On 1/31/2010 5:08:45 PM, Tom_Gutman wrote:
>>>Mathcad 11 handles this just fine, and returns the correct answer for both variables. Mathcad 13 gets it wrong.<<
>Did MC13 get an actually wrong
>answer, or did it merely
>return complex results?
I think that what Richard did is that with ZZ=Z(R,L) mathcad 11 returns the correct answer, but mathcad 14 (I don't have 13, and 14 taken by a friend) returns RL = 4.7 (untouched), like the symbolic solver in mathcad 11.
Splitting the equation (both in mc 11 & 13) into Re and Im parts make mc14 to understand what mc11 'knows' - that for numerical solvers.
>While this is highly undesirable, and often produces useless results (what do you do with a complex inductance?)
Z = R +jwL is the impedance, then:
Re(Z) = R and Im(Z) = wL
For the usual alternate current w is constant. Because u*v and 1/v for u,v complex have an easy geometrical interpretation the use of impedances are very popular to enable electricians installers to make their calculus with a minimum mathematical background and in a graphical fashion way.
So, inductances are usually taken as pure imaginary values.
Above split in ReIm parts is the 'usual', but the method in the problem is one of three:
1- Take advantages of the advanced knowledge of numerical implementation of the mathcad numerical solver (and in the case mc14 is recovered taking Re and Im of the same equation).
2- There are an implicit assumption somewhere that mc11 take -badly- and splitting equations forces mc14 to make also -badly again, but excusable and with much more sense.
3- Is wrong.
Because doing the problem by hand I get the same numerical solution, I assume that the method is valid. What I can't do is write the mysterious implicit assumption, like some Re or Im = 0, and the complex guessing for a complex equation collapses my understanding of the mc numerical solver. That for 1 and 2.
The argument that I found for 3 is better explained by the figure
Regards. Alvaro.
>>>Mathcad 11 handles this just fine, and returns the correct answer for both variables. Mathcad 13 gets it wrong.<<
>Did MC13 get an actually wrong
>answer, or did it merely
>return complex results?
I think that what Richard did is that with ZZ=Z(R,L) mathcad 11 returns the correct answer, but mathcad 14 (I don't have 13, and 14 taken by a friend) returns RL = 4.7 (untouched), like the symbolic solver in mathcad 11.
Splitting the equation (both in mc 11 & 13) into Re and Im parts make mc14 to understand what mc11 'knows' - that for numerical solvers.
>While this is highly undesirable, and often produces useless results (what do you do with a complex inductance?)
Z = R +jwL is the impedance, then:
Re(Z) = R and Im(Z) = wL
For the usual alternate current w is constant. Because u*v and 1/v for u,v complex have an easy geometrical interpretation the use of impedances are very popular to enable electricians installers to make their calculus with a minimum mathematical background and in a graphical fashion way.
So, inductances are usually taken as pure imaginary values.
Above split in ReIm parts is the 'usual', but the method in the problem is one of three:
1- Take advantages of the advanced knowledge of numerical implementation of the mathcad numerical solver (and in the case mc14 is recovered taking Re and Im of the same equation).
2- There are an implicit assumption somewhere that mc11 take -badly- and splitting equations forces mc14 to make also -badly again, but excusable and with much more sense.
3- Is wrong.
Because doing the problem by hand I get the same numerical solution, I assume that the method is valid. What I can't do is write the mysterious implicit assumption, like some Re or Im = 0, and the complex guessing for a complex equation collapses my understanding of the mc numerical solver. That for 1 and 2.
The argument that I found for 3 is better explained by the figure
Regards. Alvaro.
