Sol. Let us number the corners as 1,2,3,… and so on as shown
a(1−8−7−6),b(1−2−7−6),c(1−8−5−6),d(1−2
−3−6),e(1−4−5−6),f(1−4−3−6)
Now for each of these paths, we have identical resistances.
Let the current entering at point A be I. The current will be
distributed equally among all the three parts. Therefore, we can
say that the current is
I12=I14=I18=I/3
Now look at all paths from 2 to 6 . We have the following paths
g(2−7−6) and h(2−3−6) .
since they also have the same resistances, we can assume that
current in 2−3 and current in 2−7 must be same. Therefore,
I27=I23=12(I3)=I6
we can say that I43=I45=I85=I87=I/6
Therefore, to calculate the potential difference across 1−6
(see Fig. 5.34 ) the path A12%, we have
V1−I3R−I6R−I3R=V6⇒V1−V6=I[R3+R6+R3]=56IR
Hence, (V1−V6)I=56R
Therefore, the equivalent resistance is 56R .
I=V56R=105/6=12A
Hence, equivalent resistance is 56Ω , I12=I14=I18=4A andI43=I45=I85=I87=2A