pik1432

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@Preben Alsholm : thank you for your answer, but when it was tried, Maple gave another complaint. Here is the worksheet. 


 

restart;

with(LinearAlgebra):

TrainLoad := -10*10^6*(cos(convert(40*degrees, radians))+I*sin(convert(40*degrees, radians)));

-10000000*cos((2/9)*Pi)-(10000000*I)*sin((2/9)*Pi)

(1)

 

evalf(TrainLoad, 7);

-7660444.-6427876.*I

(2)

f1n2 := (0.03 + I*0.1515)*Ix[c1] - (0.03 + I*0.1515)*Ix[c2] + 2 * V[at1] - 55*10^3 = 0;

(0.3e-1+.1515*I)*Ix[c1]+(-0.3e-1-.1515*I)*Ix[c2]+2*V[at1]-55000 = 0

(3)

f3n4 := (1.6 + I*6.24)*Ix[c1] + (1.12 + I*2.64)*Ix[c2] + V[t] - V[at1] = 0;

(1.6+6.24*I)*Ix[c1]+(1.12+2.64*I)*Ix[c2]+V[t]-V[at1] = 0

(4)

f5n6 := (1.36 + I*4.44)*Ix[c2] + V[at2] - V[t] = 0;

(1.36+4.44*I)*Ix[c2]+V[at2]-V[t] = 0

(5)

f7n8 := (-1.12 - I*2.64)*Ix[c1] + (-3.92 - I*12.00)*Ix[c2] + V[at2] - V[at1] = 0;

(-1.12-2.64*I)*Ix[c1]+(-3.92-12.00*I)*Ix[c2]+V[at2]-V[at1] = 0

(6)

f9n10 := V[t] * conjugate(Ix[c1] - Ix[c2]) + TrainLoad = 0;

V[t]*conjugate(Ix[c1]-Ix[c2])-10000000*cos((2/9)*Pi)-(10000000*I)*sin((2/9)*Pi) = 0

(7)

polynomials := {f1n2, f3n4, f5n6, f7n8, f9n10};

{(1.36+4.44*I)*Ix[c2]+V[at2]-V[t] = 0, V[t]*conjugate(Ix[c1]-Ix[c2])-10000000*cos((2/9)*Pi)-(10000000*I)*sin((2/9)*Pi) = 0, (-1.12-2.64*I)*Ix[c1]+(-3.92-12.00*I)*Ix[c2]+V[at2]-V[at1] = 0, (0.3e-1+.1515*I)*Ix[c1]+(-0.3e-1-.1515*I)*Ix[c2]+2*V[at1]-55000 = 0, (1.6+6.24*I)*Ix[c1]+(1.12+2.64*I)*Ix[c2]+V[t]-V[at1] = 0}

(8)

variables := {Ix[c1], Ix[c2], V[at1], V[at2], V[t]};

{Ix[c1], Ix[c2], V[at1], V[at2], V[t]}

(9)

solve(polynomials, variables);

Warning, solutions may have been lost

 

 

 


 

Download conjugate_sol_PrebenAlsholm.mw

@Carl Love : Thank you for the answer. The command 'fnormal' after 'simplify(..., zero)'did what I expected:

@nm : The following was what I was able to get, after following your answer:

Thank you!

@Carl Love Your answers are highly appreciated. 

@Carl Love : Then, I got it working and was able to see the final number. 

@rlopez : I was able to move one more step forward!

@acer : Your points are absolutely true! Thank you for the extra information regarding the difference between 'simplify()' and 'is()' as well. 

@Kitonum : would you allow me to ask this following up question?


 

restart;

Is_eq := I[mu]/sqrt((omega__r^2 + omega__rK^2)*sigma^2/(sigma^2*omega__rK^2 + omega__r^2));

I[mu]/((omega__r^2+omega__rK^2)*sigma^2/(sigma^2*omega__rK^2+omega__r^2))^(1/2)

(1)

Is_eq2 := I[mu] * sqrt((1+(omega__r/(sigma * omega__rK))^2)/(1+(omega__r/(omega__rK))^2));

I[mu]*((1+omega__r^2/(sigma^2*omega__rK^2))/(1+omega__r^2/omega__rK^2))^(1/2)

(2)

simplify(Is_eq - Is_eq2);

-I[mu]*(((omega__r^2+omega__rK^2)*sigma^2/(sigma^2*omega__rK^2+omega__r^2))^(1/2)*((sigma^2*omega__rK^2+omega__r^2)/(sigma^2*(omega__r^2+omega__rK^2)))^(1/2)-1)/((omega__r^2+omega__rK^2)*sigma^2/(sigma^2*omega__rK^2+omega__r^2))^(1/2)

(3)

is(Is_eq = Is_eq2);

false

(4)

 


As far as I can see, two expressions, 'Is_eq' and 'Is_eq2' are the same, but Maple seems to think differently. 

I would appreciate if you can offer any clue regarding why Maple thinks in a different way. 

Download Qkitonum_20200626.mw

@Kitonum Perhaps, I put too much emphasis on the aesthetics of expression. 

@acer Freeze/Thaw command set would be useful!

@acer Thank you for the answer. That would be the way to differentiate the variable s from the literal 's'. 

 

@acer : The second part of your worksheet is what I tried and asked for. Thank you!

@Carl Love Your step by step explanation is appreciated. I'm still on a learning curve regarding how to use Maple, thus any detailed explanation like yours becomes a meaningful help. 

@panke: It looks more intuitive!

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