Arbitrary current with mesh analysis

  • #1
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
9,244
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
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
9,244
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
473
126
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
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
20,925
2,867
After doing all of the calculations is an answer for mesh current 1 -17/27A correct?
Nope.
 
  • #8
The Electrician
Gold Member
1,302
171
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
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
1,302
171
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
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
1,302
171
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
9,244
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
473
126
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
9,244
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
473
126
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
1,302
171
I get:
i1 = -.82759
i2 = -2
i3 = -1.7069
 
  • #19
473
126
Check -25+30*i3+5*ua+20*ib+ua=204.827 with your results, but has to be 0.
 
  • #20
The Electrician
Gold Member
1,302
171
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
473
126
It is very interesting: both results seem to be correct.
 
  • #23
cnh1995
Homework Helper
Gold Member
3,444
1,144
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
1,302
171
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
473
126
Something goes wrong in your last calculation The Electrician. See post no.17 for i3 and ib values.
 

Related Threads on Arbitrary current with mesh analysis

  • Last Post
Replies
3
Views
916
  • Last Post
Replies
4
Views
859
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
0
Views
3K
Replies
6
Views
866
Replies
2
Views
6K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
1
Views
2K
Replies
7
Views
1K
Top