Please login with a confirmed email address before reporting spam
Posted:
7 years ago
2019/04/04 5:19 GMT-4
Yet another question: I prefer converting equations into a dimensionless form in order to minimize the number of adjustable parameters. COMSOL seems to multiply the current automatically with the Faraday constant, right? Hence, in order to get the dimensionless current, I need to divide the COMSOL values with F? Otherwise I do not understand the high values of the current (should be 0.4463 in a simple cyclic voltammogram).
Lasse
Yet another question: I prefer converting equations into a dimensionless form in order to minimize the number of adjustable parameters. COMSOL seems to multiply the current automatically with the Faraday constant, right? Hence, in order to get the dimensionless current, I need to divide the COMSOL values with F? Otherwise I do not understand the high values of the current (should be 0.4463 in a simple cyclic voltammogram).
Lasse