3D_Multi Resistor Electric Circuit

Haiduke Sarafian

The Pennsylvania State University, University College, York, USA.

DOI: 10.4236/ajcm.2023.132017                

Abstract

This report addresses the issues concerning the analysis of an electric circuit composed of multiple resistors configured in a 3-Dimension structure. Noting, all the standard textbooks of physics and engineering irrespective of the used components are circuits assembled in two dimensions. Here, by deviating from the “norm” we consider a case where the resistors are arranged in a 3D structure; e.g., a cube. Although, independent of the dimension of the design the same physics principles apply, transitioning from a 2D to a 3D makes the corresponding analysis considerably challenging. In general, with no exception, depending on the used components the analysis faces with solving a set of either algebraic or differential-algebraic equations. Practically, this interfaces with a Computer Algebra System (CAS). The main objective is symbolically to identify the current distributions and the equivalent resistor (s) of cubically assembled resistors.

Keywords

3D Electric Circuit, Equivalent Resistor, Computer Algebra System, Mathematica

1. Introduction

As noted, the main objective of this report is to learn about the various aspects of electric systems in a 3D structure. For the sake of simplicity, a cube is considered with sides embodying twelve electric resistors. Shortly, the lesson learned is used in circuits embodying elements other than resistors. For the time being, the case on hand is limited to twelve different-valued ohmic resistors. One of the objectives is to seek an equivalent resistor (s), R_eq. E.g., to identify one resistor to replace the “cubic resistor” containing all twelve resistors in a circuit, R_eq, without altering the feeder and departing currents. See Figure 1 for a better understanding of the issue. For simple 2D circuits, this is a classic question with a well-known answer [1] [2] . A literature search shows no answer is available for a 3D case. This report answers. The answer is not unique because the R_eq has a value depending on the contact points of the feeder and the departing current. Since the cube has eight corners, even by selecting the feeder’s vertex the current can depart at any of the other seven vertices; providing seven pairs associated with seven different R_eq, see Figure 1 again. This report with the help of a CAS, specifically Mathematica [3] generates all the seven symbolic results. Meaning without assigning numeric values to the twelve resistors we provide a symbolic expression for all R_eq. But then of course the symbolic results are converted to numeric. While deriving the needed information we also were able to derive symbolic expressions for the currents in each of the twelve resistors. Hence, we generate information about the distribution of these currents. Graphs of these distributions for chosen numeric values of the currents are shown.

With these objectives, we craft a report answering these questions. The report is composed of three sections. In Section 2 the specifics of a 3-Dimensional case are presented. It contains the basic needed physics principles with associated Mathematica codes. The solution of the equations on hand symbolically and numerically is given.

The symbolic word intentionally is italicized emphasizing the power of Mathematica codes to solving the equations and at the end numeric are giving as well; noting Mathematica is not an Excel [4] .

And finally, the 3rd Sect. is the conclusion and we outlined the lesson learned and the future-plan.

Figure 1. Twelve different-valued resistors formed the cube are labeled R_n with n = 1, 2, …, 12. The cube is fed with an incoming current i at vertex A leaving the structure at vertex G.

2. A 3-Dimensional Cubic Circuit Embodying 12 Different Symbolic Ohmic Resistors

The 3D circuit of interest is shown in Figure 1. It is composed of twelve different-valued symbolic ohmic resistors. The resistors are the sides of the cube. Objectively one might ask why a cube and not any other 3D-shaped structure. This is addressed shortly in the upcoming report. As shown in Figure 1, the cube is being fed with a current at one of the selected vertices, e.g., vertex A, it could depart the structure at any one of the seven leftover vertices, e.g., vertex G. As pointed out, the interest is to replace the twelve resistors with one unique resistor, R_eq, without altering the incoming and the outgoing currents. Since anyone of the seven vertices leads to a different R_eq, seven different R_eqs are calculated.

