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 e , I = 4 A
The current flowing in each edge can now be read off from the
Fig