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For the below DC circuit, does the 1Ω resistor in the middle have an effect on the equivalent resistance?

enter image description here

If it does, how would you calculate the equivalent resistance? The rules for series and parallel resistors don't seem to apply, combining the two 2Ω resistors can't be done because their right sides don't join without going through the 1Ω resistor, and trying to combine the 2Ω and 3Ω pairs using the series rule leaves the ends of the 1Ω in the middle of the combine resistor. Does the circuit just need to be redrawn in some way that I missed?

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    \$\begingroup\$ rules for series and parallel resistors don't seem to apply What about delta - wye? \$\endgroup\$
    – greybeard
    Mar 8, 2023 at 20:48
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    \$\begingroup\$ Think about what electrical condition is necessary for current to flow through a resistor. Is that condition present for the 1ohm resistor? \$\endgroup\$
    – InBedded16
    Mar 8, 2023 at 20:49
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    \$\begingroup\$ Google "Wheatstone Bridge".... \$\endgroup\$
    – Kyle B
    Mar 8, 2023 at 22:15
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    \$\begingroup\$ In that specific circuit (where the series resistors are matched), the 1-ohm resistor has no effect. For the general case you need to perform a delta-wye transformation on one side. \$\endgroup\$
    – brhans
    Mar 9, 2023 at 1:55
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    \$\begingroup\$ @InBedded16 There would need to be a voltage across it, naively I figure that condition isn't met because the resistance applied to the current flowing into both sides of the 1Ω resistor is the same. But what if one of the 3Ω and 2Ω were swapped, would that still be the case? If they were swapped, the resistance going into each side of the 1Ω would be different, wouldn't that create difference in either current or voltage across the 1Ω? \$\endgroup\$
    – Gumpf
    Mar 9, 2023 at 13:15

4 Answers 4

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I will refer to this annotated version of your circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

If you look carefully, the top two resistors, R1 and R2 form a potential divider, and the bottom two resistors, R3 and R4 form another, identical potential divider.

The consequence of this is that the potentials at C and D are the same, regardless of the voltage you place across A and B. In other words, the top divider produces a potential at C, and the bottom divider, being identical, and having the same potential difference across it, produces that same potential at D.

Therefore, the potentials at C and D are the same, another way of saying the voltage across R5 is zero. How much current flows in a resistor with zero volts across it? By Ohm's law \$I=\frac{V}{R}=\frac{0}{R}=0A\$.

With no current flowing through it, you can remove R5 and this circuit's behaviour would not change at all. The following circuit would have exactly the same behaviour:

schematic

simulate this circuit

It's also very trivial to reduce it to a single, equivalent resistance.

More generally, for an H-arrangement like this, the above simplifcation will not be possible if the potentials at C and D are not equal, which would be the case if ratios \$\frac{R_1}{R_2} \ne \frac{R_3}{R_4}\$. In such cases, it is impossible to reduce using the standard conversions of series and parallel to single equivalents, but there are other tools available to convert between "delta" (a triangular arrangement) of resistors, and "wye" (Y) shaped formation. They are called "delta-wye" (Δ-Y) and "wye-delta" (Y-Δ) transformations.

This permits you to convert your "H" arrangement, into something like this:

schematic

simulate this circuit

These two circuits are functionally identical, however, as you can see on the right, there are series pairs that may now be reduced to single resistances, and those two singles will be in parallel, suitable for replacement by a single, and so on, until you end up with one single resistance between A and B.

It would have the same value as whatever you measure (with an ohmmeter) between A and B on the left, and be functionally identical in every respect.

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    \$\begingroup\$ I think this is the right next step for the questioner. I hope your time is appreciated by Gumpf. Regardless, +1. \$\endgroup\$ Mar 9, 2023 at 5:42
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Your situation is a made trivial by its symmetry. My comment and the other answers point this out. Were it less trivial, a direct tool (with limited application) called the delta-wye conversion could be applied here. And still more generally, nodal analysis (KCL) and mesh analysis (KVL) are commonly applied to problems still more complicated than this one. These (KCL and KVL) are very general/powerful tools, indeed. You need go no further to handle most of what you may find in electronics.

But there are still more powerful tools that not only handle KCL/KVL problems but far more. Everything from harmonics, probability, Dirichlet's principle, Thomson's principle, Tellegen's theorem, and even into string theory. (Electronics as only a minor bump along the road, so to speak.)

