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                                                Posted:
                            
                                7 years ago                            
                            
                                2019年4月4日 GMT+8 17:19                            
                        
                        
                                                    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