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// Exa 1.1
clc;
clear;
// Given
I = 10; // Current in Amp
dI = 0.1; // Probability of error in I (Amp)
R = 100; // Resistor value in Ohms
dR = 2; // Probability of error in R (Ohms)
// Solution
printf('The power dissipated P = I^2*R \n');
printf(' The probable error can be determined with the help of Erss(Root Sum Square) Formula, i.e Error = sqrt((dR*I^2)^2 + (2*I*R*dI)^2) \n');
PE = sqrt((dR*I^2)^2 + (dI*2*I*R)^2); // Probable error
P = I^2 * R;
printf(' Error = %d W \n',round(PE));
printf(' Power dissipated P = %d kW \n',P*10^-3);
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