Determine currents Ia, Ib and Ic

[agata78]

Homework Statement

Determine currents Ia, Ib & Ic in the network?

Homework Equations

FIRST EMF

RT = R1 + (R2 x R3) / (R2 + R3)

RT = 47 + (j100 x ( -j75)) / (j100 + (-j75))

RT = 47+ 300j (ohms)

Ohms Law then states that:

I1 = E1 / RT

I1 = 12∠0º / 83-84.48j

I1 = (996+1013.76j) / (14025.87)

Using the Current Division Laws leaves:

I2 = I1 x ( R3 ) / (R2 + R3)

I2 = ( ((996+ 1013.76j) / 14025.87) x ( -j75 )) / (j100 + -j75)

I2 = ( (2988-3041.28j) / 14025.87) ∠0º I3 = I1 x ( R2 ) / (R2 + R3)

I3 = I1 x ( j100 ) / (j100 + -j75)

I3 = ((3984+4055.04j)/ 14025) ∠0º



SECOND EMF

RT = R3 + (R1 x R2) / (R1 + R2)

RT = - j75 + ( 47 x j100 ) / ( 47 + j100 )

RT = ? Ω


Ohms Law then states that:

I4 = E2 / RT

I4 = 10∠0º / ?

I4 = ?

Using the Current Division Laws leaves:

I5 = I4 x ( R2 ) / (R1 + R2)

I5 = ? x ( j100 ) / ( 47 + j100 )

I5 = ? x ?

I5 = ? Ω
I6 = I4 - I5

I6 = ? _ ?

I6 = ? Ω


So:

IA = I1 - I5

IA = ? - ?

IA = ?∠ ?º Ω
IB = I3 - I4

IB = ? - ?

IB = ?∠ ?º Ω
IC = I2 + I6

IC = ? + ?

IC = ?∠ ?º Ω

Can someone please let me know if the information provided is answering the question correctly? Thanks!

In the diagram R3 is j75. However, it is suppose to read -j75.

 

[gneill]

Wow, looks like a lot of work taking the superposition approach. Why not write a pair of mesh equations? Two of the currents will pop out as the mesh currents, and the third is just the sum of the other two.

 

[agata78]

I spent all day calculating all theses, and everytime i have a different figure and they don't look good. I have a simillar example but with much easier numbers ( no j).
A pair of mesh equations?

 

[agata78]

Ok i will have a go!-12 +100j (I1 + I2) + 47I1= 0 and -100j (I1 + I2) +10- (-75j)I2 = 0

Am i right?

 

[agata78]

The answer for I2 = 0.4- 4jI1 and for I1= 147jI1+400I1=12-40j.

Which looks rubish and very not correct!
I need to finish this asap my time is running out!

 

[gneill]

I'd choose my mesh currents to correspond to the currents Ia and Ib . Then "walk" both loops in the directions of the mesh currents and sum the potential changes:

Loop1:
+12 - Ia47 - (Ia + Ib)j100 = 0

Loop2:
+10 - Ib(-j75) - (Ia + Ib)j100 = 0

arriving at what appears to be the same equations that you found. When I solve for Ia and Ib I get reasonable values.

If you think carrying through all the complex manipulations while you solve the equations might be tripping you up, try leaving the component values as symbols and solve the equations symbolically. Plug in complex values in the final expressions for Ia and Ib and then reduce.

 

[agata78]

12- Ia47 - (Iaj100+Ibj100) =0
12- Ia47- Iaj100- Ibj100=0
Ia47 - Iaj100= Ibj100-12
Ia(47-j100)=Ibj100-12

Ia= (Ibj100-12) / (47-j100)

Is it right so far?

 

[gneill]

12- Ia47 - (Iaj100+Ibj100) =0
12- Ia47- Iaj100- Ibj100=0
Ia47 - Iaj100= Ibj100-12 <--- you lost the sign on the Ia47
Ia(47-j100)=Ibj100-12

Ia= (Ibj100-12) / (47-j100)

Is it right so far?

Watch your algebra!

 

[agata78]

And for loop 2:

10-Ib (-75j)-(Ia+Ib)j100=0
10+Ibj75-Ia100j-Ibj100=0
Ibj75-Ibj100-Ia100j=-10
Ibj(-25)- Ia100j=-10
Ibj(-25)-100j((Ibj100-12)/ (-47-j100)) =-10

Ibj(-25) -((10000Ibj2 - 1200j)/ (-47-j100))= -10

Ibj(-25) - ( 10000(-1) Ib -1200j ) / (-47-j100) =-10

Ibj(-25) +(( 10000 Ib + 1200j) / (-47-j100)) = -10

And what next?

