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+                             // 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