Let the current distribution be as shown in the figure. Considering the close circuit ABCA and applying Kirchhoffs Second Law, we
100x- 300z + 500y = 0
or x - 5y + 3z = 0 (i)
Similarly, considering the closed loop BCDB, we have
- 300z- 100(y+ z) + 500(x- z) = 0
or 5x - y - 9z = 0 ...(ii)
Taking the circuit ABDEA, we get
- 100x - 500(x - z) + 100 - 100(x + y) = 0
or 7x + y - 5z =1 ...(iii)
The value of x, y and z may be found by solving the above three simultaneous equations or by the method of determinants as given below :
Putting the above three equations in the matrix form, we have
∴ x = 48/240 = 1/5A; y = 24/240 = 1/10A; z = 24/240 = 1/10A
Current supplied by the battery is x + y =1/5 + 1/10 =3/10 A
