Circuit analysis - current dissipated and pot. difference

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Discussion Overview

The discussion revolves around a circuit analysis problem involving mesh analysis to determine circulating currents and the potential difference across specific resistors. Participants are exploring the application of Kirchhoff's Laws in their calculations and the implications of current direction on voltage across resistors.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a mesh analysis approach to find currents I1 and I2, leading to calculations for current through resistor R1 and potential difference across R3.
  • Another participant emphasizes the importance of verifying that calculated currents satisfy Kirchhoff's Laws before proceeding with further calculations.
  • There is a question regarding the convention of current polarities when analyzing the circuit, specifically whether to consider the effect of I2 on R3 as positive or negative based on the direction of analysis.
  • A later reply suggests that consistency in polarity choice is key, regardless of whether clockwise or counterclockwise is chosen as positive.
  • Participants refine their equations based on their interpretations of current directions and polarities, leading to different values for I1 and I2.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of current polarities and the resulting equations. There is no consensus on the correct approach, as multiple interpretations and calculations are presented.

Contextual Notes

Some calculations depend on assumptions about current direction and polarity conventions, which remain unresolved. The discussion reflects variations in the application of Kirchhoff's Laws and mesh analysis techniques.

DevonZA
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Homework Statement

Homework Equations

The Attempt at a Solution



1.
upload_2015-11-14_11-50-11.png


2. i) Use mesh analysis to solve for circulating currents I1 and I2.
ii) What is the current dissipated by resistor R1?
iii) What is the potential difference across resistor R3?

3

i)
.
upload_2015-11-14_11-53-32.png


Loop ABEFA

E1-I1RI-(I1-I2)R3=0
10-10I1-40(I1-I2)=0
-50I1+40I2=-10
50I1-40I2=10 ... EQUATION 1

upload_2015-11-14_11-57-59.png


Loop BCDEB

-I2R3-I1R3-I2R2-E2=0
-40I2-40I1-20I2-I2=0
-40I1-60I2=12 ... EQUATION 2

EQ 1 X 4 = 200I1-160I2=40
EQ 2 X 5 = -200I1-300I2=60 +[/B]
= -460I2=100
I2=-0.217A

Substitute I2 into equation 1: 50I1-40(-.217)=10
I1=0.0264Aii) I=V/R
= 10/10
=1A

iii) I3=I1+I2 (KCL)
=(0.0264)+(-0.217)
= -0.1906AVR3 = I3xR3
= -0.1906x40
= -7.624V
 
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Loop BCDEB

-I2R3-I1R3-I2R2-E2=0
In their passage through R3, currents I1 and I2 are in opposite directions, so partly cancel. This means their contributions to the R3 voltage will have opposite polarities, one will have a + sign, the other a - sign.

Before using your I1 and I2 values in parts (ii) and (iii) you should have first checked that they correctly satisfy Kirchoff's Laws and your circuit. Do they?
 
Firstly - do we take exit or entrance polarities? By this I mean if we have are working in a clockwise direction would the effect of I2 on R3 be +I2R3 or -I2R3? The examples in my study guide suggest that we use exit polarities. If so then I would do the following:

Loop BCDEB (clockwise)

+I2R3-I1R3+I2R2+E2=0
40I2-40I1+20I2+I2=0
-40I1+60I2=12 ... EQUATION 2


 
Okay I think I have it now:

Loop BCDEB

+12V+40ohm(I2-I1)+2ohmI2=0
40ohmI1-60ohmI2=12V

Eq 1. x 4 200ohmI1-160ohmI2=40V
Eq. 2 x 5 200ohmI1-300ohmI2=60V -
140ohmI2=-20V
I2=-0.14A

Subst I2 into equation 1: 50ohmI1-40ohm(-.014)=10V
50ohmI1=4.4V
I1=0.088A

I3=I1-I2
= 0.088-(-0.14)
= 0.288A

 
DevonZA said:
Firstly - do we take exit or entrance polarities? By this I mean if we have are working in a clockwise direction would the effect of I2 on R3 be +I2R3 or -I2R3? The examples in my study guide suggest that we use exit polarities.
It makes no difference whether you consider clockwise to be positive or negative, just so long as you stay consistent.

If so then I would do the following:

Loop BCDEB (clockwise)

+I2R3-I1R3+I2R2+E2=0
That's right, so far.
 

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