# Determine currents Ia, Ib and Ic

• agata78
In summary: IA = - 12 + 47jIa + 10-75jIb IB = - 10 + 47jIb+ 400jIa In summary, the homework statement is to determine currents Ia, Ib, and Ic in the network. Homework equations state that: first EMF, RT, = 47 + (j100 x ( -j75)) / (j100 + (-j75)) and RT = 47+300j (ohms), ohms law then states that: I1 = E1 / RT and I1 = (996+1013.76j) / (14025.87), second EMF, RT, =
I thought i could use it:

+12 - Ia47 - (Ia + Ib)j100 = 0
+10 - Ib(-j75) - (Ia + Ib)j100 = 0 / (-1)
+12 - Ia47 - (Ia + Ib)j100 = 0
-10+ Ib(-75j) + (Ia+ Ib) j100 =0

2-Ia47 - Ib75j =0

Ia= (Ib75j+2 ) / 47what do you think?

You can do that, but when you add the equations one of the variables should be eliminated. You'll have to scale one of the equations so that the coefficients on one of the variables match accordingly.

What you mean by:

You'll have to scale one of the equations so that the coefficients on one of the variables match accordingly.

I tried to do that but my figures are not great. Maybe its not the best idea!

12-Ia47 -(Ia+Ib)j100=0
10-Ib(-75j) - (Ia+Ib) j100=0

12-Ia47-Ia100j - Ibj100=0
-Ia47-Ia100j = -12+ Ibj100
Ia47 + Ia100j =12- Ibj100
Ia(47+ 100j) = 12- Ibj100

Ia= (12- Ibj100) / (47+100j)
do you agree with me?

i went further more:
((12 - Ibj100) (47-100j)) / ((47+100j) (47-100j))
(564-1200j-4700jIB +Ibj2 (10000) ) / ((2209-J4700+ j4700- J2 (10000) )
(564-1200j-4700jIb-Ib10000 ) / ( 2209+10000)
564-1200j- Ib(4700j-10000) / 12209

and i don't know what to do next?
Should i first calculate a and b from first loop and check it out with second one?

agata78 said:
12-Ia47 -(Ia+Ib)j100=0
10-Ib(-75j) - (Ia+Ib) j100=0

12-Ia47-Ia100j - Ibj100=0
-Ia47-Ia100j = -12+ Ibj100
Ia47 + Ia100j =12- Ibj100
Ia(47+ 100j) = 12- Ibj100

Ia= (12- Ibj100) / (47+100j)
do you agree with me?
Looks okay!

i went further more:
((12 - Ibj100) (47-100j)) / ((47+100j) (47-100j))
(564-1200j-4700jIB +Ibj2 (10000) ) / ((2209-J4700+ j4700- J2 (10000) )
(564-1200j-4700jIb-Ib10000 ) / ( 2209+10000)
564-1200j- Ib(4700j-10000) / 12209
Probably not worth doing anything else with the first loop equation once you've isolated one of the variables from it. Move on to the second loop equation.

and i don't know what to do next?
Should i first calculate a and b from first loop and check it out with second one?

Isolate one variable from the first equation (you isolated Ia above). Substitute that expression for Ia into the second equation. That'll leave you with an equation in one unknown, Ib. Solve for Ib.

Can i just use for a:
Ia= (12- Ibj100) / (47+100j) or 564-1200j- Ib(4700j-10000) / 12209

Which one would be easier one?

I think that they'll turn out to be equally difficult once substituted into the second equation.

The method you proposed of adding (or subtracting) equations to eliminate one variable has merit in this regard. It accomplishes the task of obtaining one equation in one unknown for one of the variables without having to go through substitution and expansion of complex values.

For example, if you collect the terms of the two loop equations on the unknown variables:

(1) 12 + (-47 - 100j)Ia - 100jIb = 0
(2) 10 - 100j Ia - 25jIb = 0

If you multiply the second equation by -4 then you can add them and eliminate Ib, yielding an expression involving only Ia:

(1) 12 + (-47 - 100j)Ia - 100jIb = 0
(2) -40 + 400j Ia + 100jIb = 0
+ -------------------------------------------

-28 + (47 + 300j)Ia = 0

This is a much cleaner path to finding one of the currents.

Yes it looks actually quite nice and clear ( if you know what i mean)

I will take this path!

agata78 said:
Thanks, i checked it out, looks ok but quite complicated for me. but for sure i will try to purchase software to help me with calculation asap!

Scilab is free and is nearly as powerful as the more common Matlab and Mathcad:

http://www.scilab.org/

It will take you a while to learn to use it, but that effort will be worthwhile.

1 person
Ia= (1316 -8400j ) / 92209

and now i can use any loop to find Ib

Yes?

gneill said:
If you multiply the second equation by -4 then you can add them and eliminate Ib, yielding an expression involving only Ia:

(1) 12 + (-47 - 100j)Ia - 100jIb = 0
(2) -40 + 400j Ia + 100jIb = 0
+ -------------------------------------------

-28 + (-47 + 300j)Ia = 0

This is a much cleaner path to finding one of the currents.

There's a sign error here.

agata78 said:
Ia= (-1316 -8400j ) / 92209

and now i can use any loop to find Ib

Yes?

Sign error propagated to here.

D'oh! Yup, The Electrician has spotted a sign error that I made above. I dropped the "-" on the "47" when I transcribed it for the sum. Thank you T.E. This is why I like math software!

@agata78: Yes, after repairing the sign, use that value for Ia in one of the loop equations to find Ib.

Ia= 0.0142-0.009j
Ib= -0.057+ 0.0356j
i think its right, what do you think?

