Arbitrary current with mesh analysis

AI Thread Summary
The discussion revolves around mesh analysis in a circuit, focusing on the relationship between various mesh currents, particularly ib, i1, and i3. Participants clarify equations and correct errors in the calculations, emphasizing the importance of accurately representing mesh currents and their relationships. There is a debate about the correct formulation of the equations, particularly regarding the signs and substitutions used in the calculations. Ultimately, the accuracy of the results is validated through simulation, confirming that both approaches yield consistent outcomes. The conversation highlights the complexities of mesh analysis and the necessity for precision in electrical engineering calculations.
garr6120
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Given the circuit in the diagram I was wondering would I be able to make ib equal to the mesh i3.

Screen Shot 2017-09-27 at 10.54.12 PM.png


I have the equation for my Super Mesh at i1 and i3 which is:

-25+30i3+5Va+20(i3-i2)+10(i1-i2)=0 (1).

The mesh i2 is equal to -2A (2).

subbing (2) into (1) gives:
-25+50i3+5Va-30i2+10i1=0 (3).

Va=i1-i2 (4).

subbing (4) into (3) gives: -25+50i3-35i2+15i1=0 (5).

subbing (2) into (5).
45+50i3+15i1=0 (6).

3ib=i1-i3 (7).

Now I almost have one unknown. Is it possible to set ib equal to i3 so i can isolate for i1 in equation 7. If i can't how would i be able to solve it.

Also, are the steps to finding the mesh currents correct for this problem?
 
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ib is the current through the 20Ω (from right to left). I can see that resistor's current is composed of 2 mesh currents, I3 flowing R to L, and I2 flowing L to R.

Your eqn (4) is missing a 10, it's a 10Ω resistor.

The current through the 10Ω can also be equated to the sum of the currents into the node above it.
 
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NascentOxygen said:
ib is the current through the 20Ω (from right to left). I can see that resistor's current is composed of 2 mesh currents, I3 flowing R to L, and I2 flowing L to R.

Your eqn (4) is missing a 10, it's a 10Ω resistor.

The current through the 10Ω can also be equated to the sum of the currents into the node above it.
What do you mean the current of the 10Ω resistor? Do you mean mesh current 3?
 
garr6120 said:
What do you mean the current of the 10Ω resistor? Do you mean mesh current 3?
I am speaking of the two currents 3ib and ib.
 
What it is still missing is: i1=3*ib-i3 and since ib=i3+2 and ua=10*(i1-i2) you may reduce the first equation so that only i3 stays as unknown.
 
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Babadag said:
What it is still missing is: i1=3*ib-i3 and since ib=i3+2 and ua=10*(i1-i2) you may reduce the first equation so that only i3 stays as unknown.
After doing all of the calculations is an answer for mesh current 1 -17/27A correct?
 
garr6120 said:
After doing all of the calculations is an answer for mesh current 1 -17/27A correct?
Nope.
 
garr6120 said:
After doing all of the calculations is an answer for mesh current 1 -17/27A correct?

Show all your equations and we can help you find the error.
 
The Electrician said:
Show all your equations and we can help you find the error.

I have the equation for my Super Mesh at i1 and i3 which is:

-25+30i3+5Va+20(i3-i2)+10(i1-i2)=0 (1).

The mesh i2 is equal to -2A (2).

subbing (2) into (1) gives:
35+50i3+5Va+10i1=0 (3).

Va=10(i1+2) (4).
Va=10i1+20 (4)

subbing (4) into (3) gives: 135+50i3+60i1=0 (5).

3ib=i1-i3 (6).

ib=i3-i2 (7)

subbing (7) into (6)
3i3-3i2=i1-i3
i1=4i3-3i2
i1=4i3+6 (8)

subbing (8) into (5)
135+50i3+240i3+360
290i3=-495
i3=-99/58

I was wondering if i went wrong anywhere in my steps?
 
Last edited:
  • #10
garr6120 said:
I have the equation for my Super Mesh at i1 and i3 which is:

-25+30i3+5Va+20(i3-i2)+10(i1-i2)=0 (1).

The mesh i2 is equal to -2A (2).

subbing (2) into (1) gives:
35+50i3+5Va+10i1=0 (3).

Va=10(i1+2) (4).
Va=10i1+20 (4)

subbing (4) into (3) gives: 135+50i3+60i1=0 (5).

3ib=i1-i3 (6).

ib=i3-i2 (7)

subbing (7) into (6)
3i3-3i2=i1-i3
i1=4i3-3i2
i1=4i3+6 (8)

subbing (8) into (5)
135+50i3+240i3+360
290i3=-495
i3=-99/58

I was wondering if i went wrong anywhere in my steps?

You made a sign error with respect to i2 in the red part above. Fix this and recalculate.
 
  • #11
The Electrician said:
You made a sign error with respect to i2 in the red part above. Fix this and recalculate.

Well 20(-1)(-2)=40 and 10(-1)(-2)=20

So -25+40+20=35

where is the sign error
 
  • #12
garr6120 said:
Well 20(-1)(-2)=40 and 10(-1)(-2)=20

So -25+40+20=35

where is the sign error

Never mind. I had a situation of poor visibility on my laptop, and the 35 looked like 85.

