# 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, =
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.

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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.

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?

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

Am i right?

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!

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.

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?

agata78 said:
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?

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?

ooops, thank you!

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

agata78 said:
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.

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##

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

Last edited:
Ok,

Vx= 12.67 + 4.28j

Last edited:
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!

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?

Last edited:
agata78 said:
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.

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 :-(

agata78 said:
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.

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

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

Is there a way to check the answer is correct?

agata78 said:
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!

I checked wolframalpha. it says:

0.570667+ (0.0356 / j)

Its even worse!

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

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

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!

agata78 said:
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.

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

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!

agata78 said:
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).

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

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

-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?

Sorry for my bad english, still learning!

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