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Original current to be simplified with Thevenin calculations

I want to find the Thevenin equivalent circuit but because of the positioning of the nodes a and b I can not visualize it correctly and that causes me to struggle. How wil I calculate the Thevenin Resistor (Rt) and Thevenin Voltage (Vt)?

These are my calculations. But when I simulate (LTspice) my voltages and resistance from the original is not equivalent to the thevenin circuit? enter image description here

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  • \$\begingroup\$ Just start from the left and convert to a current source then merge two resistors to one then keep moving to the right merging resistances as you proceed. \$\endgroup\$ – Andy aka Mar 3 '18 at 16:30
  • \$\begingroup\$ There you go now that wasn't too hard was it! Same answer as I got BTW. \$\endgroup\$ – Andy aka Mar 3 '18 at 17:03
  • \$\begingroup\$ Looks right to me, as well. Though I get a short circuit current that is slightly closer to \$1\:\text{mA}\$. \$V_\text{TH}\approx 1.279\:\text{V}\$ and \$R_\text{TH}\approx 1.271\:\text{k}\Omega\$. \$\endgroup\$ – jonk Mar 3 '18 at 19:33
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Do it the same way you always would:

  • Calculate the open circuit voltage. With no load connected between A and B, what will be the voltage \$V_{AB}\$? That is the Thevenin voltage \$V_{th}\$.

  • Calculate the short circuit current. If A and B were shorted, what current would flow through that short? If the short circuit current is \$I_s\$ then the Thevenin resistance is \$R_{th}=V_{th}/I_s\$.

If it helps you to visualize the problem, then you can re-draw the circuit however you like, provided you maintain the same connections between the elements.

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  • \$\begingroup\$ So included my calculations i the question now. Can you go look if my visualization was correct. Especially when I calculate the parallel resistors and after that transitioning the 2.5mA source to a voltage source \$\endgroup\$ – Michelle Mar 3 '18 at 16:50
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Thanks for adding all that work. I think you followed the ideas well and got an answer within reasonable error. Good work!

I know you may not yet have been introduced to nodal analysis. But this is another way to imagine and I thought I may as well show you how your answer might be approached in a different way -- one I expect you will soon see, if you haven't already.

The way I use nodal analysis is to put all the "outgoing currents" on the left side and all of the incoming currents on the right side of each equation. You have two unknown node voltages, assuming your bottom node can be considered \$0\:\text{V}\$, so you'd need to set up the following two:

$$\begin{align*} \frac{V_a}{2200\:\Omega}+\frac{V_a}{2700\:\Omega}+\frac{V_a}{2200\:\Omega}&=\frac{5.5\:\text{V}}{2200\:\Omega}+\frac{0\:\text{V}}{2700\:\Omega}+\frac{V_b}{2200\:\Omega}\\\\ \frac{V_b}{2200\:\Omega}+\frac{V_b}{1800\:\Omega}&=\frac{V_a}{2200\:\Omega}+\frac{0\:\text{V}}{1800\:\Omega} \end{align*}$$

This solves out as \$V_a\approx 2.3258\:\text{V}\$ and \$V_b\approx 1.0466\:\text{V}\$, with the difference of \$V_{ab}\approx 1.2792\:\text{V}\$. This is also the Thevenin voltage.

This is a way of not having to go through individual steps, one at a time, transforming the circuit. You can get the answer, somewhat more directly. But it does involve the simultaneous solution of two linear equations. (Not hard, at all, though.)

The equivalent resistance can be achieved by simply shorting out the voltage supply and then looking at the resistor network from the perspective of nodes \$a\$ and \$b\$. Here, the answer is easily seen as:

$$\left[\left( 2200\:\Omega\mid\mid 2700\:\Omega\right)+1800\:\Omega\right]\mid\mid 2200\:\Omega\approx 1271.42\:\Omega$$

So there are other ways to get there.

Just a heads-up.

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