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+                             // Examle 3.18
+
+              // From the diagram (3.33a) Apply KVL to Bigger loop i.e (For I1 )
+              // Will get { 10-5(I1-2)-8I1= 0 }
+              // Using loop-circuit analysis 
+
+I1=20/13;     // Current through 8 ohm resistor
+disp(' Current through 8 ohm resistor (I1) = '+string(I1)+' Amp');
+
+
+
+
+              //   p 74         3.18 
+
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