Electrical – How to find Rth (Thevenin Equivalent R) of this circuit

circuit analysisthevenin

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Suppose I find \$V_{th}\$ = 2 V. How to find \$R_{th}\$ seen from a,b?

My effort is short ab, and then in this case, 2//4 = 4/3. Note here that, the line across \$R_L\$ is a node (short ab). So the total current is

16/(5+4/3*2) = 2.087 A.
So I get Isc = 2.087 A. Note that \$I_{sc}\$ is the current flowing into the node when I short ab.

So \$R_{th}\$ = 2/2.087 \$\Omega\$.

Am I correct?

Best Answer

The Thévenin resistance can be determined using the extra-element theorem or EET forged by Dr. Middlebrook. It is a really easy and cool way to determine transfer functions in circuits where a particular element is bothering you. To determine a resistance (or an impedance), install a test generator \$I_T\$ across the output terminals. This is the excitation or stimulus. This current source will generate a response \$V_T\$ across its terminals. What we want is \$R_{th}=\frac{V_T}{I_T}\$

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According to the EET principle, the element causing troubles here would be \$R_5\$ but you could choose any other resistance of course. We have the choice to set to infinity (open-circuit it) or replace it by a short circuit. So the selection depends on how you visualize the circuit when the extra element is replaced by a short circuit or open-circuited. We go for the second choice and determine the resistance \$R_{ref}\$ offered from the output terminals (across which \$R_{th}\$ must be determined) after setting the 16-V bias source to 0 V (a short circuit). The value is simple to obtain by inspecting the circuit: no equation here.

Then, suppress the excitation \$I_T\$ and determine the resistance \$R_d\$ “seen” from \$R_5\$’s connecting terminals. Again, no stress, it is a simple series-parallel combination obtained by inspecting the circuit.

Finally, null the output response \$V_T\$ to determine \$R_n\$. This is a degenerate case and you replace the current source by a short circuit to “see” the resistance \$R_n\$ from \$R_5\$’s connecting terminals. Same philosophy as in the above, inspect the circuit.

This is it, the final result is obtained by assembling the pieces as follows: \$R_{th}=R_{ref}\frac{1+\frac{R_5}{R_n}}{1+\frac{R_5}{R_d}}\$. The below Mathcad file shows the exact formula and a resistance computed to \$2.875 \Omega\$.

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You can see how easy it is to apply the EET which is part of the fast analytical techniques (FACTs) toolkit. You divide the circuit into small pieces that you individually solve and assemble at the end. If later on you find a mistake, no need to restart from scratch, just fix the guilty drawing and voilà. I added in the Mathcad sheet the EET formula for which \$R_5\$ is now considered infinite rather than a short circuit as initially selected. The \$R_{th}\$ expression slightly differs but results are identical.

I have added a quick SPICE simulation using a .TF primitive which instructs the simulator to run a transfer function at the output node while the modulation comes from the 16-V input source. It confirms the small-signal output resistance \$R_{th}\$ obtained using the FACTs.

Not only you did not write a line of algebra (with the risk of making mistakes) but you obtained a nice-looking low-entropy expression in which you have series and parallel arrangements. It is invaluable to see how the final result changes if one element changes to a small or large value. If you use KVL or KCL, it is very likely that you obtain an arm-long expression with all mixed-up labels providing no insight of what does what in the formula. To that respect, FACTs are difficult to beat.