Arbitrary current with mesh analysis

  • #1
garr6120
42
0
Members are reminded to use the 3 header template in the homework help forums
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?
 

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
9,242
1,073
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.
 
  • #3
garr6120
42
0
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?
 
  • #4
NascentOxygen
Staff Emeritus
Science Advisor
9,242
1,073
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.
 
  • #5
Babadag
537
147
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.
 
  • #6
garr6120
42
0
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?
 
  • #7
gneill
Mentor
20,945
2,886
After doing all of the calculations is an answer for mesh current 1 -17/27A correct?
Nope.
 
  • #8
The Electrician
Gold Member
1,344
188
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.
 
  • #9
garr6120
42
0
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
The Electrician
Gold Member
1,344
188
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
garr6120
42
0
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
The Electrician
Gold Member
1,344
188
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
NascentOxygen
Staff Emeritus
Science Advisor
9,242
1,073
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:
 
  • #15
Babadag
537
147
Your above eq.no.6 it is incorrect.
It has to be i1=3ib-i3.
Then i3= -2.9117647 A.
 
  • #16
NascentOxygen
Staff Emeritus
Science Advisor
9,242
1,073
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
Babadag
537
147
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
The Electrician
Gold Member
1,344
188
I get:
i1 = -.82759
i2 = -2
i3 = -1.7069
 
  • #19
Babadag
537
147
Check -25+30*i3+5*ua+20*ib+ua=204.827 with your results, but has to be 0.
 
  • #20
The Electrician
Gold Member
1,344
188
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
 
  • #22
Babadag
537
147
It is very interesting: both results seem to be correct.
 
  • #23
cnh1995
Homework Helper
Gold Member
3,467
1,162
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

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
The Electrician
Gold Member
1,344
188
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
Babadag
537
147
Something goes wrong in your last calculation The Electrician. See post no.17 for i3 and ib values.
 
  • #26
The Electrician
Gold Member
1,344
188
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
The Electrician
Gold Member
1,344
188
Here are two solutions:

TwoSolutions.png


Unfortunately, the second one is incorrect.
 
  • #28
Babadag
537
147
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
The Electrician
Gold Member
1,344
188
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.
 

Suggested for: Arbitrary current with mesh analysis

  • Last Post
Replies
7
Views
629
  • Last Post
Replies
4
Views
337
Replies
2
Views
153
  • Last Post
Replies
1
Views
1K
Replies
22
Views
847
Replies
2
Views
516
  • Last Post
Replies
7
Views
261
  • Last Post
Replies
2
Views
344
Replies
7
Views
526
Replies
5
Views
698
Top