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//===========================================================================
//chapter 10 example 15
clc;clear all;
//variable declaration
P = 200; //resistance in arm in Ω
Q = 200; //resistance in arm in Ω
S = 200; //resistance in arm in Ω
R = 200; //resistance in arm in Ω
p = 0.5; //power in W
r = 2; //r is internal resistance of battery in Ω
E = 24; //voltage in V
//calculations
//P = (I^2)*R; power disiipation in W
I = sqrt(p/R);
V = I*2*R; //the maximum voltage ,that can be appliedto the bridge in V
I1 = 2*I; //current through series resistor in A
//E = V+(2*I*(r+R) battery emf E
R1 = ((E-V)/I1)-r; //series resistance in Ω
//result
mprintf("current = %3.2f A",I);
mprintf("\nseries resistance = %3.2f Ω",R1);
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