 

[agata78]

ooops, thank you!

 

[agata78]

235bj +1500b= (-11280j + 24000-2 ) / (47-100j)

b= ( (-11280j + 24000-2 ) / (47-100j) ) / (235j +1500).

Its after midnight will jump to bed and do some checking on my algebra.
thank you

 

[gneill]

And for loop 2:

10-Ib (-75j)-(Ia+Ib)j100=0
10+Ibj75-Ia100j-Ibj100=0
Ibj75-Ibj100-Ia100j=-10
Ibj(-25)- Ia100j=-10
Ibj(-25)-100j((Ibj100-12)/ (-47-j100)) =-10

Ibj(-25) -((10000Ibj2 - 1200j)/ (-47-j100))= -10

Ibj(-25) - ( 10000(-1) Ib -1200j ) / (-47-j100) =-10

Ibj(-25) +(( 10000 Ib + 1200j) / (-47-j100)) = -10

And what next?

Continue banging away until you've solved for Ib. Then use that Ib to go back to a previous expression for Ia and solve for Ia.

It's a lot of work carrying out all these complex operations for every manipulation of the equations. I would suggest solving symbolically first, then plug in the values for the impedances in the final expressions. Then it's just two expressions to evaluate with complex numbers, rather than a whole series of complex expansions and shuffles over and over.

I'll think about whether there's another approach that would be less time consuming.

 

[gneill]

Okay, I'm thinking that a nodal analysis approach might be a bit more streamlined. If you take the top middle node and call its potential Vx, and if you leave the complex impedances as R2 and R3, then the node equation becomes:

$\frac{Vx - 12}{47} + \frac{Vx}{R2} + \frac{Vx - 10}{R3} = 0~~~~~$ whence:

$Vx = \frac{R2(12 R3 + 470)}{R2 R3 + 47 (R2 + R3)}$

Plug in values for R2 and R3 and solve for Vx (the only really messy expression to deal with), then:

$Ia = \frac{12 - Vx}{47}$

$Ib = \frac{10 - Vx}{R3}$

$Ic = Ia + Ib$

 

[The Electrician]

You are torturing yourself!

Once you have learned how to do calculations with complex arithmetic, and have practiced for a while, it's time to move on to modern methods using a computer or calculator. As you can see, it's easy to make mistakes when doing such an involved series of complex calculations by hand.

Your school should have Matlab or Mathcad or something similar on the school network, which can do these sorts of calculations.

Also, you can find applications for free on the web that can do complex arithmetic. You can even find complex matrix solvers.

If you plan to continue your studies in a field that requires these sorts of calculations, you might consider buying a handheld calculator that can do complex arithmetic.

Hewlett-Packard has recently introduced a new calculator, the HP Prime, and this introduction has led to a discounting in price for the older calculators. The older HP50G can do complex arithmetic, including complex matrix calculations, and can be had for under $100. This is a very powerful calculator and you would be able to use it for your entire academic career.


Edit: Here's an online calculator for complex numbers using the RPN format of HP calculators:

http://www.alpertron.com.ar/CALC.HTM

 

[agata78]

Ok,

Vx= 12.67 + 4.28j

 

[agata78]

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!

 

[agata78]

Ia = -0.014-0.09j
Ib=( -2.67-4.28j) / -75j

I didnt calculate Ic yet cause a amd be doesn't look very good!
Can you check it out for me?

 

[gneill]

Ok,

Vx= (1644000j - 257560) / 92209

Doesn't look right. Calculate the numerator and denominator of the expression separately and show your values. Maybe we can pin down where the slip occurred.

 

[agata78]

Vx= 12.67 + 4.28j
Ia = -0.014-0.09j
Ib=( -2.67-4.28j) / -75j

I found a mistake, i changed it. Is it better now? if not i will post my calculations :-(

 

[gneill]

Vx= 12.67 + 4.28j
Ia = -0.014-0.09j
Ib=( -2.67-4.28j) / -75j

I found a mistake, i changed it. Is it better now? if not i will post my calculations :-(

The numbers look good. Might as well express Ib in decimal form too.

