How to Find Potential Between A and B with R1 Present?

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Discussion Overview

The discussion revolves around finding the potential difference between points A and B in a circuit that includes a resistor R1. Participants explore various circuit analysis methods, including loop analysis and superposition, while grappling with the complexities introduced by the presence of R1 and other components.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant expresses confusion regarding the role of R1 in the circuit and suggests that without it, a simple voltage divider would suffice.
  • Another participant inquires about the methods of circuit analysis the original poster has studied, specifically mentioning loop, mesh, and node analysis, as well as Thevenin equivalents.
  • A participant describes their approach using the superposition principle and notes discrepancies in the results obtained when applying loop analysis with and without suppressing voltage sources.
  • Participants discuss the original circuit and the need for visual aids to clarify the problem, with one suggesting the use of an image hosting service to share the circuit diagram.
  • One participant proposes using the loop method without suppressing any voltage sources and shares their calculated current through R7, while encouraging the original poster to write out the loop equations.
  • Another participant provides their loop equations based on the circuit diagram and seeks confirmation on their accuracy.
  • There is a discussion about the currents calculated from the loop equations, with participants reporting different values and discussing potential algebraic mistakes in their calculations.
  • One participant describes using a matrix solver to set up and solve the equations, suggesting that this method can help avoid algebraic errors.
  • Another participant points out a potential issue in the second equation of the loop analysis, encouraging the original poster to rewrite it in a form suitable for matrix representation.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach or the resulting values for the currents. Multiple competing views and methods for solving the circuit analysis problem are presented, leading to differing results.

Contextual Notes

Participants express uncertainty regarding the accuracy of their calculations and the setup of their equations, indicating potential algebraic mistakes and the complexity of the circuit analysis involved.

Altairs
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Homework Statement


Find potential between A and B


Homework Equations





The Attempt at a Solution


The R1 is confusing me. If it hadnt been there simple voltage divider would have taken care of things quite easily...I can't make logic of how to solve it now...How do I deal with R1 ?
 

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What methods of circuit analysis have you studied so far?

Do you know how to do loop, mesh or node analysis?

Do you know about Thevenin equivalents?
 
The Electrician said:
What methods of circuit analysis have you studied so far?

Do you know how to do loop, mesh or node analysis?

Do you know about Thevenin equivalents?

Everything except the Thevenin Equivalents. The circuit I have attached is the result of applying superposition principle. Actually there was another Voltage source (V') connected between R3 and the common joint of R3,R4 and R7, with positive towards the right. I had to find the voltage across AB (One Port). SO by using the superposition principle and suppressing the V' this is the circuit I got. I applied loop analysis one by one i.e. once on this circuit and then by suppressing the V source but I don't get the right answer. And to make matters worse when I apply the loop analysis without suppressing any of the sources I get a completely different answer.

The values given are :-
V = 18V
V' = 13V
R1=R2=R3=3ohm
R4=R5=6ohm
R6=20ohm
R7=10ohm

My answers :-
1. Suppressing the V' then current through R7 = 0.09A
2. Suppressing the V then current through R7 = 0.571A
3. Solving without superposition then current through R7 = 0.125A
 
I would be better able to help you if you will post the original circuit, including the V' source.
 
here is the original circuit
 

Attachments

  • Problem1.jpg
    Problem1.jpg
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I won't be able to see the jpg until it gets approved, which could take hours.

If you'll upload it to:http://www.freeimagehosting.net/

and post the URL here, I will be able to get it sooner.
 
Here it is http://www.freeimagehosting.net/image.php?3c03009db1.jpg
 
Ok, I got it. This problem is a natural for the loop method. You don't need to suppress any voltage sources. I labeled the loop currents as I1 for the bottom left loop, I2 for the top loop, and I3, I4 for the next two loops to the right of the I1 loop.

Can you write the loop equations for those loops and show your work?

Just to inspire you, I got a current of 1.5 amps in R7, and that makes 15 volts at the A-B port. I have been known to make mistakes though. :-(
 
For loop 1:-

[tex]3(i_1 + i_4) + 6(i_1 - i_2) = 18[/tex]

For loop 2:-

[tex]6(i_2 - i_1) + 3(i_2 + i_4) + 20(i_2 + i_3) = 13[/tex]

For loop 3:-

[tex]6(i_3 - i_4) + 20(i_3 + i_2) +10i_3 = 0[/tex]

For loop 4:-

[tex]3i_4 +3(i_4 + i_1) + 3(i_4 + i_2) + 6(i_4 - i_3) = 13[/tex]
 
  • #10
here is how I made the loops. the above equations are according to this figure. http://www.freeimagehosting.net/image.php?c421c9d797.jpg
 
  • #11
You have it exactly right. Your equations give the same currents I got, even though I selected my loops slightly differently.

What currents do you get when you solve the system?
 
Last edited:
  • #12
I get i3 =0.125
 
  • #13
How are you solving it? I get:

I1 = 3.9333
I2 = 2.4
I3 = -1.5
I4 = -1.0
 
  • #14
Simlifying the equations then making i1 as the subject from last equation and substituting it in two of above and then solving simultaneously to get i2 , i3 and i4...must be some algebraic mistake
 
  • #15
I used a matrix solver after setting up the matrix and excitation vector like this; just collect the coefficients for each current:

Code: 
[  9  -6   0  3 ]  [ I1 ]     [ 18 ]
[ -6  29  20  3 ]* [ I2 ]  =  [ 13 ]
[  0  20  36 -6 ]  [ I3 ]     [  0 ]
[  3  3   -6 15 ]  [ I4 ]     [ 13 ]

then use Matlab, or a calculator with matrix capabilities to solve it.

If you have the matrix set up properly, this method guarantees no algebra mistakes! :-)

Given the magnitude of the voltage sources and the resistor values, I think .125 amps is definitely too small.
 
  • #16
which loops did you make?
 
  • #17
I used yours.
 
  • #18
Look at your 2nd equation, for example. You can collect terms and rewrite it as:

-6*I1 + 29*I2 + 20*I3 + 3*I4 = 13

and similarly for the others. Thats how you get the entries in the matrix.
 

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