// Examle 3.22

               // From the diagram (3.40b) Apply KCL to node a
               // will get { (va-0)/2+ (va-vb)/3 = 5 }............(1
               // Similarly apply KCL at node b
               // will get { (vb-va)/3+ vb-0)/4 = -6 }............(2

               // After solving these 2 equation will have

Va=2.44;                     // Voltage at node a
Vb=-8.89;                    // Voltage at node b
Vab=Va-Vb;                   // Voltage across 3 ohm resistor
disp(' Voltage across 3 ohm resistor = '+string(Vab)+' Volt');




                     //   p 80        3.22