# Find i1 (KCL)

1. May 27, 2013

### xlu2

1. The problem statement, all variables and given/known data

Find i1

2. Relevant equations
KCL
V=IR

3. The attempt at a solution
See picture. What am I doing wrong?

Also, the current through 7 ohms = 2+i1.
So for V1 (left loop), can it be written as V1=-5-0.5i1+7(2+i1)?
Then V1=-5-0.5i1+14+7i1=9+6.5i1
For V1 (right loop), can it be written as V1=5-10i1?

Last edited: May 27, 2013
2. May 27, 2013

### Staff: Mentor

I don't understand the node equation in your diagram; Why doesn't V1 appear in it? V1 should be the unknown potential at the node that determines the currents in the branches.
It doesn't look like you've assigned the correct signs to the potential changes with respect to the reference node. i1+2 flowing through the 7Ω resistor in the direction indicated should cause a potential drop, but the 5V source must cause a +5V potential rise. Similarly, the controlled source makes another potential rise of 0.5*i1 on the way from the reference node to V1.
Nope. Pay attention to the direction with which i1 passes through the 10Ω resistor. What's the resulting polarity of the potential change?