Why Is My Calculation of i1 Using KCL and Ohm's Law Incorrect?

[xlu2]

Homework Statement

Find i1

Homework Equations

KCL
V=IR

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?

Many thanks in advance!

 

[gneill]

Homework Statement

View attachment 59056

Find i1

Homework Equations

KCL
V=IR

The Attempt at a Solution

See picture. What am I doing wrong?

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.

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)?

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.

Then V1=-5-0.5i1+14+7i1=9+6.5i1
For V1 (right loop), can it be written as V1=5-10i1?

Nope. Pay attention to the direction with which i1 passes through the 10Ω resistor. What's the resulting polarity of the potential change?