Calculating Current using Kirkoff's Rules

[MasterVivi]

Homework Statement

Apply the loop rule to two of the three loops to calculate I₁,I₂,and I₃.

Picture of DC circuit involved

Homework Equations

V=IR

The Attempt at a Solution

L₁=2v+2I₂-4I₁=0 (simplified)
and
L₃=4-4I₁-5I₃=0

But if you work through the algebra you'd get
I₂=1
which would then make
I₁=1

which does not work for the junction rule of
I₃=I₁-I₂ cause that is 0.

The rules still confuse me when look at some of the resistors on the right side I can tell which way current flows, obviously did it wrong though, any help would be appreciated.

 

[Spinnor]

Don't try to figure out which way the current is going, just identify the currents and give a direction. If the direction is wrong the current will have a minus value.

I suggest you study some worked out examples, like this,

http://commons.bcit.ca/math/examples/elex/linear_algebra/backgd4.html

Found here,

https://www.google.com/search?hl=en...e3bf2c90a1398d&bpcl=35277026&biw=1024&bih=517



Good luck!

 

[OmCheeto]

3. The Attempt at a Solution
loop 1 --> 2v+2i₂-4i₁=0 (simplified)

1Ω + 3Ω ≠ 2Ω

 

[MasterVivi]

Okay, re-attempted using loop one and loop two, does, this seem more accurate?

L₁=12V-I₁(2)+I₂(1)-10V+I₂(3)-I₁(1)=0
=>
L₁=2V-4I₁+4I₂=0


L₂=-3(I₂)+10V-I₂(1)-I₃(1)-8V-I₃(2)=0
=>
L₂=2V-4I₂-5I₃=0
=>
L₂=2-4I₂-5(I₁-I₂)=0
=>
L₂=2-4I₂-5I₁+5I₂=0
=>
L₂=2-5I₁+I₂=0

So from there
I₂=5I₁-2


Finally plugging it all through
I₁=I₂+(1/2)[From the first loop equation]
=>
I₁=5I₁-2+(1/2)
=>
I₁=(3/8)

if that's true

I₂=5(3/8)-2=(-1/8) [Is a negative current okay?]

and from there

I₃=I₁-I₂=(3/8)-(-1/8)=1/2

The negative current for I₂ feels wrong, did I mess up any of the loop equation or any algebra along the way?

 

[frogjg2003]

For the second loop, it should be +5I₂, not -5I₂
As for there being a negative current, it just means it's flowing in the opposite direction from what you set it up as in the problem.

 

[MasterVivi]

Okay, think I got it

You're right it should be +5I₃ because it's with the current, just like I₂ works out in the first loop equation

L₁=2-4I₁+4I₂=0

Using this to solve for I₁

I₁=I₂+(1/2)

Now I did this wrong, so going back through the derivation of this loop equation

L₂=-3I₂+10-I₂+2I₃+I₃-8+2I₃=0
=>
2-4I₂+5I₃=0
=>
2-4I₂+5(I₁-I₂)=0
=>
L₂=1+5I₁-9I₂=0


Using that to solve for I₂

I₂=(5/9)I₁+(2/9)

Solving for the numerical value of I₁
I₁=((5/9)I₁+(2/9))+(1/2)
=>
I₁=(5/9)I₁+(13/18)
=>
(4/9)I₁=(13/18)
=>
I₁=(13/8)


Solving for the numerical Value of I₂

I₂=(5/9)(13/8)+(2/9)
=>
I₂=(9/8)


Which Finally implies

I₃=I₁-I₂
=>
I₃=(13/8)-(9/8)=(1/2)

Aye?Nay?

 

[frogjg2003]

correct
quick note: you had I₁=I₂+1/2, so you could have easily figured out I₃ right then and there. I₃ = I₁-I₂ = I₂+1/2-I₂ = 1/2

 

[OmCheeto]

correct
quick note: you had I₁=I₂+1/2, so you could have easily figured out I₃ right then and there. I₃ = I₁-I₂ = I₂+1/2-I₂ = 1/2

really?

...
L₂=1+5I₁-9I₂=0
...
I₁=(13/8)
...
I₂=(9/8)
Which Finally implies
I₃=(13/8)-(9/8)=(1/2)
Aye?Nay?

L2=1+5I1-9I2=0
i1=13/8
i2=9/8
ergo:
1+5*(13/8)-9*(9/8)=0
8/8+65/8-81/8=0
73/8-81/8=0
-8/8=0
-1=0 ?

Don't all 4 equations have to come out correctly?