PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

Direct Current and Transient Analysis 151

Observe from Figure 2.49 that when R 3 = 250 Ω, I 2 = I 3. Also observe that the current I

is the least affected, whereas the currents I 3 and I 2 are equally affected by the changes

in R 3 , over the range 0–500 Ω.

Example 2.10

Write a MATLAB program that returns in a tablelike format the resistance values (R)

versus length L of a copper wire with a diameter of √

____

300 mills (the cross-sectional area:

300 CM [circular mills]), over the range 50 ft ≤ L ≤ 500 ft in linear increments of 50 ft,

assuming that the temperature remains constant at 20°C.

ANALYTICAL Solution

Recall that the resistance R is given by

R

L

A

rho* (at a room temperature of 20°C)

where rho = ρ denotes resistivity in CM Ω/ft, and the rho of copper is 10.37 CM Ω/f t,

at 20°C; L defi nes its length in feet; A defi nes the cross-sectional area in CM (circular

mills); A (area of a circle) = pi r^2 = π (^) d^2 /4; 1000 mills = 1 in. and π/4 sq. mills = 1 CM,
then if d is given in mills, A = d^2 (in CM).
MATLAB Solution

L = 50:50:500;
Rho =10.37;
A = 300;
R=Rho.*L./A;
disp(‘’)
disp(‘Length (ft) Resit (Ohms)’); results = [L’ R’];
disp(‘’)
disp(results)
disp(‘*****’)

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Length (ft) Resit (Ohms)

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50.0000 1.7283
100.0000 3.4567
150.0000 5.1850
200.0000 6.9133
250.0000 8.6417
300.0000 10.3700
350.0000 12.0983
400.0000 13.8267
450.0000 15.5550
500.0000 17.2833

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Example 2.11
Write a program that returns in a tablelike format the resistance of a copper wire of
100 ft long with a fi xed cross-sectional area A = 1 CM as a function of temperature
variations (∆T), over the range 0–100°C, in linear increments of ∆T = 10°C.