Circuit analysis - current dissipated and pot. difference

[DevonZA]

Homework Statement

Homework Equations

The Attempt at a Solution

1.

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)
.

Loop ABEFA

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

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

 

[NascentOxygen]

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?

 

[DevonZA]

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

 

[DevonZA]

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

 

[NascentOxygen]

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.