| Author | Message | ||
     Eduardo Ghershman (edug) Member Username: edug Post Number: 4 Registered: 08-2005 |
Dear Mr. Nelson. In your program resistor.pde,where this the height z to calculates the area of the resistor? Thanks in advance, Eduardo | ||
| Robert G. Nelson (rgnelson) Moderator Username: rgnelson Post Number: 445 Registered: 06-2003 |
The example I posted was a 2D model, so the z thickness was incorporated into the resistivity number. | ||
| Eduardo Ghershman (edug) Member Username: edug Post Number: 5 Registered: 08-2005 |
Dear Mr. Nelson. I modify the program "resistor.pde",to calculate de resistance , title '1000 OHM RESISTOR' title 'Resistor Model' select contours = 10 errlim = 1.0e-4 variables v definitions rho {sheet resistivity} sigma = 1/rho {conductivity} Va {applied potential} equations div(-sigma*grad(v)) = 0 boundaries Region 1 rho = 100 Va = 1 start (0,0) natural(v) = 0 line to (10,0) value(v) = Va line to (10,1) natural(v) = 0 line to (0,1) value(v) = 0 line to finish monitors contour(v) contour((sigma*grad(v)^2)/Va^2) as 'Resistive Heating(1/R)' surface((sigma*grad(v)^2)/Va^2) as 'Resistive Heating(1/R)' plots grid(x,y) contour(v) as "Potential" contour(magnitude(-sigma*grad(v))) as "Current Density" contour(magnitude(grad(v))) as 'Field Strength' contour((sigma*grad(v)^2)/Va^2) as 'Resistive Heating(1/R)' integrate report 1.0/integral((sigma*grad(v)^2)/Va^2) as 'Resistance' surface((sigma*grad(v)^2)/Va^2) as 'Resistive Heating(1/R)' integrate report 1.0/integral((sigma*grad(v)^2)/Va^2) as 'Resistance' end The value of the resistance,and de current density are good but the power disipation is 0.0001 instead of 0.001,why? Thanks in advance, Eduardo |