Spice uses a subset of the following method. So it's worth getting exposed even if you never directly use them.

A circuit is a directed graph. In your case, something like this:

enter image description here

In the above, I've removed the battery and simply show the battery current entering and leaving. (Rank-Nullity Theorem explains why. For another day.)

If the nodes are columns and the edges are rows then the connection matrix \$A\$ and conductance matrix \$C\$ look like this:

$$\begin{align*} \begin{smallmatrix} \fbox{A}&N_0&N_1&N_2&N_3\\ \hline a:\vphantom{\frac1{R_b}}&\hfill -1&\hfill 0&\hfill 1&\hfill 0\\ b:\vphantom{\frac1{R_b}}&\hfill -1&\hfill 0&\hfill 0&\hfill 1\\ c:\vphantom{\frac1{R_b}}&\hfill 0&\hfill 0&-1&\hfill 1\\ d:\vphantom{\frac1{R_b}}&\hfill 0&\hfill 1&\hfill -1&\hfill 0\\ e:\vphantom{\frac1{R_b}}&\hfill 0&\hfill 1&\hfill 0&\hfill -1 \end{smallmatrix} && \begin{smallmatrix} \fbox{C}&a&b&c&d&e\\ \hline a:& \frac1{R_a}& 0& 0& 0& 0\\ b:& 0&\frac1{R_b}&\hfill 0&0&0\\ c:&0& 0&\frac1{R_c}& 0&0\\ d:&0&0&0&\frac1{R_d}& 0\\ e:& 0& 0& 0&\hfill 0&\frac1{R_e} \end{smallmatrix} \end{align*}$$

For the connection matrix \$A\$ on the left, a -1 says that the arrow is pointing away from that node and a 1 says that the arrow is pointing into that node. An edge must point into and out of two nodes, so the sum on a row must be 0. (You don't have to use the same directions I did. You can choose them, arbitrarily.) For the conductance matrix \$C\$, all that is done is to list the edge conductances down the diagonal. The rest is all zero.

We can arbitrarily set \$V_0=1\:\text{V}\$ and \$V_1=0\:\text{V}\$ and try to find out what \$I\$ becomes. Then it's easy to work out the resistance.

From KCL (in the rowspace and right nullspace) and Ohm's Law (in the columspace and left nullspace) we know that \$A^T\,C\,A\,\left[\begin{smallmatrix}1 \\ 0\\v_2\\v_3\end{smallmatrix}\vphantom{\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}}\right]=\left[\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}\right]\$. We can break up the matrix product, \$A^T\,C\,A\$, into four quadrants that map out as \$\left[\begin{smallmatrix}P&Q^T\\Q&R\end{smallmatrix}\right]\$. From that, we have \$\left[\vphantom{\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}}\begin{smallmatrix}P&& Q^T\\\\\\Q&& R\end{smallmatrix}\right]\,\left[\begin{smallmatrix}1 \\ 0\\v_2\\v_3\end{smallmatrix}\vphantom{\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}}\right]=\left[\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}\right]\$ and then we can apply the Schur complement concept to find \$\left[\vphantom{\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}}P-Q^T\,R^{-1}\,Q\right]\left[\vphantom{\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}}\begin{smallmatrix}\hfill 1\\\\\\\\\hfill 0\end{smallmatrix}\right]=\left[\vphantom{\begin{smallmatrix}\hfill I \\ -I\\\hfill 0\\\hfill 0\end{smallmatrix}}\begin{smallmatrix}\hfill I\\\\\\\\-I\end{smallmatrix}\right]\$.

(By using the Schur complement, the matrix work was reduced as we don't care about the voltages, \$V_2\$ and \$V_3\$, and so don't need to solve the entire system of KCL equations to find what we needed. We can focus on the bits of interest. This ability to narrow the focus is difficult to see when just applying KCL and/or KVL.)

That will solve for \$I\$ (and \$-I\$, which we don't need.) Since the voltage applied was \$1\:\text{V}\$, the resistance will be \$R=\frac{1\:\text{V}}{I}\$.

