We construct a cube frame from $1 \Omega$ resistance wires, that is, a cube where only the edges are present. We pass a current of strength $i$ through the cube. What will be the current strengths along the individual edges, and what will be the resulting resistance of our cube frame, that is, with what resistance single wire could we replace it if we apply the electrical voltage to a) the two endpoints of one of its body diagonals, b) the two endpoints of one of its face diagonals?
We state that at a branch point, the sum of the currents flowing in the individual branches is equal to the current strength flowing in the main branch: $i=i_{1}+i_{2}+i_{3}+\ldots$, and the current strengths flowing in two parallel branches are inversely proportional to the resistances of the branches: $i_{1}: i_{2}=r_{2}: r_{1}$. The resultant resistance of series-connected conductors is equal to the sum of the individual resistances: $r=r_{1}+r_{2}+r_{3}+\ldots$, and the reciprocal of the resultant resistance of parallel-connected conductors is equal to the sum of the reciprocals of the individual resistances:

$$
\frac{1}{r}=\frac{1}{r_{1}}+\frac{1}{r_{2}}+\frac{1}{r_{3}}+\cdots
$$

a) Let's place the body diagonal vertically, whose two endpoints we apply the voltage to, these two points being $A$ and $H$, with voltages $V_{1}$ and $V_{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e0900ab86453995282b3g-1.jpg?height=363&width=463&top_left_y=542&top_left_x=820)

Three more vertices are at the same height $(B, C$ and $D$, and also $E, F$ and $G$), and their voltages are also the same, $V_{2}$ and $V_{4}$, which follows from the fact that equally constructed, equally resistive conductors connect them to both the $A$ and $H$ points. Between the points with voltages $V_{1}$ and $V_{2}$, 3 parallel-connected $1 \Omega$ conductors lead, whose resultant resistance is $1 / 3 \Omega$, and the same is true for the conductors connecting the points with voltages $V_{3}$ and $V_{4}$. Six conductors connect the points with voltages $V_{2}$ and $V_{3}$, their resultant resistance is $1 / 6 \Omega$. The total resistance of the cube frame is composed of a $1 / 3 \Omega$ segment, a $1 / 6 \Omega$ segment connected in series to it, and another $1 / 3 \Omega$ segment connected in series, so it is $5 / 6 \Omega$.

The current strength in the branches starting from points $A$ and $H$ is $i / 3$, and in the other branches, it is $i / 6$.

b) Apply voltage to the endpoints of the $A G$ face diagonal and place this face diagonal vertically. Now the voltages at points $C, D, E$ and $F$ will be the same, because the voltage drop from point $A$ to these points is the same as from these points to point $G$. Let the voltage at point $A$ be $V_{1}$, at point $G$ be $V_{5}$, and the common voltage at points $C, D, E$ and $F$ be $V_{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e0900ab86453995282b3g-1.jpg?height=355&width=526&top_left_y=1405&top_left_x=797)

The current can travel from point $A$ to the points with voltage $V_{3}$ in two ways: directly along the $1 \Omega$ conductors $A C$ and $A F$, or from $A$ to $B$ and then branching to $B D$ and $B E$. The resultant resistance of this branching is $1 / 2 \Omega$, and with the $A B$ conductor segment, it is $\frac{3}{2} \Omega$. The resultant resistance of the $A C, A B \searrow D$ and $A F$ conductors can be calculated as follows:

$$
\frac{1}{r}=1+\frac{2}{3}+1=\frac{8}{3}
$$

so $r=3 / 8 \Omega$, and the total resistance of the cube frame is twice this: $3 / 4 \Omega$.

No current flows in the $C D$ and $E F$ branches $(i=0)$. Considering that the resultant resistance of $A C$ and $A F$ is $1 / 2 \Omega$, while that of the $A B \backslash D$ branch is $3 / 2 \Omega$, a total of $3 / 4 i$ flows through $A C$ and $A F$, and $1 / 4 i$ through $A B \backslash D E^{-\mathrm{n}}$. In the $A C$ and $A F$ branches, the current strength is $3 / 8 i$ each, and in the $A B$ branch, it is $1 / 4 i$, of which $1 / 8 i$ flows through the $B D$ branch, and the other $1 / 8 i$ through the $B E$ branch. Using the symmetry, the current strengths in the lower part branches can also be determined.