Solving rectangular form simultaneous equations?

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The discussion revolves around solving three simultaneous equations for currents I1, I2, and I3, which involve complex coefficients. Users seek online software tools to assist in solving these equations, as manual calculations have proven challenging. The equations can be simplified by separating real and imaginary components, allowing for easier manipulation. Participants confirm that combining terms with complex coefficients is valid, reinforcing the mathematical principles of complex number addition and multiplication. The conversation emphasizes the importance of correctly interpreting and rearranging the equations to facilitate solving.
alex.daciz
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Hi, I have the following 3 simultaneous equations to solve for the currents I1, I2 and I3:

I1(0-j5) + I1(5+j5) - I3(5+j5) - 20 = 0

I2(5+j5) + I2(0-j10) + I3(0-j10) - 20 = 0

I1(0-j5) + I3(10+j0) - I2(5+j5) = 0

Does anyone have the link to any online software I could use to solve the problem? I have tried solving it by hand but seem to get a bit stuck. Can the I1(0-j5) + I1(5+j5) terms be combined to form 2I1(5) = 10I2 ? Likewise I2(5+j5) + I2(0-j10) = 2I2(5-j5) is this possible?
 
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The complex coefficients of I2 are just constants. (5+j5)I2 + (0-j10)I2 = (5-j5)I2

The system of equations with complex values can be split up be equating the real part of the coefficients to the real part of the RHS. Similarly for the complex part of the coefficients.

for example, the first equation from the OP can be written:
0 * I1 +5 * I1 - 5 * I3 = 20

-j5 * I1 + j5 * I1 - j5 * I3 = 0

If the first equation is written correctly (it does not contain an I2 term), then

5* I1 - 5 * I3 = 20
-j5* I3 = 0
 
alex.daciz said:
Hi, I have the following 3 simultaneous equations to solve for the currents I1, I2 and I3:

I1(0-j5) + I1(5+j5) - I3(5+j5) - 20 = 0

I2(5+j5) + I2(0-j10) + I3(0-j10) - 20 = 0

I1(0-j5) + I3(10+j0) - I2(5+j5) = 0

Does anyone have the link to any online software I could use to solve the problem? I have tried solving it by hand but seem to get a bit stuck. Can the I1(0-j5) + I1(5+j5) terms be combined to form 2I1(5) = 10I2 ? Likewise I2(5+j5) + I2(0-j10) = 2I2(5-j5) is this possible?

Sure.

(A + jB) + (C + jD) = (A+C) + j(B+D)

and

(A + jB)(C + jD) = AC - BD + j(AD+BC)
 

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