To simplify given circuits is an art of electrical engineers. Often it turns out that a **complicated problem** seen from a different point of view is extremely **easy**. The resistance cube ("**R-cube**") is one such example. The solution we will find can also be used to calculate the impedances of more complicated elements like the "**C-cube**" or the oscillatory "**RLC-cube**".

## Problem Statement

Find the total resistance \(R_{c}\) of the **resistance cube** ("R-cube") shown below with respect to opposing edges. You may want to draw a two-dimensional circuit diagram first. There is a lot of redundancy in the cube since all resistances are equal to \(R\) - maybe you can find a way to simplify the circuit.

When you have successfully calculated the resistance of the “R-cube”, you can easily calculate the impedance \(Z_{c,C}\) of the “**C-cube**”, where all resistances are exchanged with equal capacitances \(C\). Assume some time-harmonic voltage source connected to the opposing edges. Can you also find the impedance \(Z_{c,RLC}\) of the “**RLC-cube**” as shown below with resistance \(R\), inducance \(L\) and capacitance \(C\)?