Parallel, serial or something else?

The whole network consists of four fully connected knots, which are labeled with an index $i =0,1,2,3$ .

• Each knot has a corresponding electric potential $U_0$ , $U_1$ , $U_2$ and $U_3 = 0$ (ground)

• The current $I_{ij}$ between two knots follows Ohm's law $I_{ij} = \frac{U_i - U_j}{R}$

• For the points $i =1$ and $2$ Kirchhoff's current law states $\begin{aligned} I_{10} + I_{12} + I_{13} &= \frac{U_1 - U_0}{R} + \frac{U_1 - U_2}{R} + \frac{U_1}{R} = \frac{1}{R} \left( 3 U_1 - U_0 - U_2 \right) = 0 \\ I_{20} + I_{21} + I_{23} &= \frac{U_2 - U_0}{R} + \frac{U_2 - U_1}{R} + \frac{U_2}{R} = \frac{1}{R} \left( 3 U_2 - U_0 - U_1 \right) = 0 \end{aligned}$ This is a linear equation system $\left( \begin{array}{cc} 3 & - 1 \\ -1 & 3 \end{array} \right) \left( \begin{array}{c} U_1 \\ U_2 \end{array} \right) = \left( \begin{array}{c} U_0 \\ U_0 \end{array} \right)$ for the two unknown $U_1$ , $U_2$ with the solution $U_1 = U_2 = \frac{1}{2} U_0$

• For the point $i = 0$ the sum of all outgoing currents corresponds to the total current $I_\text{tot} = I_{01} + I_{02} + I_{03} = \frac{U_0}{2 R} + \frac{U_0}{2 R} + \frac{U_0}{R} = \frac{2}{R} U_0$

• The quotient of total voltage and current yields the total resistance $R_\text{tot} = \frac{U_0}{I_\text{tot}} = \frac{R}{2}$

By symmetry of the circuit the rightmost resistor can be removed for further analysis of the circuit as there will be no current in that resistor . After removal ,the circuit becomes a simple connection of series and parallel resistors