Find IA, IB, IC, I1, I2, V0, VA Solutions

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Discussion Overview

The discussion revolves around solving a circuit analysis problem involving mesh analysis to find the currents IA, IB, IC, and the voltages V0, VA. Participants are exploring the application of Kirchhoff's Voltage Law (KVL) and supermesh techniques in their attempts to derive the correct equations and solutions.

Discussion Character

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

Main Points Raised

  • One participant presents initial equations derived from KVL but encounters discrepancies when checking results with Ohm's law.
  • Another participant points out an error in the application of KVL, suggesting that the term IA-IB is a current, not a voltage, and recommends using a supermesh approach.
  • A participant attempts to reformulate the problem using supermesh analysis but finds that their equations lead to cancellation of variables, indicating potential redundancy in their equations.
  • There is a suggestion that only two mesh equations are needed along with the supernode constraint, as adding a third mesh equation may duplicate information.
  • Participants express confusion about how to proceed with the equations, particularly regarding the number of equations needed to solve for three variables.
  • One participant is reminded to include the supermesh constraint equation to resolve the system of equations.

Areas of Agreement / Disagreement

Participants generally agree on the need to use supermesh techniques and KVL, but there is no consensus on the correct formulation of the equations or the approach to take in solving for the variables. Confusion remains regarding the number of equations required to solve the problem.

Contextual Notes

Participants express uncertainty about the correct application of mesh analysis and the implications of the controlled current source in their equations. There are indications of missing assumptions and potential dependencies on the definitions used in the problem.

Who May Find This Useful

This discussion may be useful for students and practitioners involved in circuit analysis, particularly those learning about mesh analysis and KVL in electrical engineering contexts.

jdawg
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Homework Statement


Find IA, IB, IC, I1, I2, V0, VA

Homework Equations

The Attempt at a Solution


(V0/10) = IA-IB

For IA: 10v + 10(IA+IC) + (IA-IB)=0
11(IA) - (IB) + 10(IC) = -10v

For IB: -(IB-IA) - 20(IB+IC) - 10(IB) = 0
IA - 31(IB) - 20(IC) = 0

For IC: 10(IA+IC) - 20(IC+IB) - 10(IC) = 0
10(IA) - 20(IB) - 20(IC) = 0

I then put these equations into a matrix (Sorry I don't know how to put them into a matrix on here) and used an online matrix calculator to get the currents:

IA= -0.4A
IB=(18/55)A
IC=(-29/55)A

So then I plugged these values into this formula:
(V0/10) = IA-IB
And got V0= -7.3V

But when I used v=iR to see if I would get the same thing I got

v=(-29/55)(10)= -5.27v

What am I doing wrong?
 

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Hi jdawg. Are you able to replace your pdf with a simple jpeg?
 
jdawg said:

Homework Statement


Find IA, IB, IC, I1, I2, V0, VA

Homework Equations

The Attempt at a Solution


(V0/10) = IA-IB

For IA: 10v + 10(IA+IC) + (IA-IB)=0
11(IA) - (IB) + 10(IC) = -10v
Your term: IA-IB is a current, not a voltage. That won't work for a KVL statement.

The source in the center of the circuit is a current source. It has a value ##V_o / 10kΩ##, which you'll note is a current value. Lying between two meshes as it is, your instinct should be to form a Supermesh to avoid dealing with its unknown potential change in your KVL equations.

Fig1.png
 
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Oh ok, I retried the problem using super meshes:

First I did a super mesh around the whole circuit:
10+10(IC)-10(IB) = 0
IC = IB - 1 equation1

Then I did a second super mesh around meshes IA and IB excluding IC:
10 - 10(IA+IC) - 20(IC+IB) - 10(IB) = 0
10(IA) + 30(IB) +30(IC) = 10 equation2

And then around IC:
-10(IC) - 10(IC+IA) - 20(IC+IB) = 0
IA +2(IB) +4(IC) = 0 equation3

I then plugged equation 1 into equation 2 and equation 3 and IB ended up canceling out. I'm not sure what I did wrong this time.
 
jdawg said:
Oh ok, I retried the problem using super meshes:

First I did a super mesh around the whole circuit:
10+10(IC)-10(IB) = 0
IC = IB - 1 equation1

Then I did a second super mesh around meshes IA and IB excluding IC:
10 - 10(IA+IC) - 20(IC+IB) - 10(IB) = 0
10(IA) + 30(IB) +30(IC) = 10 equation2

And then around IC:
-10(IC) - 10(IC+IA) - 20(IC+IB) = 0
IA +2(IB) +4(IC) = 0 equation3

I then plugged equation 1 into equation 2 and equation 3 and IB ended up canceling out. I'm not sure what I did wrong this time.
You only need two mesh equations plus the supernode constraint equation. Writing a third mesh equation doesn't provide new information, it just duplicates other information from the other two equations. That's why you found things cancelling out.

So, you previously wrote the required constraint equation relating IA, IB, and IC (via the controlled current supply). Use that in place of one of your mesh equations. Usually the approach is to combine the two loops with the shared current into a supermesh, and write regular mesh equations for any other loops.
 
Ok so I kept my super mesh 2 and IC equations. I'm still a little confused, so now I just write an equation for either loop IA or IB?
 
jdawg said:
Ok so I kept my super mesh 2 and IC equations. I'm still a little confused, so now I just write an equation for either loop IA or IB?

You can't write equations for loops IA or IB because you don't have a potential for the controlled current source. So you combine IA and IB into a supermesh as you've done, and then write mesh equations for what's left: that's loop IC.

Fig2.png
 
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So now I have IA+3(IB)+3(IC) = 1 from the super mesh and IA+ 2(IB)+4(IC) = 0 for the IC loop. How do I solve for the three variables with only two equations? Sorry for asking so many questions!
 
jdawg said:
So now I have IA+3(IB)+3(IC) = 1 from the super mesh and IA+ 2(IB)+4(IC) = 0 for the IC loop. How do I solve for the three variables with only two equations? Sorry for asking so many questions!
You've forgotten the supermesh constraint equation.
 
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Thanks so much for all your help!
 

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