P 2.4-10 The voltage source shown in

Figure P 2.4-10 is an adjustable dc voltage source. In other words, the voltage v_s is a constant voltage, but the value of that constant can be adjusted. The tabulated data were collected as follows. The voltage, v_s, was set to some value, and the voltages across the resistor, v_a and v_b, were measured and recorded. Next, the value of v_s was changed, and the voltages across the resistors were measured again and recorded. This procedure was repeated several times. (The values of v_s were not recorded.) Determine the value of the resistance, R.

Figure P 2.4-10

This is a simple circuit so we can find v_b from Ohm’s law

We get i also from Ohm’s law

Combining these

Solving for R

So we could just choose one set of values, but what was the point of taking several? Better choice would be to plot v_b vs v_a, find the slope and use the slope in place of so the equation becomes

Using MATLAB we can find the best line and its slope. From MATLAB we find the slope is 0.6.

MATLAB Plot and Code follows:

%Program to find slope of best line for Homework Problem 2.4-10.

%Version 2018-12-28 D.W. Donovan

clear all;

va = [11.75 7.5 5.625 10 4.375]';

vb = [7.05 4.5 3.375 6 2.625]';

x = va;

y = vb;

a(:,2) = va;

a(:,1) = ones(size(va,2));

c = a\vb;

b = c(1);

m = c(2);

inew = [min(va) : (max(va) - min(va))/1000 : max(va)]';

vnew = m*inew +b;

R = 40*m;

ls = ['Model Slope is ' num2str(m) ' '];

ly = ['Model Y-intercept is ' num2str(b) ' (V)'];

lv = ['R = (40 \Omega)(slope) = ' num2str(R) ' \Omega'];

ans = {'Slope ' m; 'Y-intercept ' b;

'R = ' R };

ans

figure

hold on

plot(va, vb, 'k *','MarkerSize',20)

plot(inew, vnew, 'k-', 'LineWidth', 3)

plot(max(va), min(vb), '.w')

plot(max(va), min(vb), '.w')

plot(max(va), min(vb), '.w')

tt1 = 'PH 320 Homework Problem 2.4-10';

tt2 = 'Voltage vs voltage for a Circuit Element';

ttn = 'D.W. Donovan -- ';

tnl = '\newline';

ttf = [tt1 tnl tt2 tnl ttn date];

xl = 'Voltage, v, (V)';

yl = 'Voltage, v, (V)';

sp = 1;

axxmin = min(x)-sp;

axxmax = max(x) + sp;

axymin = min(y) - sp;

axymax = max(y) + sp;

title (ttf,'FontSize', 16)

xlabel(xl, 'FontSize', 16)

ylabel(yl, 'FontSize', 16)

legend ('Raw Data', 'Model Curve', ls, ly, lv, 'Location','SE')

legend ('boxoff')

axis([axxmin axxmax axymin axymax])

%{

ans = 'Slope ' [ 0.6000]

'Y-intercept ' [-1.2545e-15]

'R = ' [ 24.0000]

%}

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This page last updated on December 28, 2018