has terminals A and B placed) and the current source \$J=0\$. I then calculated the voltage between A and B: $$V_0=\frac{R_2}{R_2+R_4}E$$ Then I did the same but put \$E=0\$ and removed it so I got: $$V_1=J\left(R_3+\frac{R_2 \times R_4}{R_2+R_4}\right)$$ I then added them together to get the Thevenin voltage \$V_t=v_0+v_1\$. And the Thevenin's resistance should be $$R_t=R_3+\frac{R_2 \times R_4}{R_2+R_4}$$ Am I even close?

\$\endgroup\$
5
  • \$\begingroup\$ electronics.stackexchange.com/a/682618/319836 should help answer your question. \$\endgroup\$
    – RussellH
    Commented Sep 25, 2023 at 14:05
  • \$\begingroup\$ The left-side node is \$V_\text{L}=\frac{E-J\cdot R_1}{1+\frac{R_1}{R_3}}\$ and the right-side node is \$V_\text{R}=\frac{E+J\cdot R_2}{1+\frac{R_2}{R_4}}\$. What else do you need? \$\endgroup\$ Commented Sep 25, 2023 at 16:07
  • \$\begingroup\$ I just tried to do it so, that I first removed the R1 resistor and the current source J=0. I then calculated the voltage between where the R1 would be and got: V0=R2/(R2+R4)E. Then I did the same but put E=0 and removed it so I got: V1=J(R3+((R2*R4)/(R2+R4))). I then added them together to get the Thevenin voltage Vt=v0+v1. And the Thevenin's resitance should be Rt=R3+((R2*R4)/(R2+R4)). Am I even close? \$\endgroup\$
    – JohnnyB
    Commented Sep 25, 2023 at 16:40
  • \$\begingroup\$ The resistance seen by E? That's just E divided by the currents in R1 and R2. This value will depend upon E. So it's not fixed. For the schematic I see, and from the perspective seen by E, find: \$R_{_\text{T}}=\frac{R_1\cdot R_2+R_1\cdot R_4+R_2\cdot R_3+R_3\cdot R_4}{R_1+R_2+R_3+R_4+\frac{J}{E}\cdot\left(R_2\cdot R_3-R_1\cdot R_4\right)}\$. Clearly, the denominator has a term that depends upon the applied source. \$\endgroup\$ Commented Sep 26, 2023 at 11:56
  • \$\begingroup\$ But you need to clarify what you are really on about here. I thought it was about: \$I_1=\frac{J+\frac{E}{R_3}}{1+\frac{R_1}{R_3}}\$. I'm probably missing the point and just need a kick in the head about it. \$\endgroup\$ Commented Sep 26, 2023 at 12:09

1 Answer 1

1
\$\begingroup\$

I don't think the schematic editor here includes your voltage and current source symbols. And it's unusual to use J for a current source, my experience. But I'll use your variable naming, regardless.

I understand from here I'm supposed to solve the current flowing through \$R_1\$ ... using Thevenin's Theorem

Wikipedia's page on Thévenin's theorem discusses the theorem. But it doesn't really provide an algorithm for getting \$I_1\$. For that, you may use any of these commonly applied methods:

  1. Superposition Theorem: see this Wikipedia page.
  2. KCL (aka nodal.)
  3. KVL (aka mesh.)

I don't know which of these match your "using Thevenin's Theorem" requirement. Perhaps any or all of them? Or just the first? No idea.

Let's use the first. The algorithm is fairly easy to follow:

Turn off all the sources in the schematic.
Set SUM = 0
For each source,
   Turn on this source.
   Compute the value of I(R1) and add this to SUM.
Your answer is the value of SUM.

The meaning of "turn off" is that for voltage sources you short across them and for current sources you open them up so there's no connection across them.

In your case, when the above algorithm is applied you will get the following two circuits:

schematic

simulate this circuit – Schematic created using CircuitLab

The right side of the left schematic can be ignored for its analysis to find \$I_{R_1}\$. The only question there is to ask what share of \$I_1\$ will be taken up by \$R_1\$. And that's just its conductance as a portion of the total conductance. This works out to \$\frac1{1+\frac{R_1}{R_3}}\$. So the current in \$R_1\$ will be that portion times \$J\$ (in this case.) So here we can say:

$$I_{{R_1}_a}=J\cdot \frac1{1+\frac{R_1}{R_3}}$$

The right schematic is very simple. Here:

$$I_{{R_1}_b}=\frac{E}{R_1+R_3}=\frac{E}{R_3}\cdot \frac{1}{1+\frac{R_1}{R_3}}$$

So the sum of these two is the final answer:

$$I_{R_1}=I_{{R_1}_a}+I_{{R_1}_b}=\left(J+\frac{E}{R_3}\right)\cdot \frac{1}{1+\frac{R_1}{R_3}}=\frac{J+\frac{E}{R_3}}{1+\frac{R_1}{R_3}}$$

Which is the same answer I gave you in comments.

This can also be solved using KCL or KVL. But I'm not sure that's what you wanted.

Meanwhile, the resistance seen by \$E\$ is just \$E\$ divided by the currents in \$R_1\$ and \$R_2\$. You can use the same algorithm above to solve for \$I_{R_2}\$. With both those in hand, the rest is easy. Feel free to get some paper and work it out. But the result is this:

$$R=\frac{R_1\cdot R_2+R_1\cdot R_4+R_2\cdot R_3+R_3\cdot R_4}{R_1+R_2+R_3+R_4+\frac{J}{E}\cdot\left(R_2\cdot R_3-R_1\cdot R_4\right)}$$

Note that the denominator has a term which depends upon the applied source. So it's not a fixed value.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.