You're all good. i3 does indeed equal -99/58
 
  • #13
garr6120 said:
I was wondering if i went wrong anywhere in my steps?
The customary check is to substitute your provisional 3 answers back into the 3 equations. If all equations are satisfied, then conclude you have the correct solution to those equations. That works out faster than waiting for someone to look over your calculations...and running the risk of having them miss something anyway. :oldfrown:
 
  • #14
NascentOxygen said:
running the risk of having them miss something anyway. :oldfrown:

o:)
 
  • #15
Your above eq.no.6 it is incorrect.
It has to be i1=3ib-i3.
Then i3= -2.9117647 A.
 
  • #16
Babadag said:
Your above eq.no.6 it is incorrect.
It has to be i1=3ib-i3.
Then i3= -2.9117647 A.
OP's eqn 6 is correct. You were wrong in post #5
 
  • #17
It may be but this is the solution. Check this with first equation and see it fits.
i3=-2.9117647 A
i2=-2
ib=i3-i2=-0.9117647
i1=3*ib-i3=3*(-0.9117647)+2.9117647=0.1764706 A
ua=10*(i1-i2)=10*(2.1764706)=21.764706 V
SUM=-25+30*i3+5*ua+20ib+ua=0
 
  • #18
I get:
i1 = -.82759
i2 = -2
i3 = -1.7069
 
  • #19
Check -25+30*i3+5*ua+20*ib+ua=204.827 with your results, but has to be 0.
 
  • #20
Babadag said:
Check -25+30*i3+5*ua+20*ib+ua=204.827 with your results, but has to be 0.

OK, let's evaluate all the terms using my values for i1, i2 and i3, using the fact that ua = 10*(i1-i2), ib = i3-i2
Listing the results, I get:
-25 = -25
30*i3 = -51.2068965516
5*ua = 58.620689655
20*ib = 5.8620689656
ua = 11.724137931

Adding all these up, I get zero, not 204.827
 
  • #21
Perhaps @garr6120 has been given the solution?
 
  • #22
It is very interesting: both results seem to be correct.
 
  • #23
Babadag said:
It is very interesting: both results seem to be correct.
I ran it through a simulation, and the results agree with @The Electrician's solution.
Screenshot_20171003-153549.png

The Electrician said:
OK, let's evaluate all the terms using my values for i1, i2 and i3, using the fact that ua = 10*(i1-i2), ib = i3-i2
Listing the results, I get:
-25 = -25
30*i3 = -51.2068965516
5*ua = 58.620689655
20*ib = 5.8620689656
ua = 11.724137931

Adding all these up, I get zero, not 204.827
 
  • #24
Babadag said:
It is very interesting: both results seem to be correct.

Check -25+30*i3+5*ua+20*ib+ua, has to be 0.

OK, let's evaluate all the terms using your values for i1, i2 and i3, using the fact that ua = 10*(i1-i2), ib = i3-i2
Listing the results, I get:

-25 = -25
30*i3 = 87.352941
5*ua = 108.82353
20*ib = 98.235294
ua = 21.764706

These add up to 291.176471, not zero
 
  • #25
Something goes wrong in your last calculation The Electrician. See post no.17 for i3 and ib values.
 
  • #26
Babadag said:
Something goes wrong in your last calculation The Electrician. See post no.17 for i3 and ib values.

I missed the minus sign on i3 in post #17. You had i3=-2.9117647 A.
I think it helps to leave spaces in appropriate places like this: i3 = -2.9117647 A

So I'll redo the calculation:

Check -25+30*i3+5*ua+20*ib+ua, has to be 0.

OK, let's evaluate all the terms using your values for i1, i2 and i3, using the fact that ua = 10*(i1-i2), ib = i3-i2
Listing the results, I get:

-25 = -25
30*i3 = -87.352941
5*ua = 108.82353
20*ib = -18.235294
ua = 21.764706

These add up to .000001, essentially zero.

For any value of i1, a value of i3 = -27/10 - (6*i1)/5 gives a value of zero. However, this is only the solution to one equation, and not a solution to the complete network. It is also necessary to satisfy the equation 3ib = i1-i3, but you assumed i1=3*ib-i3, which is incorrect.
 
  • #27
Here are two solutions:

TwoSolutions.png


Unfortunately, the second one is incorrect.
 
  • #28
It has to be a conventional rule in order to state if i1=3ib-i3 or 3ib=i1-i3
since i1 and i3 are virtual [ghosts] but 3ib it is a concrete one.
Since I am only an engineer –master it does not help here, I think-and not a professor
I get the professors’ opinion.:smile:
 
  • #29
Babadag said:
It has to be a conventional rule in order to state if i1=3ib-i3 or 3ib=i1-i3
since i1 and i3 are virtual [ghosts] but 3ib it is a concrete one.
Since I am only an engineer –master it does not help here, I think-and not a professor
I get the professors’ opinion.:smile:

The branch current 3ib is clearly composed of some combination of the loop currents i1 and i3.

The loop current i1 is in the same direction as the 3ib source current, and i3 is in the opposite direction, thus 3ib = i1 - i3.

Post #23 indicates that simulation gives the same result as assuming 3ib = i1 - i3.
 

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