If I have a configuration of 5 resistors as follows:


simulate this circuit – Schematic created using CircuitLab

How can I calculate the expected total resistance between A and B, given only values for R1-R5?

edit: okay, here's what I've tried so far.

I intended to find the expected voltage drops across each resistor, and from there I knew I could easily find the total resistance of the circuit.

First, I observed three loops in the diagram: R1,R3,R4; R3,R5,R2; and R1,R2,R5,R4. I remember from my old electronics class that the voltage drop around each loop in a circuit should always equal zero, as follows: $$ v1+v2-v4-v5=0\\ v1+v3-v4=0\\ v3+v5-v2=0 $$ (Assuming that current is flowing from A to B, I further assume that it is flowing down through r3. If this is wrong, then the value for v3 would just have the opposite sign, and all would be well).

The sum of currents flowing into a point should equal the current flowing out, so I have these equations as well: $$ \frac{v1}{r1}=\frac{v2}{r2}+\frac{v3}{r3}\\ \frac{v5}{r5}=\frac{v3}{r3}+\frac{v4}{r4}\\ \frac{v1}{r1}+\frac{v4}{r4}=\frac{v2}{r2}+\frac{v5}{r5} $$

So now I have 6 equations in 5 unknowns... and if these equations are correct, then there should be one redundant equation. However, plugging them into a cas such as maple cannot find any unique values for v1-v5 in this system (an infinite number, in fact), suggesting that there are at least 2 redundant equations above.

Does this mean that there is no unique resistance for this circuit fragment, or what have I done wrong?

  • 1
    \$\begingroup\$ electronics.stackexchange.com/questions/137914/… \$\endgroup\$
    – diverger
    Nov 22, 2014 at 10:33
  • \$\begingroup\$ You are probably closer than you think. Because of symmetry, there should be at least two solutions, that are mirrors of each other. R3 however, should be unique. \$\endgroup\$ Nov 22, 2014 at 18:22
  • \$\begingroup\$ What I find is an infinite number of solutions, dependent upon the voltage drop across one of the resistors, which suggests that the net resistance is dependent on how much voltage is applied to it. Considering this is a circuit fragment composed of just resistors, that seems highly unlikely to me. \$\endgroup\$
    – Mark
    Nov 22, 2014 at 18:37
  • \$\begingroup\$ I recommend this method to calculate the resistance: physics.stackexchange.com/a/19296 it works for arbitrary networks of resistors \$\endgroup\$
    – Paul
    Oct 18, 2016 at 14:11

3 Answers 3


In this case I would use Y-delta transformation. As shown in the schematic below:


simulate this circuit – Schematic created using CircuitLab

This should give you one unique solution for an equivalent resistance. It's probably an enormous equation, so luckily you have a CAS.

  • \$\begingroup\$ Thank you. I didn't know about that technique before, so I read about it on Wikipedia and added a link \$\endgroup\$
    – Mark
    Nov 22, 2014 at 22:50

OK, Mark claimed that this is not homework. As a weekend bonus and for common interest here a complete rewrite with full calculation example.

The example circuit

Example circuit

Calculation of the resistance between A & B

  • First let's assume a voltage over A & B, say 10V.
  • Now we convert the two voltage dividers R1 & R2 and R3 & R4 to voltage sources with a series resistor. The voltage is simply the output voltage of the divider without load. The new resistance is the parallel circuit of the old ones.

U_R12 = 10V / ( 3kΩ + 7kΩ ) * 7kΩ = 7V

U_R34 = 10V / ( 3kΩ + 7kΩ ) * 3kΩ = 3V

R_R12 = 1 / ( 1 / 3kΩ + 1 / 7kΩ ) = 2.1kΩ

R_R34 = 1 / ( 1 / 3kΩ + 1 / 7kΩ ) = 2.1kΩ

The new equivalent circuit:

Equivalent circuit

  • We calculate the current through the resistors,
  • the voltage over the substitute resistors
  • and the voltages U1 and U2 (with respect to ground)

I_R5 = ( 7V - 3V ) / ( 2.1kΩ + 1kΩ + 2.1kΩ ) = 0.77mA

U_RS = 2.1kΩ * 0.77mA = 1.62V

U1 = 7V - 1.62V = 5.38V

U2 = 3V + 1.62V = 4.62V

The voltages U1 and U2 are the same for both circuit variants and we can switch back to the example variant.

  • We calculate the currents through R1 and R3,
  • the total current through A
  • and the total resistance

I_R1 = ( 10V - 5.38V ) / 3kΩ = 1.54mA

I R3 = ( 10V - 4.62V ) / 7kΩ = 0.77mA

I_A = 1.54mA + 0.77mA = 2.31mA

R_Tot = 10V / 2.31mA = 4.33kΩ

  • \$\begingroup\$ It's not homework, and I don't have a teacher... I haven't had a teacher on this subject since 1980. I'm just somebody who likes to find solutions to problems like this. \$\endgroup\$
    – Mark
    Nov 22, 2014 at 17:57
  • \$\begingroup\$ @Mark - Ok i believe you, have a look at the new variant. \$\endgroup\$
    – Kitana
    Nov 22, 2014 at 21:12

If you are interested in a formula:


simulate this circuit – Schematic created using CircuitLab

enter image description here

In terms of conductance:

enter image description here

  • \$\begingroup\$ Can you please tell me the name of this formula? So that I could try to solve few problems based on this. And I'll google it to see how this resistors are arranged in this formula (If you mention the title or name) \$\endgroup\$
    – user105009
    Dec 10, 2016 at 17:10
  • \$\begingroup\$ I don't know if it has a name. I deduced it myself. \$\endgroup\$ Dec 10, 2016 at 17:14
  • \$\begingroup\$ Can you describe how you proceeded? Resistors arrangement? \$\endgroup\$
    – user105009
    Dec 10, 2016 at 17:20
  • \$\begingroup\$ The resistor arrangment is the one shown in my picture (resistor labels are different from the labels used in the question). The equivalent resistance is measured between points A and B. I used only ohm's law and kirchoffs laws. The demonstration is quite lengthy, took several pages of my notebook \$\endgroup\$ Dec 10, 2016 at 23:17
  • \$\begingroup\$ I see!!! I always used Y-delta transformations. As your answer is quiet unlike I was curious to learn it. \$\endgroup\$
    – user105009
    Dec 11, 2016 at 7:20

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