Needing help with calculating currents

  • Thread starter Thread starter astenroo
  • Start date Start date
  • Tags Tags
    Currents
AI Thread Summary
The discussion revolves around solving a system of five equations with five unknowns related to currents in a DC circuit. The user initially struggles with the complexity of the equations but receives guidance on using matrix methods to simplify the problem. It is suggested to express the currents as a vector and formulate the equations into a matrix equation, which can then be solved using matrix inversion. While the user considers sticking with a brute-force method for now, they acknowledge the need to learn matrix techniques for future problems. The conversation emphasizes the importance of maintaining clear notes to verify solutions regardless of the method used.
astenroo
Messages
47
Reaction score
0

Homework Statement


The currents in a DC circuit is to be calculated, and I end up with a system of 5 equations and 5 unknowns. This is where my headache starts.


Homework Equations



(1) I_{1}=I_{2} + I_{3}
(2) I_{2}=I_{4} + I_{5}
(3) 6 - 2I_{1} - I_{3} = 0
(4) 4.5 - 5I_{2} - 3I_{5} + I_{3} = 0
(5) 3I_{5} - 3I_{4} -3 = 0


The Attempt at a Solution



Now, I have no problem solving equations with two unknowns. But 5?
Anyway, I gave it a try. I tried to substitute (2) in (1) and substituting the result into (3) and then substituting (2) into (4), resulting in a new system with three unknowns

(3') -3I_{3} - 2I_{4} - 2I_{5} = -6
(4') I_{3} - 5I_{2} - 8I_{5} = -4,5
(5') -3I_{5} + 3I_{5} = 3

I've been on this problem for the last 6 hours and I still haven't figured out how to solve it.
I've even tried using a matrix (never done one before today), and it all goes wrong, since I apparently don't know how to use a matrix :P
 
Physics news on Phys.org
You are on the right track with matrices. Write the current as a vector C=(I1,I2,I3,I4,I5), and then you can write the equations as a matrix equation M * C = B, where B is a constant vector which has the constant coefficients in it. Then you invert the matrix M, and can write the current C as C = Inverse(M) * B. I carried it out and got an answer for the currents using Mathematica. Can you try to write down the matrix M and the constant vector B?
 
phyzguy said:
You are on the right track with matrices. Write the current as a vector C=(I1,I2,I3,I4,I5), and then you can write the equations as a matrix equation M * C = B, where B is a constant vector which has the constant coefficients in it. Then you invert the matrix M, and can write the current C as C = Inverse(M) * B. I carried it out and got an answer for the currents using Mathematica. Can you try to write down the matrix M and the constant vector B?

Thanks for the quick reply!
i have to get back on that one as soon as I find instructions on how to set up said matrices... The one I tried was using it on (3'), (4') and (5'). Apparently that is a no go ;)
 
Matricies are cleaner and if you have the time to figure it out, that is the preferable way to go. However, this "brute-force" way will work. The best part is that if you keep clean notes, simply plug your answers back in and you will know for sure if you have the right answer regardless how you get there.

Except for a couple of typos in your subscripts, your equations look right. If you substitute out I5 from (3') and (4') using (5'), you will end up with two equations (3'') and (4'') with two unknowns, I3 and I4. Solve for those and then plug them back into (5') and you have I5. Plug these back into (2) then (1) and you'll be done.
 
dulrich said:
Matricies are cleaner and if you have the time to figure it out, that is the preferable way to go. However, this "brute-force" way will work. The best part is that if you keep clean notes, simply plug your answers back in and you will know for sure if you have the right answer regardless how you get there.

Except for a couple of typos in your subscripts, your equations look right. If you substitute out I5 from (3') and (4') using (5'), you will end up with two equations (3'') and (4'') with two unknowns, I3 and I4. Solve for those and then plug them back into (5') and you have I5. Plug these back into (2) then (1) and you'll be done.

Oh yes! Indeed, it is exactly as solving systems with two equations and 2 unknowns. Now I noticed the typos :) I might stick around using the brute-force method for now, but I will eventually have to learn matrices also :) Thank you for the help.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top