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//===========================================================================
//chapter 5 example 14
clc;clear all;
//variable declaration
V = 250; //voltage in V
RA = 100; //resistance in Ω
RB = 400; //resistance in Ω
x = 0.005; //error in measuring voltage in
//calculations
I = V/(RA+RB); //current flowing through resistance in A
VB = I*RB; //potential drop acreoss resitance in V
//Req = RA+((r*RB)/(r+RB))
//Ieq =V/Req = V/ RA+((r*RB)/(r+RB))
//Ieq = (V*(r+RB))/((RA*(r+RB))+(r*RB))
//V1 = Ieq*(r*RB)/(r+RB)
// V1 = (V*(r+RB))*(r*RB))/((r+RB)*((RA*(r+RB))+(r*RB)))
//V1 = (V*r*RB)/((r+RB)*((RA*(r+RB))+(r*RB)))
//V1 = (200*r)/(80+r)
V1 = VB*(1-x); //voltage measured with 0.5% error
r = (V1*80)/(200-V1); //solving equations we get minimum resistance in Ω
//result
mprintf("minimum resistance = %3.2f Ω",r);
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