Irrespective of the chosen pair, one needs to apply Kirchhoff’s laws; the nodal and the loop laws to generate adequate information addressing the issue of interest. In short, the nodal law states the net current at any vertex is null, and the loop law says in any closed loop the net emf equals the net voltage drop across the ohmic resistors. Although these laws are sound, in practice they lead to multiple-coupled linear equations that are cumbersome to solve numerically let alone symbolically! Here is where the project is tied to a CAS, specifically Mathematica.

By applying the mentioned laws, we form the needed equations.

Analysis

The currents split along the sides of the cube, applying the nodal law this gives eight separates but intertwined linear equations. The currents are labeled i_n as shown in Figure 1. Conventionally, the incoming currents are positive, and the outgoing ones are negative. The nodes are labeled and shown in Figure 1.

$\text{nodeA}=i-i_1-i_4-i_9$ ;

$\text{nodeB}=i_1-i_2-i_{10}$ ;

$\text{nodeC}=i_2+i_3-i_{11}$ ;

$\text{nodeD}=i_4-i_3-i_{12}$ ;

$\text{nodeE}=i_9-i_5-i_8$ ;

$\text{nodeF}=i_5+i_{10}-i_6$ ;

$\text{nodeG}=i_6+i_7+i_{11}$ ;

$\text{nodeH}=i_8+i_{12}-i_7$ ;

Applying the loop law gives another set of needed intertwined equations. The loops are labeled and shown in Figure 1.

$\text{loopABFEA}=R_1{i}_1+R_{10}{i}_{10}-R_5{i}_5-R_9{i}_9$ ;

$\text{loopDCGHD}=R_3{i}_3+R_{11}{i}_{11}-R_7{i}_7-R_{12}{i}_{12}$ ;

$\text{loopBCGFB}=R_2{i}_2+R_{11}{i}_{11}-R_6{i}_6-R_{10}{i}_{10}$ ;

$\text{loopADHEA}=R_4{i}_4+R_{12}{i}_{12}-R_8{i}_8-R_9{i}_9$ ;

$\text{loopABCDA}=R_1{i}_1+R_2{i}_2-R_3{i}_3-R_4{i}_4$ ;

With the mentioned set of twelve equations, one might try solving them symbolically. This requires a non-singular 12 × 12 determinant, its matrix is,

$\left(\begin{array}{cccccccccccc}-1& 0& 0& -1& 0& 0& 0& 0& -1& 0& 0& 0\\ 1& -1& 0& 0& 0& 0& 0& 0& 0& -1& 0& 0\\ 0& 1& 1& 0& 0& 0& 0& 0& 0& 0& -1& 0\\ 0& 0& -1& 1& 0& 0& 0& 0& 0& 0& 0& -1\\ 0& 0& 0& 0& -1& 0& 0& -1& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& -1& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& -1& 1& 0& 0& 0& 1\\ R_1& 0& 0& 0& -R_5& 0& 0& 0& -R_9& R_{10}& 0& 0\\ 0& 0& R_3& 0& 0& 0& -R_7& 0& 0& 0& R_{11}& -R_{12}\\ 0& R_2& 0& 0& 0& -R_6& 0& 0& 0& -R_{10}& R_{11}& 0\\ 0& 0& 0& R_4& 0& 0& 0& -R_8& -R_9& 0& 0& R_{12}\\ R_1& R_2& -R_3& -R_4& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right)$

Applying Mathematic the determinant becomes an expression with 384 symbolic terms, containing all kinds of terms. The expression is long, it fills pages. We avoid its display.

However, if one assumes identical resistors with value R, the matrix becomes,

$\left(\begin{array}{cccccccccccc}-1& 0& 0& -1& 0& 0& 0& 0& -1& 0& 0& 0\\ 1& -1& 0& 0& 0& 0& 0& 0& 0& -1& 0& 0\\ 0& 1& 1& 0& 0& 0& 0& 0& 0& 0& -1& 0\\ 0& 0& -1& 1& 0& 0& 0& 0& 0& 0& 0& -1\\ 0& 0& 0& 0& -1& 0& 0& -1& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& -1& 0& 0& 0& 1& 0& 0\\ 0& 0& 0& 0& 0& 0& -1& 1& 0& 0& 0& 1\\ R& 0& 0& 0& -R& 0& 0& 0& -R& R& 0& 0\\ 0& 0& R& 0& 0& 0& -R& 0& 0& 0& R& -R\\ 0& R& 0& 0& 0& -R& 0& 0& 0& -R& R& 0\\ 0& 0& 0& R& 0& 0& 0& -R& -R& 0& 0& R\\ R& R& -R& -R& 0& 0& 0& 0& 0& 0& 0& 0\end{array}\right)$

