# Finding 2 Unknown Currents and Voltage with Resistance

1. Oct 29, 2013

### RainbowTuba

1. The problem statement, all variables and given/known data
Find V, I1, I2 and the power of the 6Ω resistor:

Known:
R1=6Ω
R2=3Ω
Iknown1=8
Iknown2=2
RTotal=2Ω

2. Relevant equations
P=V2/R
P=IV
V=IR

3. 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.

2. Oct 29, 2013

### Staff: Mentor

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?

3. Oct 29, 2013

### RainbowTuba

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

4. Oct 29, 2013

### Staff: Mentor

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....

5. Oct 29, 2013

### RainbowTuba

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...