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//===========================================================================
//chapter 10 example 23
clc
clear all
//variable declaration
V = 3; //battery voltage in volts
Rm = 60; //resistance in Ω
Ifm = 1.2; //full-scale deflection meter current in mA
Rh = 1500; //half-scale deflection resistance in Ω
V1 = 0.3; //at 10 % drop in battery voltage in V
//calculations
If = V/Rh; //battery current for full-scale deflection in A
If1 = If*10^3; //battery current for full-scale deflection in mA
Ish = If1-Ifm; //current through zero adjuster resistor i.e,shunt resistor in mA
Rsh = (Ifm*Rm)/Ish; //resistance in Ω
Rse = Rh-((Rsh*Rm)/(Rsh+Rm)); //current limiting resistor i.e,series resistor
V3 = V-V1; //voltage in V
If2 = V3/Rh; //battery current at full-scale deflection in A
If21 = If2*10^3; //battery current at full-scale deflection in mA
Ish1 =If21-Ifm; //current through shunt resistor in mA
Rsh1 = (Ifm*Rm)/Ish1; //shunt resistor in Ω
Rh1 = Rse+((Rm*Rsh1)/(Rm+Rsh1)); //total internal circuit resistance in Ω
e =((Rh-Rh1)/(Rh1))*100; //percentage error in %
//calculation
mprintf("resistance = %3.2f Ω",Rsh);
mprintf("\ncurrent limiting resistor = %3.2f Ω",Rse);
mprintf("\nshunt resistor = %3.2f Ω",Rsh1);
mprintf("\npercentage error = %3.3f percentage ",e);
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