Solut ions (^261)
CD: RAB = R/2, where R is the resistance of
an edge. Therefore,
2U

R '
where U is the applied voltage.
The total current can be changed in two ways:
(1) if we remove one of the edges AD, AC, BC or
BD, the change in the current will be the same;
(2) if we remove the edge AB, the change will be
different. In the first case, the change in the current
will be Al = —(2/5) U/R = —1/5, and in the
second case, the total resistance will be R, and
hence Al = — U/R = —//2 = —"max.
3.25. It follows from symmetry considerations
that the potentials of points C and D are equal.
Therefore, this circuit can be replaced by an equiv-
alent one (we combine the junctions C and D,
c(D)
Fig. 212
Fig. 212). The resistance between points A and B
of the circuit can be determined from the formulas
for parallel and series connection of conductors:
R12 (1112+R) = 3 // ___ 3 R
RC(D)B— R/2-FR/2+R 4 2 8 '