Further,
at the corners A¢, B and D, the incoming current I must split equally
into the two outgoing branches In this manner, the current in all
the 12 edges of the cube are easily written down in terms of I, using
Kirchhoff’s first rule and the symmetry in the problem Next take a closed loop, say, ABCC¢EA, and apply Kirchhoff’s second
rule:
–IR – (1/2)IR – IR + e = 0
where R is the resistance of each edge and e the emf of battery Thus,
e = 5
2 I R
The equivalent resistance Req of the network is
5
3
6
Req
R
=εI
=
For R = 1 W, Req = (5/6) W and for e = 10 V, the total current (= 3I) in
the network is
3I = 10 V/(5/6) W = 12 A, i