blob: b540a958af21a37fdb1b1641b22149bcb0e0776a (
plain)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
|
//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)
|
