I am learning the method of Fast Analytical Circuit Techniques (FACTs), based on the Extra Element Theorem of the late Prof. Middlebrook. In the future, I intend to teach an undergraduate-level course based on these techniques.

I’ve been looking for an example that could really showcase the power of FACTs, and I believe I found the perfect one: the following unbalanced bridge circuit, for which the equivalent resistance must be calculated (i.e. \$V_{in}/I_{in}\$, where \$I_{in}\$ applied to the bridge terminals marked \$V_{in}\$ and ground in the figure.)


simulate this circuit – Schematic created using CircuitLab

Solution with the EET

The solution is easy enough when you apply the Extra Element Theorem. I'll evidently consider \$R_5\$ as the extra element. Recall that:

$$ Z_{eq} = \left. Z_{eq} \right\rvert_{R_5 \to \infty} \frac{1 + \frac{Z_n}{R_5}}{1 + \frac{Z_d}{R_5}}. $$

\$\left. Z_{eq} \right\rvert_{R_5 \to \infty}\$ is obtained by taking out \$R_5\$ from the circuit and calculating the equivalent resistance seen from the input terminals (\$V_{in}\$ and ground). When this is done, it is easily seen by inspection that:

$$ \left. Z_{eq} \right\rvert_{R_5 \to \infty} = (R_1 + R_3) || (R_2 + R_4). $$

\$Z_d\$ is obtained by zeroing out the excitation source (\$I_T\$ in this case, which becomes an open circuit) and calculating the equivalent resistance seen from \$R_5\$'s terminals. Again, it's easily seen by inspection that:

$$ Z_d = (R_1 + R_2) || (R_3 + R_4). $$

Finally, \$Z_n\$ is obtained by null double injection: leaving \$I_{in}\$ in place, a test current source \$I_T\$ is applied into \$R_5\$'s terminals such that \$V_{in}\$ is nulled, and the equivalent resistance seen by \$I_T\$ is calculated. Since \$V_{in} = 0\$, \$R_1\$ and \$R_2\$'s terminals connected to \$V_{in}\$ are shorted to ground instead, revealing that:

$$ Z_n = (R_1 || R_3) + (R_2 || R_4). $$

Combining the results, we find that the equivalent resistance is given by:

$$ Z_{eq} = ((R_1 + R_3) || (R_2 + R_4)) \frac{1 + \frac{(R_1 || R_3) + (R_2 || R_4)}{R_5}}{1 + \frac{(R_1 + R_2) || (R_3 + R_4)}{R_5}}. $$

The question: alternate solutions?

I'm looking for the best "elementary" circuit analysis technique that undergraduate students might reasonably have heard about, that could be used to analyze this circuit, short of the EET itself. I just want to make sure I won't be surprised by a different way of solving the circuit that I hadn't heard before, ruining the intended dramatic effect of showing how easy it is to analyze circuits with the EET (never mind the meaningful low-entropy expressions obtained with it), compared with everything else they're familiar with

Alternate solutions: mesh analysis

For now, I tried mesh analysis, with the three meshes that can be immediately identified from the figure. Note that the left-hand side mesh coincides with \$I_{in}\$, so only two currents remain to be solved. Writing out these two equations (naming the upper mesh \$I_2\$ and the lower mesh \$I_3\$, with current flowing clockwise inside each mesh), I get the following equations:

$$ I_2 (R_1 + R_2 + R_5) - I_3 R_5 = R_1 I_{in} $$


$$ -I_2 R_5 + I_3 (R_3 + R_4 + R_5) = R_3 I_{in} $$

From these, I get the following expression for \$I_2\$:

$$ I_2 = \frac{R_1 + \frac{R_3 R_5}{R_3 + R_4 + R_5}}{R_1 + R_2 + R_5 - \frac{R_5^2}{R_3 + R_4 + R_5}} I_{in}, $$

and for \$I_3\$:

$$ I_3 = \frac{R_3 + \frac{R_1 R_5}{R_1 + R_2 + R_5}}{R_3 + R_4 + R_5 + \frac{R_5^2}{R_1 + R_2 + R_5}} I_{in} $$

Finally, writing out the equation for the remaining mesh:

$$ V_{in} = R_1 (I_{in} - I_2) + R_3 (I_{in} - I_3) $$

Plugging the expressions for \$I_2\$ and \$I_3\$ and performing some algebra on these unwieldy expressions, the following result is obtained:

$$ R_{eq} = R_1 + R_3 - \frac{R_1^2 (R_3 + R_4 + R_5) + 2 R_1 R_3 R_5 + R_3^2 (R_1 + R_2 + R_5)}{(R_3 + R_4 + R_5) (R_1 + R_2 + R_5) - R_5^2}. $$

I should mention it took me nearly about two hours to write and debug this expression, whereas the EET solution took about two minutes and was correct on the first try.

