Solved Problem on Resistors

We have 12 equal resistors with a value equal to R placed on the edges of a cube, as shown in the figure. Calculate the equivalent resistance between points A and G that form one of the main diagonals of the cube.

Solution

Point A is a node of the circuit, at this point, the current is divided by the resistors placed between points A and B, A and D, and between A and E since all resistors have the same value R. The voltage drop between each of these points is the same, so points B, D, and E represent the same point of the circuit \( B\equiv D\equiv E \), the three resistors "leave" the point in common A and "arrive" at the common point \( B\equiv D\equiv E \). Therefore these three resistors are in parallel (Figure 1).

Figure 1

The three resistors placed between the points C and G, F and G, H and G are also traveled by the same current that is at point G, points C, F and H represent the same point as the circuit \( C\equiv F\equiv H \). The resistors "leave" the common point \( C\equiv F\equiv H \) and "arrive" at the common point G. These are also in parallel (Figure 2).

Figure 2

The other resistors are all placed between the common points \( B\equiv D\equiv E \) and \( C\equiv F\equiv H \), they are all in parallel (Figure 3).

Figure 3

The cube circuit is equivalent to a flat circuit forming by three parallel resistors, in series with six parallel resistors and series with three more parallel resistors (Figure 4)

Figure 4

Let's call R₁ the equivalent resistance between points A and \( B\equiv D\equiv E \), and R₃ the equivalent resistance between the points \( C\equiv F\equiv H \) and G, as these parts of the circuit are equal we have that R₁ = R₃. The equivalent resistance for an association and equal resistors connected in parallel is given by

\[ \bbox[#99CCFF,10px] {R_{eq}=\frac{R}{n}} \]

for n = 3

\[ R_{1}=R_{3}=\frac{R}{3} \]

Note: We could determine the equivalent resistance by applying the general expression for resistors association in parallel

\[ \begin{gather} \frac{1}{R_{eq}}=\sum _{i=1}^{n}{\frac{1}{R_{i}}}\\ \frac{1}{R_{1}}=\frac{1}{R_{3}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}\\ \frac{1}{R_{1}}=\frac{1}{R_{3}}=\frac{3}{R}\\ R_{1}=R_{3}=\frac{R}{3} \end{gather} \]

Between the points \( B\equiv D\equiv E \) and \( C\equiv F\equiv H \) we have six equal resistors in parallel, we will call the equivalent resistance between these R₂ points, applying the expression for association in parallel resistors of equal value with n = 6

\[ R_{2}=\frac{R}{6} \]

Note: Or applying the general expression for resistors association in parallel

\[ \begin{gather} \frac{1}{R_{eq}}=\sum_{i=1}^{n}{\frac{1}{R_{i}}}\\ \frac{1}{R_{2}}=\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R}\\ \frac{1}{R_{2}}=\frac{6}{R}\\ R_{2}=\frac{R}{6} \end{gather} \]

Thus the circuit is reduced to the following

Figure 5

The equivalent resistance of the R_eq circuit will be the sum of resistors in series

\[ \bbox[#99CCFF,10px] {R_{eq}=\sum _{i=1}^{n}{R_{i}}} \]

\[ \begin{gather} R_{eq}=R_{1}+R_{2}+R_{3}\\ R_{eq}=\frac{R}{3}+\frac{R}{6}+\frac{R}{3} \end{gather} \]

multiplying and dividing the first and third terms on the right-hand side of the equation by 2

\[ \begin{gather} R_{eq}=\frac{R}{3}\times\frac{2}{2}+\frac{R}{6}+\frac{R}{3}\times\frac{2}{2}\\ R_{eq}=\frac{2R+R+2R}{6} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {R_{eq}=\frac{5R}{6}} \]

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