Example (1):

Determine the current x in the 4-ohm resistance in the circuit shown in Fig. (a) below.

Solution:

The assumed distribution of current is shown in Fig. (b). Applying Kirchhoff’s laws to different closed loops, we get

Circuit EFADE:

-2y + 10z + (x – y – 6) = 0     

or         x – 3y + 10z = 6          …(i)

Circuit ABCDA :

-2(y + z + 6) – 10 + 3(x – y – z – 6) –10z = 0

or         3x – 5y – 15 z = 40      …(ii)

Circuit EDCGE :

-(x – y – 6) – 3(x – y – z – 6) – 4x + 24 = 0

or         8x – 4y – 3z = 48         …(iii)

Multiplying equation (i) by 5 and equation (ii) by 3 and then subtracting equation (ii) from equation (i),we get

-4 x + 95 z = -90

4 x – 95 z = 90    …(iv)

Next, multiplying equation (ii) by 4 and equation (iii) by 5 and subtracting equation (iii) from equation (ii), we get

     -28 x – 45 z = -80

or    28 x + 45 z = 80                                                                      …(v)

Multiplying equation (iv) by 45 and equation (v) by 95 and adding the two, we get : 2840 x = 11650

or    284 x = 1165          

               x = 1165 / 284 =  4.1 A