blob: 4639df2323bbc78d95d27fa2af55e09fa973b6ee (
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
25
26
27
28
29
30
31
32
33
|
// Example 12.4
// Computation of (a) Required resistance of a noninductive diverter that will
// bypass 27 percent of the total armature current(b) Power rating of the
// diverter
// Page No. 494
clc;
clear;
close;
// Given data
Rs=0.00306; // Shunt generator resistance rating
Is=0.73; // Shunt generator current rating
Id1=0.27; // Armature winding resistance
Pload=170000; // Load of power
VT=250; // Shunt generator voltage rating
Id2=680; // No load voltage
Rd=0.27; // Resistance drop
// (a) Required resistance of a noninductive diverter that will bypass
// 27 percent of the total armature current
Rd=Rs*Is/Id1;
// (b) Power rating of the diverter
Ia=Pload/VT;
Pd=((Id1*Id2)^2)*Rd;
//Display result on command window
printf("\n Required resistance of a noninductive diverter = %0.5f Ohm ",Rd);
printf("\n Power rating of the diverter = %0.0f W ",Pd);
|
