Determine currents Ia, Ib and Ic

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    Currents Ib Ic
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The discussion revolves around determining the currents Ia, Ib, and Ic in a complex electrical network using mesh current analysis. Participants share their calculations and methods, including the use of Ohm's Law and current division principles, while also expressing frustrations with complex arithmetic. Suggestions are made to solve symbolically before substituting values to simplify calculations. The conversation highlights the importance of accuracy in algebra and the potential benefits of using software or calculators for complex calculations. Ultimately, the participants aim to arrive at correct values for the currents while navigating the challenges of the calculations involved.
  • #51
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:

attachment.php?attachmentid=63055&d=1382041960.png


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:

attachment.php?attachmentid=63056&d=1382041960.png


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

attachment.php?attachmentid=63057&d=1382041960.jpg


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

attachment.php?attachmentid=63058&d=1382041960.jpg


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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  • #52
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?
 
  • #53
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:
  • #54
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
 
  • #55
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.
 
  • #56
Ib= 0.057j + 0.0356

Still i need minus inside, i checked a few times and it is plus!
 
  • #57
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.
 
  • #58
but i don't understand why i have to multiply by +j

minus divide minus is plus
and i divide -25 both sides
 
  • #59
I did multiply top and bottom by +J.

Ib= 0.057- 0.0356j
 
  • #60
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.
 
  • #61
I did and it works.
The answer is :

Ib= 0.057- 0.0356j
 
  • #62
agata78 said:
I did and it works.
The answer is :

Ib= 0.057- 0.0356j

Okay! :smile:

So you'll have to analyze the logic of whatever method you used before in order to spot where the sign got lost.
 
  • #63
Thank you! Two more to go!
 

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