Its determinant reduces to one term only with a value of −384R⁵. This ensures the existence of a non-zero solution.

Here we would like to comment on solving this set of intertwined linear equations. Solving these equations symbolically is a monumental task. If one assigned numeric value to the resistors the equations easily may be solved with numeric-based software, e.g. even Excel. Here the issue is sought for the solution symbolically. To our surprise, Mathematica does the job. As one may imagine the solution is extremely long filling pages, even Mathematica output is abbreviated. We avoid showing the result, but noting the problem is solved. Here is the code,

Solve[{nodeA = = 0, nodeB = = 0, nodeC = = 0, nodeD = = 0, nodeE = = 0, nodeF = = 0, nodeH = = 0, loopABFEA = = 0, loopDCGHD = = 0, loopBCGFB = = 0, loopADHEA = = 0, loopABCDA = = 0}, {i₁, i₂, i₃, i₄, i₅, i₆, i₇, i₈, i₉, i₁₀, i₁₁, i₁₂}];

To verify the accuracy of the solutions, i.e., the currents, one may compare the output vs. the intuitive expectation. For identical resistors at any corner of the cube, the incoming currents should split evenly amongst the other two sides. The code EquaRs sets the resistors equal to R. E.g., for R = 1.0 Ω we get,

EquaRs = {R₁ → R, R₂ → R, R₃ → R, R₄ → R, R₅ → R, R₆ → R, R₇ → R, R₈ → R, R₉ → R, R₁₀ → R, R₁₁ → R, R₁₂ → R}

The code is given in the next line. The output is the currents labeled i_n for n = 1, 2, …, 12.

soliEqualR = Solve[{nodeA = = 0, nodeB = = 0, nodeC = = 0, nodeD = = 0, nodeE = = 0, nodeF = = 0, nodeH = = 0, loopABFEA = = 0, loopDCGHD = = 0, loopBCGFB = = 0, loopADHEA = = 0, looBCDA = = 0}/.EquaRs, {i₁, i₂, i₃, i₄, i₅, i₆, i₇, i₈, i₉, i₁₀, i₁₁, i₁₂}]

{{i₁ → i/3, i₂ → i/6, i₃ → i/6, i₄ → i/3, i₅ → i/6, i₆ → i/3, i₇ → i/3, i₈ → i/6, i₉ → i/3, i₁₀ → i/6, i₁₁ → i/3, i₁₂ → i/6}}

As one notices e.g., at vertex A, the incoming current i splits evenly amongst the connected sides, 1, 4, and 9. This is also true at any of the seven vertices. Assuming the numeric value of the feeder current is 1.0 Amp, the value of all twelve currents is shown in Figure 2.

In Figure 2, the horizontal axis is the Indices associated with the corresponding currents. e.g., 4 on the horizontal axis corresponds to i₄ with the value of 0.33 Amp.

The shown solution should not be confined to equal resistors. Here we consider a case where the numeric values of all resistors are different. The listing specialNonEqualR is one such case. Applying the given code in the text, Solve gives the numeric values of all twelve currents. The graph of these currents is shown in Figure 3. All the currents are associated with the value of the feeder current i, as expected. For instance, by setting the i = 1 Amp the numeric values are depicted in Figure 3.

