Need help to find I in a circuit (mesh analysis?)

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Discussion Overview

The discussion revolves around solving for the current I in a circuit using mesh current analysis. Participants explore the setup of mesh equations and the challenges faced in solving them. The context is primarily homework-related, focusing on the application of mesh analysis in circuit theory.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant expresses uncertainty about how to approach the problem, considering mesh current analysis but feeling unprepared due to lack of instruction.
  • Another participant encourages sharing the initial steps taken to facilitate assistance, suggesting that mesh analysis is a viable method.
  • A participant outlines their assignment of currents to different loops and presents the equations derived from mesh analysis, detailing the steps taken to solve them simultaneously.
  • There is a discussion about the implications of obtaining a negative current, with one participant clarifying that it indicates the actual current direction is opposite to the assumed direction.
  • Another participant points out that the assigned direction of current I in the diagram is opposite to the assumed mesh current, leading to the conclusion that I = -I1.

Areas of Agreement / Disagreement

Participants generally agree on the validity of using mesh analysis, but there is no consensus on the correctness of the calculations or the interpretation of the negative current result. The discussion remains unresolved regarding the specific solution steps and outcomes.

Contextual Notes

Participants express uncertainty about the correctness of their equations and calculations, and there are indications of missing assumptions or potential errors in the setup of the mesh analysis.

asdf12312
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Homework Statement


http://tinypic.com/r/3478nle/6
3478nle.jpg

need to find the current I

Homework Equations


was thinking about using mesh current analysis, but our teacher hasn't gone over it yet.

The Attempt at a Solution


no clue, totally stumped on what to do. was thinking mesh analysis and i attempted that, but i had no idea what i was doing after a while.
 
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asdf12312 said:

Homework Statement


http://tinypic.com/r/3478nle/6
need to find the current I

Homework Equations


was thinking about using mesh current analysis, but our teacher hasn't gone over it yet.

The Attempt at a Solution


no clue, totally stumped on what to do. was thinking mesh analysis and i attempted that, but i had no idea what i was doing after a while.

Mesh analysis looks like a fine idea. Why don't you show how you got started so we can see how to help you when you get stuck?
 
well i assigned currents I1, I2, and I3 to the right, left, and top loops respectively. moving in clockwise direction this what i got:

right loop:
3V+21(I1)-6(I2)-6(13)=0

left loop:
3V-6(I1)+21(I2)-6(I3)=0

top loop:
-6(I1)-6(I2)+21(I3)=0

now the trouble I'm having is the part where I'm supposed to solve all 3 equations simultaneously. my method is getting I1=I2=I3 in the bottom two equations and then plugging them into eq#1 so i can solve for I1 (only current i need to find).

cancelling out I3:
3.5(3V-6(I1)+21(I2)-6(I3)=0)
-6(I1)-6(I2)+21(I3)=0
--
10.5V-21(I1)+73.5(I2)-21(I3)=0
-6(I1)-6(I2)+21(13)=0
--
-27(I1)+67.5(I2)=-10.5
67.5(I2)=27(I1)-10.5
I2=0.4(I1)-0.16

cancelling out 12:
3V-6(I1)+21(I2)-6(I3)=0
3.5(-6(I1)-6(I2)+21(I3)=0)
--
3V-6(I1)+21(I2)-6(I3)=0
-21(I1)-21(12)+73.5(I3)=0
--
-27(I1)+67.5(I3)=-3V
67.5(I3)=27(I1)-3
I3=0.4(I1)-0.044

substituting into eq#1:
21(I1)-6(I2)-6(13)=-3
21(I1)-6(0.4(I1)-0.16)-6(0.4(I1)-0.044)=-3
21(I1)-2.4(I1)+0.96-2.4(I1)+0.26=-3
16.2(I1)=-4.22
I1=-0.26A

but i know i did this wrong...cause i got a negative current.
 
Nothing wrong with a negative current, it just means that your assumed current direction for the loop is the opposite of the actual current direction. Your assumed direction was a clockwise mesh current, right? So the actual current is counterclockwise.

Note the direction that was assigned to I in the diagram; It's the opposite of your assumed mesh current. So I = -I1 ...
 

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