# Solving for 2 unknowns, 2 equations in rectangular form

## Homework Statement

Find I(a) and I(b) for the equaions below:
90 = -j5(I(a)) + j25(I(b))
60 = (20+j5)(I(a)) - j5(I(b))

## Homework Equations

Equations in rectangular form

## The Attempt at a Solution

I've tried solving this for one variable and then plugging the answer into the second equation, but it gets extremely messy. Is there an easier way to solve for the 2 unknowns? Would it help to convert into polar form?

## The Attempt at a Solution

tiny-tim
Homework Helper
hi stau40! Find I(a) and I(b) for the equaions below:
90 = -j5(I(a)) + j25(I(b))
60 = (20+j5)(I(a)) - j5(I(b))

I've tried solving this for one variable and then plugging the answer into the second equation, but it gets extremely messy.

it shouldn't … show us what you've done I tried to solve for I(b) first and ended up with the equation:
I(b) = (60-(20+j5)(I(a))/-j5)
and then plugged that into the first equation and found:
90 = (60-(20-j1)I(a))(j25) - j5(I(a))
At this point it seemed like I was doing something wrong because of the complexity so that's when I started to double check, but didn't find anything.

tiny-tim
Homework Helper
I tried to solve for I(b) first and ended up with the equation:
I(b) = (60-(20+j5)(I(a))/-j5)
and then plugged that into the first equation and found:
90 = (60-(20-j1)I(a))(j25) - j5(I(a))

hmm … dunno how you got that complicated bit start again, and just add 5 times the second equation to the first 90 = -j5(I(a)) + j25(I(b))
60 = (20+j5)(I(a)) - j5(I(b))

… what do you get? I have followed your instructions and end up with:
I(a) = (390/(100+j25) = (390/(103.08angle(14.04)) = 3.78angle(14.04)

I then plug this number into the first equation and get:
90 = -j5(3.78angle(14.04)) + j25(I(b))

I'm trying to convert -j5 into an angle to multiply, but arctan of -5/0 obviously can't be calculated. Am I doing something wrong?

tiny-tim
Homework Helper
I have followed your instructions and end up with:
I(a) = (390/(100+j25)

(shouldn't that be j20 ?)

stop converting to polar! just use 1/(A + jB) = (A - jB)/(A - jB)(A + jB) = (A - jB)/(A2 + B2) Using that formula I end up with I(a) = .01 + j20

I then plug I(a) into the second equation and get:
90 = .01-j100 + j25(I(b))

Do I need to convert to polar form now in order to solve for I(b)?

tiny-tim
Homework Helper
Using that formula I end up with I(a) = .01 + j20

how?? Do I need to convert to polar form now in order to solve for I(b)?

what is it with you and polar form? no!!

I get I(a) by:
(100-j20)/(100^2 + 20^2) = .0096-j20

I keep switching to polar form because I thought it was easier to multipy/divide when the numbers were in polar form rather then rectangular form?

tiny-tim
Homework Helper
I get I(a) by:
(100-j20)/(100^2 + 20^2) = .0096-j20

hmm … the .0096 looks right …

but how did the j20 manage to sail through all that unscathed? I keep switching to polar form because I thought it was easier to multipy/divide when the numbers were in polar form rather then rectangular form?

yes it is sometimes

but i think you need to get used to doing it the rectangular way first, without mistakes (btw, i'd do 390/(100 + j20) by saying it's 390/20(5+j) = (5-j)390/20(25+1) )