Alternate solutions: nodal analysis

The other obvious alternative is nodal analysis, which I didn't try out at first since an extra equation will be required. I started working it out by nodal analysis, but stopped as soon as I realized how complicated the expressions were getting -- much more so than by mesh analysis. I estimate it would take at least twice as long to work out the algebra as it took using mesh analysis.

Any other alternate solutions?

To restate the question: are there any other ways to analyze this circuit other than the EET, mesh and nodal analysis, that might rival the simplicity of the EET?

  • \$\begingroup\$ Parenthesis in your equations would help a lot. Unless I want to take the time to go through your equations, I'm having order of operations issues. \$\endgroup\$ Commented Mar 4, 2020 at 18:04
  • 1
    \$\begingroup\$ The material I'm reading assumes || takes precedence over +, so parentheses shouldn't be necessary. At the risk of making the equations a tiny bit less readable, I'll edit the questions to include parentheses. \$\endgroup\$
    – swineone
    Commented Mar 4, 2020 at 18:08
  • \$\begingroup\$ One very important fact owing to the EET and the FACTs generally speaking is the low-entropy format (read well organized and factored form) you obtain in one shot. Besides, determining transfer function by inspection and having the ability to come back to an intermediate sketch for a correction is priceless. Good luck at teaching these facts, they are truly a skill EEs must acquire. \$\endgroup\$ Commented Mar 4, 2020 at 20:48
  • \$\begingroup\$ Merci Cristophe ! (By the way, did you get my latest email about problem 2.10?) \$\endgroup\$
    – swineone
    Commented Mar 4, 2020 at 20:51

2 Answers 2


I don't know if I would consider the nodal analysis method to be that much more complicated than mesh analysis, but you do have to apply two key strategies:

  1. Excite your circuit using a current source instead of a voltage source. This will eliminate the extra equation you need.
  2. Solve in terms of conductance instead of resistance (\$G = \frac{1}{R}\$).

I'm going to denote node a the left side of the bridge and node b the right side of the bridge. Assume current flows through the resistors top-left to bottom-right (I really should include a circuit diagram with these assumptions and labels).

Kirchhoff's current laws:

$$ I_{in} - I_{1} - I_{2} = 0\\ I_{1} - I_{3} - I_{5} = 0\\ I_{2} - I_{5} - I_4 = 0 $$

Ohm's laws:

$$ I_1 = G_1 (V_{in} - V_a)\\ I_2 = G_2 V_{in} - V_b\\ I_3 = G_3 V_a\\ I_4 = G_4 V_b\\ I_5 = G_5 (V_a - V_b) $$

Substituting in,

$$ I_{in} - G_1 (V_{in} - V_a) - G_2(V_{in} - V_b) = 0\\ G_1 (V_{in} - V_a) - G_3 V_a - G_5 (V_a - V_b) = 0\\ G_2 (V_{in} - V_b) + G_5 (V_a - V_b) - G_4 V_b = 0 $$

Taking the last equation, solve for \$V_b\$: $$ V_b = \frac{G_2 V_{in} + G_5 V_a}{G_2 + G_4 + G_5} $$ Take the second equation, plug in \$V_b\$ and solve for \$V_a\$: $$ V_a = \frac{G_1 + \frac{G_2 G_5}{G_2 + G_4 +G_5}}{G_1 + G_3 + G_5 - \frac{G_5^2}{G_2 + G_4 + G_5}} V_{in} $$ Re-write \$V_b\$ as $$ V_b = \frac{G_2 + G_5 \frac{G_1 + \frac{G_2 G_5}{G_2 + G_4 +G_5}}{G_1 + G_3 + G_5 - \frac{G_5^2}{G_2 + G_4 + G_5}}}{G_2 + G_4 + G_5} V_{in} $$ Plug everything into the first equation, solve for \$\frac{V_{in}}{I_{in}}\$: $$ I_{in} = G_1 V_{in} \left(1 - \frac{G_1 + \frac{G_2 G_5}{G_2 + G_4 +G_5}}{G_1 + G_3 + G_5 - \frac{G_5^2}{G_2 + G_4 + G_5}}\right) + G_2 V_{in} \left(1 - \frac{G_2 + G_5 \frac{G_1 + \frac{G_2 G_5}{G_2 + G_4 +G_5}}{G_1 + G_3 + G_5 - \frac{G_5^2}{G_2 + G_4 + G_5}}}{G_2 + G_4 + G_5}\right)\\ \frac{V_{in}}{I_{in}} = \frac{1}{G_1 \left(1 - \frac{G_1 + \frac{G_2 G_5}{G_2 + G_4 +G_5}}{G_1 + G_3 + G_5 - \frac{G_5^2}{G_2 + G_4 + G_5}}\right) + G_2 \left(1 - \frac{G_2 + G_5 \frac{G_1 + \frac{G_2 G_5}{G_2 + G_4 +G_5}}{G_1 + G_3 + G_5 - \frac{G_5^2}{G_2 + G_4 + G_5}}}{G_2 + G_4 + G_5}\right)} $$ The last step is have fun simplifying this expression, which is certainly unpleasant (or just leave it as is).