Its the same answer as previous one. But previous one had Ia= -0.014-0.09j and Ib= 0.057 + 0.0356j( not sure about that minus at front on Ia and plus on front of Ib)

Last edited:
Did you account for the sign error that The Electrician spotted? Both terms of Ia should turn out to be negative, and 8400/92209 is 0.0911.

I'm seeing Ib as +0.0571 - 0.0356j.

I changed every where for minus and now its correct!

I put the value for Ia into this equation
10-Ibj25-Iaj100=0
I think it doesn't have to be the first 2 loops!
?

anyway thank you for help.
I have two more questins to answer until tomorrow!

at the end it wasnt difficult but it took me almost two days to do that

Just to show how easy it can be.

Solving circuits usually ends up being a problem of solving a system of simultaneous equations. There is a way to write down a system of simultaneous equations that shows them in a very compact form. That way is to use matrix notation. Your two mesh equations can be shown in matrix form like this:

Solving simultaneous equations involves elimination of variables until a single equation in one unknown is obtained. That equation is solved for that unknown, and the result is then substituted in the remaining equations. Then another variable is solved for, and the process is continued until you have all the unknowns solved.

Once in this form, a linear solver can be used, which does all the elimination and back substitution for you, avoiding much mistake prone, tedious, algebra:

Here's how it looks on the HP50G calculator in matrix form:

And, after pressing one button, here's the solution:

It would be worth your while to learn how to do this. Don't forget the free software Scilab, but a handheld calculator is very convenient.

agata78 said:
at the end it wasnt difficult but it took me almost two days to do that

You've had enough practice doing it by hand. Learn to use a mathematical software or calculator!

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Yeap its correct, i got the same!
Ia= -0.0142-0.009jbut Ib

10-Ibj25 - Ia100
10- Ibj25- ((-1316-8400j)/ 92209 ) x j100
10- Ibj25- (- 131600j- 840000j2) / 92209
-Ibj25= (-131600j + 840000 / 92209 ) -10
-Ibj25= (-131600j+840000-922090)/ 92209
-Ibj25= (-131600j- 82090) / 92209
-Ibj25= -1.427j - 0.8902
Ib= (-1.427j - 0.8902) / (-25)

Ib= 0.057j + 0.0356j

Can you check my calculations please?

agata78 said:
Yeap its correct, i got the same!
Ia= -0.0142-0.0911jbut Ib

10-Ibj25 - Ia100
10- Ibj25- ((-1316-8400j)/ 92209 ) x j100
10- Ibj25- (- 131600j- 840000j2) / 92209
-Ibj25= (-131600j + 840000 / 92209 ) -10
-Ibj25= (-131600j+840000-922090)/ 92209
-Ibj25= (-131600j- 82090) / 92209
-Ibj25= -1.427j - 0.8902
Ib= (-1.427j - 0.8902) / (-25j)

Ib= 0.057j + 0.0356j

Can you check my calculations please?

You left out a "j" in your last step.

Don't forget the numerical error involving -.0911 that gneill already mentioned.

See how easy it is to make a mistake? Just imagine trying to do this during an exam without making a single mistake!

Last edited:
Ib supposed to be= +0.0571 - 0.0356j.

mine is Ib= 0.057j + 0.0356j
Can you check my calculations:

10-Ibj25 - Ia100
10- Ibj25- ((-1316-8400j)/ 92209 ) x j100
10- Ibj25- (- 131600j- 840000j2) / 92209
-Ibj25= (-131600j + 840000 / 92209 ) -10
-Ibj25= (-131600j+840000-922090)/ 92209
-Ibj25= (-131600j- 82090) / 92209
-Ibj25= -1.427j - 0.8902
Ib= (-1.427j - 0.8902) / (-25j)

Ib= 0.057j + 0.0356j

agata78 said:
Ib= (-1.427j - 0.8902) / (-25j)

Ib= 0.057j + 0.0356j

You've made some kind of error in this last step. The sign in front of the 0.0356 is in fact minus if you do the complex arithmetic correctly.

This kind of error won't happen if you use a calculator that can do complex arithmetic.

Ib= 0.057j + 0.0356

Still i need minus inside, i checked a few times and it is plus!

agata78 said:
Ib supposed to be= +0.0571 - 0.0356j.

mine is Ib= 0.057j + 0.0356j
Can you check my calculations:

10-Ibj25 - Ia100
10- Ibj25- ((-1316-8400j)/ 92209 ) x j100
10- Ibj25- (- 131600j- 840000j2) / 92209
-Ibj25= (-131600j + 840000 / 92209 ) -10
-Ibj25= (-131600j+840000-922090)/ 92209
-Ibj25= (-131600j- 82090) / 92209
-Ibj25= -1.427j - 0.8902
Ib= (-1.427j - 0.8902) / (-25j) <---- at this point, multiply top and bottom by +j. What do you get?

Ib= 0.057j + 0.0356j

You've got a sign issue occurring in whatever steps you take between the line indicated above and your final line.

but i don't understand why i have to multiply by +j

minus divide minus is plus
and i divide -25 both sides

I did multiply top and bottom by +J.

Ib= 0.057- 0.0356j

agata78 said:
but i don't understand why i have to multiply by +j

minus divide minus is plus
and i divide -25 both sides

You want to clear the j from the bottom. So multiply top and bottom by j. Try it.

I did and it works.

Ib= 0.057- 0.0356j

agata78 said:
I did and it works.

Ib= 0.057- 0.0356j

Okay!

So you'll have to analyze the logic of whatever method you used before in order to spot where the sign got lost.

Thank you! Two more to go!

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