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+
+
+                             // Examle 4.4
+
+               // From the diagram (4.6a)
+               // Using Superpositon theorem
+
+V=10;                         // Voltage source
+I1=(V/(2+4+6));               // When 10-V voltage source is on { by ohm's law }
+
+                              // we have to find Is= ?
+                              // When Is-A Current source is on
+                              // will have { I2= -(2/3)Is }
+                              // given that I1+I2= 0
+                              //  there for 5/6 - (2/3)Is= 0
+Is=(5*3)/(6*2);               // Source current
+disp(' The value of source current (Is)  = '+string(Is)+' Amp');
+
+
+
+
+
+           //   p 108          4.4