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Line 21 giving me troubles

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azaandria
azaandria 2021 年 2 月 7 日
コメント済み: David Hill 2021 年 2 月 7 日
Could someone help please? Having errors with line 21 for "Time".
%This program will calculate the approximate time for electrons to travel
% from cathode to anode of a rectifier tube.
%
clear
clc
%Get user input
min_V=input('Enter the min limit of accelerating voltage:');
max_V=input('Enter the max limit of accelerating voltage:');
r1=input('Enter the radius of the inner tube (cathode) in mm:');
r2=input('Enter the radius of the outer tube (anode) in mm:');
%
%
V=(min_V:1:max_V);
%Range of voltage 1-10, increments 1
z=log(r2./r1);
%Natural Logarithm of R2/R1
q=1.60206*10^-19;
%Charge of electron in coulombs
m=9.1083*10^-31;
%Mass of electron in kilograms
time=z*r1*(sqrt((2*m)/(q.*V)))*(1+(z/3)+(z^2/10)+(z^3/42)+(z^4/216));
%Time for electron to travel from cathode to anode (in sec)
%
%
%
%
%Display output as required
my_results=[V',time'];
disp(' ');
disp('Voltage Time');
disp('-------- ------');
disp('(volts) (seconds)');
disp(my_results);
%
%End of file

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David Hill
David Hill 2021 年 2 月 7 日
編集済み: David Hill 2021 年 2 月 7 日
time=z*r1*(sqrt((2*m)./(q*V)))*(1+(z/3)+(z^2/10)+(z^3/42)+(z^4/216));%just need ./ since V is an array
  2 件のコメント
azaandria
azaandria 2021 年 2 月 7 日
Thank you. That did help partially. But something is still wrong with my formula. Its calculating 0.00 seconds for every volt. I will review my calculations. Thanks!
Enter the min limit of accelerating voltage:
1
Enter the max limit of accelerating voltage:
10
Enter the radius of the inner tube (cathode) in mm:
2
Enter the radius of the outer tube (anode) in mm:
4
Voltage Time
-------- ------
(volts) (seconds)
1.0000 0.0000
2.0000 0.0000
3.0000 0.0000
4.0000 0.0000
5.0000 0.0000
6.0000 0.0000
7.0000 0.0000
8.0000 0.0000
9.0000 0.0000
10.0000 0.0000
David Hill
David Hill 2021 年 2 月 7 日
It is working. Time is of the order of 10e-6, so the display is zero. Try:
plot(V,time);

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