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//===========================================================================
//chapter 10 example 11
clc;clear all;
//variable declaration
P = 1000; //resistance in arm AC in Ω
Q = 1000; //resistance in arm AD in Ω
S = 100; //resistance in arm CB in Ω
R = 101; //resistance in arm BD in Ω
Rg = 50; //galvanometer resistance in Ω
E = 2; //voltage in V
//calculations
R1 = (Q*S)/P; //resistance required in arm BD for balance bridge
dR = R-R1; //the deviation from balanced condition in Ω
Eth = E*(((R1+dR)/(R1+dR+S))-(P/(P+Q))); //thevenin's open circuit voltage in V
Rth = (((R1+dR)*S)/(R1+dR+S))+((P*Q)/(P+Q)); //thevenin's equivalent resistance of bridge in Ω
Ig = Eth/(Rth+Rg); //galvanometer current in A
//result
mprintf("galvanometer current = %3.3f uA",(Ig*10^6));
mprintf("\nsince the point B is at higher potential with respect to point A ,current will floe from terminal A");
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