AET Netzwerk Hausübung -- Sage

Uq1=10; Uq2=1; Iq3=4; R=(1,20,24,50,80,65,35,80,75,99,68); #Element 0 willkürlich auf 1 gesetzt, für korrekte indizes

R10_6 = (1/R[10]+1/R[6])^(-1); print R10_6.n(); Uq3=Iq3*R[3]; print Uq3.n();

23.1067961165049
200.000000000000

RD = matrix([ [R[2]+R10_6+R[8],0,0,0,0], [0,R[9]+R[5]+R10_6+R[8]+R[7],0,0,0], [0,0,R[1]+R[5]+R10_6+R[8]+R[7],0,0], [0,0,0,R[4]+R[5]+R10_6,0], [0,0,0,0,R[3]+R[8]+R[7]] ]); RL = matrix([ [0,(R10_6+R[8]),(R10_6+R[8]),R10_6,R[8]], [0,0,(R[5]+R10_6+R[8]+R[7]),(R[5]+R10_6),(R[8]+R[7])], [0,0,0,(R[5]+R10_6),(R[8]+R[7])], [0,0,0,0,(0)], [0,0,0,0,0] ]); RM = RD+RL+RL.transpose(); print RM.n(digits=5);

[ 122.11  98.107  98.107  23.107  75.000]
[ 98.107  342.11  243.11  88.107  155.00]
[ 98.107  243.11  263.11  88.107  155.00]
[ 23.107  88.107  88.107  168.11 0.00000]
[ 75.000  155.00  155.00 0.00000  205.00]

RM == (RM.inverse()).inverse();

True

U = vector([Uq2,0,-Uq1,-Uq1,-Uq3]); print U;

(1, 0, -10, -10, -200)

I=RM.inverse()*U; print I.n(digits=5);

(0.36957, 0.32351, 1.1014, -0.85707, -2.1881)

I_R2=I[0]; print "I_R2: ", abs((I_R2*10^3).n(digits = 9)),"[mA]"; U_R2=I_R2*R[2]; print "U_R2: ", abs(U_R2.n(digits = 9)),"[V]";

I_R2:  369.571931 [mA]
U_R2:  8.86972634 [V]

I_R9=I[1]; print "I_R9: ", abs((I_R9*10^3).n(digits = 9)),"[mA]"; U_R9=I_R9*R[9]; print "U_R9: ", abs(U_R9.n(digits = 9)),"[V]";

I_R9:  323.505225 [mA]
U_R9:  32.0270173 [V]

I_R4=I[3]; print "I_R4: ", abs((I_R4*10^3).n(digits = 9)),"[mA]"; U_R4=I_R4*R[4]; print "U_R4: ", abs(U_R4.n(digits = 9)),"[V]";

I_R4:  857.069027 [mA]
U_R4:  68.5655221 [V]

#leitwerte Rechnen G = list(); for i in range(0,11): G.append( 1/R[i] );

Iq2 = Uq2/R[2];

GM = matrix([ [G[4]+G[6]+G[10]+G[3]+G[8],-G[6]-G[10],-G[3],-G[8]], [-G[6]-G[10],G[5]+G[6]+G[10]+G[2],0,-G[2]], [-G[3],0,G[1]+G[9]+G[3]+G[7],-G[7]], [-G[8],-G[2],-G[7],G[2]+G[7]+G[8]] ]); GM.n(digits = 4);

[ 0.08911 -0.04328 -0.02000 -0.01333]
[-0.04328   0.1003   0.0000 -0.04167]
[-0.02000   0.0000  0.09260 -0.01250]
[-0.01333 -0.04167 -0.01250  0.06750]

GM == GM.transpose();

True

I2 = vector([-Iq3,Iq2-Uq1*G[5],Iq3-Uq1*G[9],-Iq2]); I2.n(digits=5);

(-4.0000, -0.11218, 3.8990, -0.041667)

U2=GM.inverse()*I2; U2.n(digits=6);

(-68.5655, -46.9062, 22.0270, -39.0364)

U_R2_2 = (U2[1]-U2[3]-Uq2); I_R2_2 = U_R2_2 * G[2];

print "U_R2: ",(abs(U_R2_2).n(digits=6)),"[V]"; print "I_R2: ",(abs(I_R2_2*10^3).n(digits=6)),"[mA]";

U_R2:  8.86973 [V]
I_R2:  369.572 [mA]

print "U_R9: ",(abs((U2[2]+Uq1)).n(digits=6)),"[V]"; print "I_R9: ",(abs((U2[2]+Uq1)*G[9]*10^3).n(digits=6)),"[mA]";

U_R9:  32.0270 [V]
I_R9:  323.505 [mA]

print "U_R4: ",(abs((U2[0])).n(digits=6)),"[V]"; print "I_R4: ",(abs((U2[0])*G[4]*10^3).n(digits=6)),"[mA]";

U_R4:  68.5655 [V]
I_R4:  857.069 [mA]

#Test ob beide Verfahren das selbe Ergebnis für die R2 Werte liefern print "Maschen-Strom-Verfahren:\t", abs(I_R2); print "Knoten-Potential-Verfahren:\t", abs(I_R2_2); print "Differenz: ", (abs(I_R2)-abs(I_R2_2)).n(digits=20); print "Gleicher Wert: ", (abs(I_R2) == abs(I_R2_2));

Maschen-Strom-Verfahren:	4970095949/13448250616
Knoten-Potential-Verfahren:	4970095949/13448250616
Differenz:  0.00000000000000000000
Gleicher Wert:  True

AET Netzwerk Hausübung

146 days ago by MrWhite