As far as difficulty goes:

  • Probably comparable with mesh analysis, though personally I think it's easier just because I'm much more familiar with nodal analysis. Nodal analysis also doesn't suffer from the disadvantage of mesh analysis requiring circuits to be planar (though in this case, this disadvantage doesn't matter).
  • probably harder than FACTs/EET? I'm not really familiar with this method myself, though if you were able to solve it in a few minutes, that's likely faster than the ~10-15 minutes it took me to solve it using nodal analysis, not including simplifying the last expression. I will note though that the nodal analysis method is very mechanical, i.e. I didn't devote a ton of clever reasoning or brain power to finding this solution.

Using the Y-Delta transform

An alternative method you didn't state involves using the Y-\$\Delta\$ transform.

Redraw the circuit as (replaced R1, R3, R5 with its \$\Delta\$ equivalent):


simulate this circuit – Schematic created using CircuitLab

The \$\Delta\$ resistances are: $$ R_a = \frac{R_3 R_5 + R_5 R_1 + R_1 R_3}{R_3}\\ R_b = \frac{R_3 R_5 + R_5 R_1 + R_1 R_3}{R_5}\\ R_c = \frac{R_3 R_5 + R_5 R_1 + R_1 R_3}{R_1} $$ Then you just need parallel and serial combinations to get an equivalent resistance. $$ Z_{eq} = R_b || ((R_a || R_2) + (R_c || R_4)) $$

  • \$\begingroup\$ I think FACTs are easier to deal with for several reasons: 1. you don't have to solve a system of several unknowns, just inspect or solve simple intermediate circuits 2. if you make a mistake then you can go back to the failing sketch and fix it. If one of your equations is wrong and you realize it at the end of the calculation, you have to restart from scratch and 3. you naturally obtain with FACTs nice parallel-series arrangements which let you easily guess how the result evolves if one element becomes large or small. \$\endgroup\$ Commented Mar 4, 2020 at 21:13
  • \$\begingroup\$ I don't doubt that FACTs would make solving this problem easier for the reasons you stated (especially since I'm pretty sure I've made a mistake somewhere in my nodal analysis that I have to now go and try to fix). I was giving the OP some optional solution techniques which a student might recognize the procedures for performing (or miss-performing). \$\endgroup\$ Commented Mar 4, 2020 at 23:52
  • \$\begingroup\$ I think your good example is part of the learning path for understanding circuits analysis. You first need to "suffer" long expressions and sometimes algebraic paralysis to realize that another way is necessary. The FACTs represent a useful solution in many cases, whether we talk about active or passive circuits. What I've shown in my book is the association with a simulator like SPICE which gives the ability to verify and fix each intermediate sketch and progress towards success even with extremely complicated circuits. This is the "divide and conquer" strategy promoted by Dr. Middlebrook. \$\endgroup\$ Commented Mar 5, 2020 at 7:40

This is just an idea and may be badly incorrect....

Consider the source V(in) as split into two identical sources and then use impedance superposition (I’ve made that name up) to short one of those two new sources to ground. Then ditto the other invented source. For each instant, calculate the impedance seen by the one remaining source.

Then combine those impedances in parallel? Maybe: -

enter image description here

And, just in case anyone wants to compare results (because the formulas are so different), may I suggest that R1 is 1 ohm and R5 is 5 ohm with the others in between.

I get a combined impedance of 1.789 ohms for what it's worth.


It's clearly a stupid idea because it's obvious when you do a simulation: -

enter image description here

I've left this answer up because at least it's a way to compare formulas numerically.

The right answer is 1/0.417647 = 2.39437 ohms


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