How do you work out simultaneous eqns w/ complex numbers & phasor

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To solve simultaneous equations involving complex numbers and phasors, it's recommended to convert phasor forms into complex impedance using a+jb notation. For example, 20∠0 becomes 20+0j, and 20∠π/2 becomes 0+20j. Rearranging the equations to group coefficients of I1 and I2 simplifies the process, allowing for standard methods of solving with complex arithmetic. If a calculator cannot handle complex numbers, consider using software or a different model that supports complex arithmetic, like the TI-86. Understanding these methods will aid in preparing for exams effectively.
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I'm having trouble figuring out to get the answers from the 2 equations. The phasors and complex numbers confuse me. Do I need to change the phasor form? How do I go about doing this thanks! (Not homework question I am trying to figure this for my exam!)
 

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There are lots of ways to do this sort of problem ...
For complicated meshes, you may find it easier to convert to complex impedences and use a+jb type notation.
So 20∠0 would be just 20+0j, 20∠π/2 would be 0+20j (j=√-1). Then you can solve the simultaneous equations normally.

For simple problems you can usually just add the phasors like vectors.
 
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Rearrange your equations so that coefficients of I1 and I2 are together like this:

(5-j2)*I1 + (-3)*I2 = 30
(-5)*I1 + (3+j8)*I2 = 0

Now you can use a standard method to solve them but with complex arithmetic: http://www.mathportal.org/calculators/system-of-equations-solver/system-2x2.php

Will your exam be a take home? If so, you can find solvers on the internet to do the job; search for "simultaneous complex linear equations calculator". If you will be solving systems like this at school, you should get an app for your laptop or get a calculator that can solve simultaneous complex equations.
 
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The Electrician said:
Rearrange your equations so that coefficients of I1 and I2 are together like this:

(5-j2)*I1 + (-3)*I2 = 30
(-5)*I1 + (3+j8)*I2 = 0

Now you can use a standard method to solve them but with complex arithmetic: http://www.mathportal.org/calculators/system-of-equations-solver/system-2x2.php

Will your exam be a take home? If so, you can find solvers on the internet to do the job; search for "simultaneous complex linear equations calculator". If you will be solving systems like this at school, you should get an app for your laptop or get a calculator that can solve simultaneous complex equations.
No unfortunately it isn't a take home exam, I have a Texas instrument Graphic Calculator but I don't think it can solve complex simultaneous equations.

But thanks for the great help!
 
Simon Bridge said:
There are lots of ways to do this sort of problem ...
For complicated meshes, you may find it easier to convert to complex impedences and use a+jb type notation.
So 20∠0 would be just 20+0j, 20∠π/2 would be 0+20j (j=√-1). Then you can solve the simultaneous equations normally.

For simple problems you can usually just add the phasors like vectors.
I will try it again using your method. Thanks for the help!
 
whatphysics said:
No unfortunately it isn't a take home exam, I have a Texas instrument Graphic Calculator but I don't think it can solve complex simultaneous equations.

But thanks for the great help!

What model is your TI calculator?
 
In "the good old days" FORTRAN implicitly knew about complex numbers.

So if you can make a program that can solve simultaneous equations with real numbers, you just declare these real numbers as complex numbers, and FORTRAN will do the job. All calculations ( addition, division, etc. ) will be done with complex numbers.
 
The Electrician said:
What model is your TI calculator?

TI-84 Plus Pocket SE
 
It might be worth your while to get a calculator that can do complex arithmetic. The TI-86 can do complex arithmetic, including matrices with complex arithmetic.

I see them for sale on eBay from $20 to $35.
 
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I, usually, use the excel complex system.
However, at first you have to rearrange the equation as The Electrician said.
a*I1+b*I2=V1 where a=5-j2; b=-3+j0 ;c=-5+j0; d=3+j8; V1=30+j0; V2=0
c*I1+d*I2=V2
1) define a,b,c,d,V1,V2
2) put a=COMPLEX(Re,Im) where Re=5 and Im=-2 in your case.
Do the same with b,c, d,V1 and V2.
3) define DetGen and put IMSUB(IMPRODUCT(a,d),IMPRODUCT(b,c))
4) define DetI1 and set IMSUB(IMPRODUCT(V1,d),IMPRODUCT(V2,b))
5) define DetI2 and put IMSUB(IMPRODUCT(V2,a),IMPRODUCT(V1,c))
6) I1=IMDIV(DetI1,DetGen); I2=IMDIV(DetI2,DetGen)
 
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