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Consider a four terminal resistor as shown here

enter image description here

There exists a matrix resistance equation of the form $V_i=R_{ij}I_j$ for some $4\times 4$ matrix $R$ for each such four terminal resistor. I want to connect many such resistors and consider a circuit of these like the one here

enter image description here

In this figure, the left side has $V=0$, the right side has $V_i^L$ as some constant independent of $i$. Similarly, the bottom side is grounded and the top has all $V_j^H$ as some constant independent of $j$. I should thus be able to model this as an effective single four terminal resistor with some effective resistance matrix $R^{eff}_{ij}$.

Essentially, find the effective resistance matrix of the circuit, but the catch is that Kirchoff's laws are not obvious anymore. Can anyone help me figure out what are the equivalents of Kirchoff's voltage and current laws here?

EDIT: Not able to fully follow guilio_hep's argument but maybe a simpler example will help. Given matrices $Z^{1}$ and $Z^{2}$, what is the effective $Z$ when they are connected in series. The figure is shown below. The superscripts $1$, $2$ and $eff$ denote which four terminal resistor we mean. The subscripts from $1$ to $4$ are as before.

enter image description here

I know that my $Z^{eff}$ has only four parameters, so given the voltages on each terminal, I can first find the four currents through $Z^{eff}$ terminals and from there I can get $Z^{eff}$.

We have 12 equations below. Indeed the 12 unknowns are the four currents $I^1_j$ and $I^2_j$ and $I^{eff}_j$. Is this correct now?

$\sum_{j=1}^{4}V^{1}=0$

$\sum_{j=1}^{4}V^{2}=0$

$\sum_{j=1}^{4}I^{1}=0$

$\sum_{j=1}^{4}I^{2}=0$

$I^1_3=-I^2_1$

$I^1_1=I^{eff}_1$

$I^2_3=I^{eff}_3$

$I^1_2+I^2_2 = I^{eff}_2$

$I^1_4+I^2_4 = I^{eff}_4$

$V^{1}_1+V^{1}_4 = V^{2}_1+V^{2}_4 = V^{eff}_1+V^{eff}_4$

$V^{1}_1+V^{1}_2+V^{2}_1+V^{2}_2 = V^{eff}_1+V^{eff}_2$

Obviously, any equations in $V^{1}$ and $V^{2}$ can be rewritten in terms of the corresponding $I^{1}$ and $I^{2}$ using the given $Z^1$ and $Z^2$

By the way, the original reference for this paper and the figures is http://arxiv.org/pdf/cond-mat/0508229.pdf

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    $\begingroup$ Can you not simply solve for the resistance in each branch (4 in total) and the apply Kirchoff in the conventional way? Although you may have a 4x4 matrix, there can only be four independent variables so I believe you can represent the resistor with just four resistors in either a diamond or star configuration. $\endgroup$ – Floris May 9 '15 at 14:12
  • $\begingroup$ You lost me where you said that Kirchoff's laws weren't obvious any more. $\endgroup$ – CuriousOne May 10 '15 at 7:38
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First of all, notice that (for the initial 4 terminal resistor), the port potential $V_i$ is equal to the $v_i$ in you first image, but the port current $I_j$ is equal to $-i_j$ or $i_{(j+1) mod 4}$ and what you call resistance matrix $R$ is generally said impedence matrix $Z$ (because impedance parameters are calculated under open circuit conditions $Z_{nm} = {V_n \over I_m } \bigg|_{I_k = 0 \text{ for } k \ne m}$) and, due to rotational symmetry, can be represented as
\begin{bmatrix} a & b & c & d \\ d & a & b & c \\ c & d & a & b \\ b & c & d & a \end{bmatrix} In this case, KVR imply $\sum_{i=1}^{4}V_i=0$ and KCR imply $\sum_{j=1}^{4}I_j=0$
In principle, Kirchhoff’s voltage law/rule (KVL or KVR) states that the algebraic sum of all voltages around a closed path (or loop) is zero and Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering a node (or a closed boundary) is zero. enter image description here

You can model the $N*M$ network as a 4 terminal if you join together the nodes on each side (left=1, top=2, right=3, bottom=4) getting 4 ports (top-left=1, right-top=2, bottom-right=3, left-bottom=4).
You have 2 ports $L$ (top-left) and $H$ (bottom-right) with potential $V^L$ and $-V^H$ and current $\sum_{i=1}^{N}I_i^L$ and $\sum_{i=1}^{M}I_i^H$ respectively.

Compared to the initial 4 terminal resistor, if both the left and the bottom side are grounded, it is like grounding $i_1$ and $i_4$ and setting $v_4=0$, $v_1=V^H$ and $v_3=-V^L$. Obviously, from KVR, $v_2=V^L-V^H$ and, by mirror symmetry, $R_{i,j}^{eff}=R_{(i+2) mod 4,(j+2) mod 4}^{eff}$.
Now, inside the network, you can apply KVR at each 4 terminal resistor component and at every adiacent resistors in a loop as well as KCR at each node and at every closed boundary of resistors.
It is possible to consider also admittance parameters $Y_{i,j}$ since you are composing a sort of $ \pi $ and T topology (though it is indeed a box topology ...)
enter image description here enter image description here
with admittance and impedance
$ Y = \begin{bmatrix} Y_1 + Y_2 & -Y_2\\ -Y_2 & Y_2 + Y_3\\ \end{bmatrix}$ and $ Z = \begin{bmatrix} Z_1 + Z_2 & Z_2\\ Z_2 & Z_2 + Z_3\\ \end{bmatrix}$
so one should be able to eliminate the internal currents and voltages using systematic elimination of variables (simple example in this video).

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  • $\begingroup$ I think $I_i^j$ is not a current flowing along an arc of that circle, rather it is the current entering our resistor through that terminal. So indeed, $I^1_3=-I^2_1$, right? And, it is also true that $\sum_{j=1}^{4}I_j=0$ for each node (current conservation). I've edited the equations I proposed (does it make sense now?) $\endgroup$ – user1936752 May 10 '15 at 14:43

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