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-rw-r--r--1316/CH2/EX2.7/example2_7.sce34
-rw-r--r--1316/CH2/EX2.7/resultexample2_7.txt9
2 files changed, 43 insertions, 0 deletions
diff --git a/1316/CH2/EX2.7/example2_7.sce b/1316/CH2/EX2.7/example2_7.sce
new file mode 100644
index 000000000..4f13b9be6
--- /dev/null
+++ b/1316/CH2/EX2.7/example2_7.sce
@@ -0,0 +1,34 @@
+//Chapter 2
+//Example 2.7
+//Page 65
+
+clear;
+clc;
+
+R1 = 10000.00;
+R2 = 10000.00;
+R3 = 1000;
+R4 = 950.00;
+R5 = 50.00;
+Del_R3 = 1;
+V =10.00;
+
+//Calculation of Current required
+printf("First,for the nominal resistance values given, \n the bridge is set at anull with I=0,because\n")
+
+//Calculation of Va
+x = (V * R3)/(R1 + R3)
+printf("Va = %f V \n",x)
+
+//Calculation of Vb
+printf("With I=0 we get \n")
+y = ((V) * (R4+R5))/(R2 + R4 + R5)
+printf("Vb = %f V \n",y)
+
+//Calculation of Va for change in R3
+printf("When R3 increases by 1 ohm to 1001 ohm, Va becomes")
+z = (V * (R3+1000))/(R1 + (R3+1000))
+printf("Va = %f V \n",z)
+printf("which shows that the voltage at b must increase by 0.0008 V to renull the bridge.\nThis can be proved by current,from equation and Delta V=0,found from \n")
+I = 0.0008/50;
+printf("I = %f A",I)
diff --git a/1316/CH2/EX2.7/resultexample2_7.txt b/1316/CH2/EX2.7/resultexample2_7.txt
new file mode 100644
index 000000000..041ea5956
--- /dev/null
+++ b/1316/CH2/EX2.7/resultexample2_7.txt
@@ -0,0 +1,9 @@
+ First,for the nominal resistance values given,
+ the bridge is set at anull with I=0,because
+Va = 0.909091 V
+With I=0 we get
+Vb = 0.909091 V
+When R3 increases by 1 ohm to 1001 ohm, Va becomesVa = 1.666667 V
+which shows that the voltage at b must increase by 0.0008 V to renull the bridge.
+This can be proved by current,from equation and Delta V=0,found from
+I = 0.000016 A \ No newline at end of file