SKEDSOFT

For the cycle VYXV, i1R1 i2R2 = E,
For the cycle VYZV, i2R2 i5R5 i6R6 = 0,
For the cycle VWZV, i3R3 i5R5 i7R7 = 0,
For the cycle VYWZV, i2R2 – i4R4 i5R5 i7R7 = 0,

The equations arising from Kirchhoff’s first law are :

For the vertex X, i0 – i1 = 0,
For the vertex V, i1 – i2 – i3 i5 = 0,
For the vertex W, i3 – i4 – i7 = 0
For the vertex Z, i5 – i6 – i7 = 0

These eight equations can now be solved to give the eight currents i0, ...... i7.

For example, if E = 12, and if each wire has unit resistance (that is, Ri = 1 for each i), then the solution is as given in Fig. (3.91).