In Fig.5.52, the resistances are connected as shown. Given `R_1 = 10 OmegaR_2 = 20 Omega.` Determine the equivalent resistance between points (i) A and D and (ii) B and C

Explanation

(i) Calculation of equivalent between A and D:
There exists parallel axis of symmetry. The points across the
parallel axis of symmetry can be treated as equipotntial points.
Mehod 1: Points B and C, and E and F are at the same
potential, so the circuit can be redrawn as shown in fig. 5.23.
Thus, the equivalent resistance is `10 Omega`
Method 2: the points with charge symmetrical to parallel axis
are equipotential points. The resistances connected between
these points can be removed. A step- by-step procedure for the
calculation of equivalent resistance is shown in Fig. 5.54
Method 3: We can fold the circuit about the parallel axis of
symmetry. The step -by - step procedure is shown in Fig. 5.55.
ii. Calculation of equivalent resistnace between B and C :
There exists perpendicular axis of symmetry about line x-y
Hence, the points symmetrical about x-y are equipotential,
i.e., points (E and F) and (B and C) are equipotential. Thus,
the resistance connected across these points can be removed. Also points A and D are lying on the perpendicular axis of symmetry. Hence, points A and D are also equipotential. The
resistances conneted between A and D can also be removed.
The circuit reduces as shown, and the steps - by step procedure
is given in fig. 5.56