 

[agata78]

ok,

100j(12 (-75) +470 ) / (100j(-75j) +47 (100j+(-75j))
100j(-900j +470) / -7500j2 +47 (100j-75j)
-90000j2 + 47000j / 7500 +(47 x 25j)
90000+47000j / 7500 + 1175j
(90000+47000j)(7500-1175j) / (7500+1175j ) (7500-1175j)
(675000000-105750000j+352500000j-55225000j2) / ( 56250000-1380625j2)
(675000000+55225000-246750000j ) / 56250000+1380625
730225000+246750000j / 57630625
12.67+ 4.28j

 

[agata78]

Ib= -0.17-0.237j
Ic= Ia+ IB
Ic= 4.3156j+ 12.727

 

[agata78]

Is there a way to check the answer is correct?

 

[gneill]

Ib= -0.17-0.237j
Ic= Ia+ IB
Ic= 4.3156j+ 12.727

Something went wrong when you finished off Ib. Check the math. Try out the online calculator!

 

[agata78]

I checked wolframalpha. it says:

0.570667+ (0.0356 / j)

Its even worse!

 

[agata78]

I don't really know how to use online calculators. when i was at school we couldn't use them at all!

 

[gneill]

You had:

Ib=( -2.67-4.28j) / -75j

Now,

(-2.67)(75j) = -200.25j
(-4.28j)(75j) = 321
(-75j)(75j) = 5625

So,

321/5626 = 0.057
-200.25j/5625 = -0.036j

and the result is then 0.057 - 0.036j

 

[agata78]

Ok i can see to get rid off -75j you time 75j each side!

Thank you.

It was much easier way to calculate then my first one. specially i spent long hours to draw circuits.

Thanks again!

 

[gneill]

I don't really know how to use online calculators. when i was at school we couldn't use them at all!

I went through all the stages myself:
1. Tables provided only (trig and log tables)
2. Slide rules okay, no calculators
3. Hand calculators okay, no computers

I can do the math by hand, having drilled on it for so long, but my production would drop dramatically. Now I use Mathcad for just about everything and the calculator stays in its case.

I suggest that you find an app that you can live with and get comfortable with it. Using humans as typo/algebra checkers is not practical.

 

[agata78]

My course work says use mesh current analysis to determine currents. Its a right answer but maybe my tutor won't be hapy for using different analysis

 

[agata78]

Im sorry but i wasted your time and mine too. I didnt mention in task that i need to use mesh -current analysis. I need to come back to previous idea you gave me!

 

[gneill]

Im sorry but i wasted your time and mine too. I didnt mention in task that i need to use mesh -current analysis. I need to come back to previous idea you gave me!

No worries. Solve symbolically for the currents, then hammer away at the complex math. Watch for shortcuts (compare the denominators of the symbolic expressions for Ia and Ib).

 

[agata78]

I will try to come back to this then and see where i go:

Loop1:
+12 - Ia47 - (Ia + Ib)j100 = 0

Loop2:
+10 - Ib(-j75) - (Ia + Ib)j100 = 0

 

[agata78]

12- Ia47 - (Iaj100+Ibj100) =0
12- Ia47- Iaj100- Ibj100=0
-Ia47 - Iaj100= Ibj100-12 x(-1)
Ia(47+j100)=-Ibj100+12

Ia= (12- Ibj100) / (47+j100)


do i have to calculate 2 loop after when i found Ib ?? and to calculate Ib do i have to calculate it from Loop1 or loop2?
When i was in Secondary School in Poland, we used 2 equations together:

ex, a+ 2bx=2
3a + bx = 2 / x(-2)

and then first equation stay the same
but second

a+ 2bx=2
-6a-2bx = -4
and then adding them together!

-5a = -2
a= 2/5
and then coming back to one of two equations to calculate b!


Can i use this way to calculate a and b?

 

[agata78]

Sorry for my bad english, still learning!

 

[agata78]

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

and then adding them together

2-Ia47 - Ib75j =0

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

 

[gneill]

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.

 

[agata78]

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!

 

[agata78]

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?

 

[gneill]

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.

 

[agata78]

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

Which one would be easier one?

 

[gneill]

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.

 

[agata78]

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

I will take this path!

 

[The Electrician]

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.

 

[agata78]

Ia= (1316 -8400j ) / 92209

and now i can use any loop to find Ib

Yes?

 

[The Electrician]

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.

Ia= (-1316 -8400j ) / 92209

and now i can use any loop to find Ib

Yes?

Sign error propagated to here.

 

[gneill]

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.

 

[agata78]

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)

 

[gneill]

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.

 

[agata78]

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