Finding 2 Unknown Currents and Voltage with Resistance

[RainbowTuba]

Homework Statement

Find V, I₁, I₂ and the power of the 6Ω resistor:

Known:
R₁=6Ω
R₂=3Ω
I_known1=8
I_known2=2
R_Total=2Ω

Homework Equations

P=V²/R
P=IV
V=IR

The Attempt at a Solution

To be honest, I've been staring at the problem for hours not knowing where to begin. I thought I'd start with a KCL equation to find I₁ and I₂, but I'm left with too many unknowns to continue on.
I tried using a point above the 6Ω resistor as a node to find the KCL: Using V₁ as my voltage at the node.
-8 -I₁+I₂=0
= (V₁-V)/3 = 8+(V₁-V)/6
= V₁=V+48
But I don't believe I'm setting up the equation right in the first place.

 

[berkeman]

Homework Statement

Find V, I₁, I₂ and the power of the 6Ω resistor:

Known:
R₁=6Ω
R₂=3Ω
I_known1=8
I_known2=2
R_Total=2Ω

Homework Equations

P=V²/R
P=IV
V=IR

The Attempt at a Solution

To be honest, I've been staring at the problem for hours not knowing where to begin. I thought I'd start with a KCL equation to find I₁ and I₂, but I'm left with too many unknowns to continue on.
I tried using a point above the 6Ω resistor as a node to find the KCL: Using V₁ as my voltage at the node.
-8 -I₁+I₂=0
= (V₁-V)/3 = 8+(V₁-V)/6
= V₁=V+48
But I don't believe I'm setting up the equation right in the first place.

Welcome to the PF.

This equation of yours: -8 -I1+I2=0 is incomplete.

Instead, what can you say when you have 2 current sources in parallel? What is the net current source that drives the parallel resistor combination?

 

[RainbowTuba]

Welcome to the PF.

This equation of yours: -8 -I1+I2=0 is incomplete.

Instead, what can you say when you have 2 current sources in parallel? What is the net current source that drives the parallel resistor combination?

So I can find an equivalent resistance of the two, which would equal 2, and the net current source would be 10A.

I figured since the unknown current is a factor, simplifying the circuit would combine the two, is this incorrect?

So would I use
-2-8 -I1+I2=0
instead?

 

[berkeman]

So I can find an equivalent resistance of the two, which would equal 2, and the net current source would be 10A.

I figured since the unknown current is a factor, simplifying the circuit would combine the two, is this incorrect?

So would I use
-2-8 -I1+I2=0
instead?

The two current sources drive their current through the resistors in opposite directions, so they don't add.

And once you have the correct net current driving through the resistors, combine the two resistors into the equivalent resistance to find Vout. Then use Vout and the individual resistance values to...

 

[RainbowTuba]

The two current sources drive their current through the resistors in opposite directions, so they don't add.

And once you have the correct net current driving through the resistors, combine the two resistors into the equivalent resistance to find Vout. Then use Vout and the individual resistance values to...

Yes they do, so instead of -10 it would be -6. Combining the resistance and capacitance would lessen the unknowns by one, so R=2 and I is our unknown

So:
-6 = I
which would make V=IR=(-6)2=-12, my assumed direction of I was wrong, so V=12, am I on the right track?

I feel like I did something wrong...