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+//Chapter 2
+//Example 2.9
+//Page 67
+
+clear;
+clc;
+
+R1 = 5000;
+R2 = 5000;
+R3 = 1000;
+R4 = 990;
+R5 = 10;
+Vx =10;
+Pot = 0.012;
+
+//Finding Current necessary to null the bridge
+printf("First an examination of the resistancs show \n that the bridge is nulled when I = 0 and Vx = 0\n from the equation (Vx)+((R3*V)/(R1+R3))-(V*(R4+R5)/(R2+R4+R5))-(I*R5)=0")
+x=(R3*Vx)/(R1+R3);
+printf("(R3*V)/(R1+R3))=%.3f V\n",x)
+y=(Vx*(R4+R5)/(R2+R4+R5))
+printf("(V*(R4+R5)/(R2+R4+R5))=%.3f V",y)
+z = Pot/R5*1000;
+printf("\n Thus,we can use equation Vx-IR5=0: \n 12mV-10I=0 \n Thus I = %0.1f mA",z)
+