**Appendix A**

The easiest way to solve the network of Figure 3 is to use the so-called symmetrical components. One will notice that, apart from the current source, the network shown in Figure 3 is symmetrical.

Hence, the network can be represented as the superposition of two networks (*a* and *b*) shown in Figure A1. On the right-hand side, opposite current sources were introduced so that the total current remained zero. The network of Figure A1a is perfectly symmetrical so that no current can flow through the resistor *R'*. Hence, the network of Figure A1a is equivalent to the network of Figure A1c. The network of Figure A1b is antisymmetrical, which means that the middle of the resistor *R'* is always at zero potential. Consequently, this network is equivalent to the network shown in Figure A1d. At last, we just have to solve the networks of Figure A1c,d. Both are simple *RC* networks that can be easily solved by inspection. Immediately, one finds the time constants of Equation (3). Adding the two solutions gives rise to the relation in Equation (2).

**Figure A1.** Equivalent thermal network represented as the superposition of a symmetrical and an antisymmetrical network.