The matrix equation helps yield the following general solution for \$R\$:

$$R=\frac{R_{a} R_{b} R_{c} + R_{a} R_{b} R_{d} + R_{a} R_{b} R_{e} + R_{a} R_{c} R_{e} + R_{a} R_{d} R_{e} + R_{b} R_{c} R_{d} + R_{b} R_{d} R_{e} + R_{c} R_{d} R_{e}}{R_{a} R_{c} + R_{a} R_{d} + R_{a} R_{e} + R_{b} R_{c} + R_{b} R_{d} + R_{b} R_{e} + R_{c} R_{d} + R_{c} R_{e}}$$

But if you simplify things by saying that \$R_{a}=R_{b}\$ and \$R_{d}=R_{e}\$ then the above reduces to \$\frac12\left(R_a+R_d\right)\$ and you can see that \$R_c\$ is no longer any part of the solution answer. That means it doesn't matter what value it takes. It has no impact.

As already mentioned, the above is part of how Spice programs work.

A symbolic program like Sympy can also be used:

ra,rb,rc,rd,re = symbols('ra,rb,rc,rd,re',real=True)
A = Matrix([[-1,0,1,0],[-1,0,0,1],[0,0,-1,1],[0,1,-1,0],[0,1,0,-1]])
C = Matrix([[1/ra,0,0,0,0],[0,1/rb,0,0,0],[0,0,1/rc,0,0],[0,0,0,1/rd,0],[0,0,0,0,1/re]])
M = A.transpose()*C*A
P = M[[0,1],[0,1]]
Q = M[[2,3],[0,1]]
R = M[[2,3],[2,3]]
simplify(1/(((P-Q.transpose()*R.inv()*Q)*Matrix([[1],[0]]))[0]))

(ra*rb*rc + ra*rb*rd + ra*rb*re + ra*rc*re + ra*rd*re + rb*rc*rd + rb*rd*re + rc*rd*re)/(ra*rc + ra*rd + ra*re + rb*rc + rb*rd + rb*re + rc*rd + rc*re)

simplify(1/(((P-Q.transpose()*R.inv()*Q)*Matrix([[1],[0]]))[0]).subs({rb:ra,re:rd}))

ra/2 + rd/2
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    \$\begingroup\$ Very nice answer! \$\endgroup\$ Mar 9, 2023 at 8:00
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If the voltage source is 5V, then the voltage divider 2R+3R generates 3V at each side of the 1 Ohm Resistor.

There is no current flow possible, because the voltage difference between this two points is zero. If you replace the 1 Ohm resistor with 10MOhm or 0 Ohm would not make any difference.

=> Resistance is futile. :-)

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Basic idea

This symmetrical circuit configuration can be thought of as a balanced bridge. Its main property is that current does not flow through and there are not voltage across the bridge element in the middle. It arises from the fact that one of its arms (voltage divider with transfer ratio K) is a copy of the other arm (voltage divider with the same transfer ratio K) and, as a result, the voltage difference across the bridge element is zero. So there is no voltage difference, no current flowing.

Short-circuited bridge

This arrangement was very popular in the 19th century.

Measuring instrument

Then they had sensitive ammeters (galvanometers) but no voltmeters. Let's use CircuitLab to examine such an old bridge by varying the voltage V and observing the ammmeter readings (note that we are interested in the equivalent resistance of the circuit, but the simulator requires that it be powered and grounded). Our expectations are met - it always shows zero current. A more subtle way is to use, besides the convenient live DC simulation, the DC sweep simulation varying the voltage V, for example, from 0 to 10 V.

schematic

simulate this circuit – Schematic created using CircuitLab

Circuit transformation

Let's now transform this circuit configuration in three steps to obtain the equvalent resistance.

Step 1: R1, R2, R3, R4

schematic

simulate this circuit

Step 2: R1||R3, R2||R4

schematic

simulate this circuit

Step 3: R1||R3 + R2||R4

schematic

simulate this circuit

Open-circuited bridge

This arrangement is very popular nowadays.

Measuring instrument

We have perfect voltmeters - op-amps, ADS, etc. Let's examine such a modern bridge by varying the voltage V and observing the voltmeter readings. Again, our expectations are met - it always shows zero voltage. As above, besides the convenient live DC simulation, we can use the DC sweep simulation varying the voltage V, for example, from -10 to 10 V (see the graph below). Let's do it!

schematic

simulate this circuit

Bridge graph bipolar

Circuit transformation

As above, we can transform and this circuit configuration in three steps to obtain the equvalent resistance.

Step 4: R1, R2, R3, R4

schematic

simulate this circuit

Step 5: R1+ R2, R3 + R4

schematic

simulate this circuit

Step 6: (R1+ R2)||(R3 + R4)

schematic

simulate this circuit

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