Figure 2. Values of the currents vs. the labeled index of the currents.

specialNonEqualR = {R₁ → 1, R₂ → 2, R₃ → 3, R₄ → 4, R₅ → 5, R₆ → 6, R₇ → 7, R₈ → 8, R₉ → 9, R₁₀ → 10, R₁₁ → 11, R₁₂ → 12}

{{i₁ → (1,507,727 i)/2,870,812, i₂ → (8721 i)/29,596, i₃ → (170,671 i)/2,870,812, i₄ → (671,897 i)/2,870,812, i₅ → (380,987 i)/2,870,812, i₆ → (1,042,777 i)/2,870,812, i₇ → (811,427 i)/2,870,812, i₈ → (310,201 i)/2,870,812, i₉ → (172,797 i)/717,703, i₁₀ → (330,895 i)/1,435,406, i₁₁ → (254,152 i)/717,703, i₁₂ → (250,613 i)/1,435,406}}

The other objective of this report is to replace the twelve different valued resistors shown in Figure 1 with one, R_eq. As mentioned, this depends on the chosen pair. For instance, if the cube is being fed at vertex A, this can be paired with seven vertices requiring seven different R_eq, we are reporting all seven. Having symbolic expressions for all R_eq and because these are super lengthy expressions, we avoid including them in the report. But for special numeric cases, these are included.

Here we give the code calculating symbolic R_eq for the $\overline{\text{AH}}$ pair see Figure 1. For a special case of identical resistors, this is,

(R₁i₁ + R₂i₂ − R₃i₃ + R₁₂i₁₂)/i/.soliEqualR/.EquaRs (*AH*)

{R/2}

And for R = 1 Ω this gives 0.5 Ω. Exercising the same procedure for all seven pairs gives values that are depicted in Figure 4 and Figure 5.

Figure 3. The description is the same as the caption of Figure 2. Along the horizontal axis, e.g., 4 is the R₄ = 4 Ω and its corresponding ordinate is 0.234.

Figure 4. Display of the R_eq for the chosen pair e.g., along the horizontal axis 7 means $\overline{\text{AH}}$ that connects A to H via multiple routes described in the text all with the values of 0.5 Ω.

Figure 5. The caption is the same as in Figure 4. The difference between the two Figures is the values of the resistors. All the resistors are different given in the listing specialNonEqualR.

3. Conclusion and Comment

We set out crafting this report addressing the issues concerning electric circuits in a 3D structure. One of the issues of interest is the calculation of R_eq. That is instead of placing a cube with twelve different-valued resistors in a circuit to place one resistor, equivalent-resistor without altering the input and output currents. Addressing this issue led to the evaluation of not one but seven different possible R_eq. This report has identified all seven not only numerically but symbolically. The latter has been achieved merely because of the deployed CAS, specifically Mathematica. As an unintentional side product, we also generated symbolic expressions for the currents running in all twelve sides of the cube irrespective of the values of the resistors. This project keenly is tied with a CAS, otherwise, it couldn’t have been completed. The literature search reveals no similar report whatsoever. Motivated by the outcome the author will report shortly the R_eqs for structures other than a cube. We have provided the needed Mathematica codes. An interested individual may reproduce our results and potentially generalize it on a need basis. Concerning the embedded codes references [5] [6] [7] are resourceful.

Acknowledgements

The author acknowledges the John T. and Page S. Smith Professorship funds for completing and publishing this work.

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

References

[1]	Bauer, W. and Westfall, G, (2011), University Physics with Modern Physics, McGraw Hill, New York.
[2]	Haillday, D., Resnick, R. and Walker, J. (2013) Fundamental of Physics Extended. 12th Edition, Wiley John & Sons, Inc., New Jersey.
[3]	Mathematica V13.1 (2022) http://Wolfram.com
[4]	Excel, Standard Package in Microsoft 365. http://Microsoft/Excel
[5]	Wolfram, S. (2003) Mathematica Book. 5th Edition, Cambridge University Press, Cambridge.
[6]	Sarafian, H. (2019) Mathematica Graphics Examples. 2nd Edition, Scientific Research, Publishing, Wuhan.
[7]	Sarafian, H. (2015) Mathematica Graphics Example Book for Beginners, Scientific Research Publishing, Wuhan.