Solving rectangular form simultaneous equations?

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SUMMARY

This discussion focuses on solving a system of three simultaneous equations involving complex currents I1, I2, and I3. The equations are expressed in rectangular form and include complex coefficients. Participants clarify that the equations can be simplified by separating real and imaginary parts, allowing for easier manipulation. The discussion also highlights the importance of correctly combining terms and provides foundational rules for handling complex numbers.

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  • Understanding of complex numbers and their operations
  • Familiarity with simultaneous equations
  • Knowledge of electrical circuit theory, specifically current analysis
  • Experience with mathematical software for solving equations
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  • Research online tools for solving complex simultaneous equations
  • Learn about the application of complex numbers in electrical engineering
  • Study methods for separating real and imaginary parts in equations
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Electrical engineers, physics students, and anyone involved in solving complex simultaneous equations in circuit analysis will benefit from this discussion.

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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