Answer:
The plot for percentage error as a function of fractional displacement ( [tex]\frac{R_{L} }{R_{C} }[/tex]) for the values of 0.1,1.0,10.0 is shown in image attached below.
Explanation:
Electrical loading non linearity error (percentage) is shown below.
[tex]E=\frac{(\frac{v_{o} }{v_{r} }-\frac{Q}{Q_{max} } )}{\frac{Q}{Q_{max} } }[/tex]×[tex]100[/tex]
where Q= displacement of the slider arm
[tex]Q_{max}=[/tex] maximum displacement of a stroke
[tex]\frac{v_{o} }{ v_{r} } =[/tex][tex]\frac{(\frac{Q}{Q_{max} }(\frac{R_{L} }{R_{C} } ) )}{(\frac{R_{L} }{R_{C} } ) +(\frac{Q}{Q_{max} })-(\frac{Q}{Q_{max} })^{2} }[/tex]
here [tex]R_{L}=load resistance[/tex]
[tex]R_{C}=[/tex]total resistance of potentiometer.
Now the nonlinearity error in percentage is
[tex]E=\frac{(\frac{(\frac{Q}{Q_{max} }(\frac{R_{L} }{R_{C} } ) )}{(\frac{R_{L} }{R_{C} } ) +(\frac{Q}{Q_{max} })-(\frac{Q}{Q_{max} })^{2} }-\frac{Q}{Q_{max} } )}{\frac{Q}{Q_{max} } }[/tex]×[tex]100[/tex]
The following attached file shows nonlinear error in percentage as a function of [tex]\frac{R_{L} }{R_{C} }[/tex] displacement with given values 0.1, 1.0, 10.0. The plot is drawn using MATLAB.
The MATLAB code is given below.
clear all ;
clc ;
ratio=0.1 ;
i=0 ;
for zratio=0:0.01:1 ;
i=i+1 ;
tratioa (1,i)=zratio ;
E1(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zrtio)*100 ;
end
ratio=1.0 :
i=0 ;
for zratio=0:0.01:1 ;
i=i+1 ;
tratiob (1,i)=zratio ;
E2(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zratio)*100 ;
end
ratio=10.0 :
i=0 ;
for zratio=0:0.01:1 ;
i=i+1 ;
tratioc (1,i)=zratio ;
E3(1,i)=((((zratio*ratio)/(ratio+zratio-zratio^2))-zratio)/zrtio)*100 ;
end
k=plot(tratioa,E1,tratiob,E2,tratioc,E3)
grid
title({non linear error in % as a function of R_L/R_C})
k(1). line width = 2;
k(1).marker='*'
k(1).color='red'
k(2).linewidth=1;
k(2).marker='d';
k(2).color='m';
k(3).linewidth=0.5;
k(3).marker='h';
k(3).color='b'
legend ('location', 'south east')
legend('R_L/R_C=0.1','R_L/R_C=1.0','R_L/R_C=10.0')
