Calculating current in circuit with multiple emf's using Kirchhoff's

[Spudnik]

Homework Statement

Find I_1, I_2 and I_3 if the switch, k, is closed using Kirchhoff's Rules.

Homework Equations

V=IR

The Attempt at a Solution

I've had multiple attempts at this and this one is the one that came closest to a solution.

Setting up 3 equations:
1) I_1 = 1_2 + I_3 => I_1 - I_2 - I_3 =0
2) 12V + I_3*(4 Ohm) - I_1*(7 Ohm) = 0
3) 12V - 9V - I_2*(5 Ohm) - I_1*(7 Ohm) = 0

2) 12 = 7*I_1 - 4*I_3 => 12 = 7*(I_2 + I_3) - 4*I_3 => 12 = 7*I_2 + 3*I_3
3) 3 = 7*I_1 + 5*I_2 => 3 = 7*(I_2 + I_3) + 5*I_2 => 3 = 12*I_2 + 7*I_3

Getting rid of I_3:

84 = 49*I_2 + 21*I_3
09 = 36*I_2 + 21*I_3
---------------------------
75 = 13*I_2

I_2 = 5.75

By substituting in, I get

I_1 = -3.7
I_3 = -9.5

Is this anywhere near correct, I don't know why, but somehow this seems fishy to me :S

 

[collinsmark]

1) I_1 = 1_2 + I_3 => I_1 - I_2 - I_3 =0

I think your first equation is not correct. Pay close attention to the direction of the arrows. See how the sum together at either center top node, or center bottom node. You'll find that

I₂ = I₁ + I₃

(not, I₁ = I₂ + I₃ that you originally had.)

[Edit: Oh, and Spudnik, welcome to Physics Forums!]

 

[Spudnik]

OK, I redid it with the new formula and I got

I₁=0.25
I₂=0.25
I₃=-2.5625

Clearly this does not fit the initial equation I₂=I₁+I₃

It's strange though because those are the same values I got for I₁ and I₂ for part A) where it asked for the currents with the switch open...

 

[Spudnik]

Double posting because I figured it out:

I₁= 72/83 A
I₂= -51/83 A
I₃= -123/83 A

Using
I₂=I₁+I₃
12+4I₃-7I₁=0
12-9-5I₂-7I₁=0

Setting them up for a matrix:

I₁ + I₂ + I₃ = Voltage
+1 -1 +1 =0
-7 +0 +4 =-12
-7 -5 +0 =-3

Letting the calculator do the rest for you: priceless.