# I'm having a problem solving for voltage between two points

#### barrett50

Joined Feb 1, 2016
14
I'm having some problems finding the value for Vx. So far I have found the mesh current equations for all of the loops, and all that's left is Vx. Can someone help me find a starting point for this?

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#### shteii01

Joined Feb 19, 2010
4,644
I'm having some problems finding the value for Vx. So far I have found the mesh current equations for all of the loops, and all that's left is Vx. Can someone help me find a starting point for this?
The fact that you can not find Vx tells me that you don't have mesh currents.
Nice try though.

#### WBahn

Joined Mar 31, 2012
26,398
So... are we supposed to use our crystal balls to see these mesh current equations that you claim to have found for all the loops?

#### barrett50

Joined Feb 1, 2016
14
The fact that you can not find Vx tells me that you don't have mesh currents.
Nice try though.
I didn't say I have the currents, I said I have the equations. The problem comes in at the end of my third KCL that is

8(Iy-I3) + 6(Iy-I2)+3Vx

My problem is that I don't know what I'm supposed to substitute in for Vx.

#### shteii01

Joined Feb 19, 2010
4,644
I didn't say I have the currents, I said I have the equations. The problem comes in at the end of my third KCL that is

8(Iy-I3) + 6(Iy-I2)+3Vx

My problem is that I don't know what I'm supposed to substitute in for Vx.
What is that equal to?

#### barrett50

Joined Feb 1, 2016
14
So I labeled the first box I1, upper mid box is I2, lower mid is I3, and right box is I4. There is a supermesh at I2 and I3. I go clockwise

I1 = -4 (A)

KCL at SuperMesh
3(I3-I1)-2+5(I2)+6(I2-I4)+8(I3-I4)

KCL at I4 is
8(Iy-I3) + 6(Iy-I2)+3Vx

KVL for the dependent current source is
Iy=(I2-I3)/7

#### shteii01

Joined Feb 19, 2010
4,644
Anyway. Vx=2+3(whatever currents pass through the 3 Ohm resistor)

#### WBahn

Joined Mar 31, 2012
26,398
So I labeled the first box I1, upper mid box is I2, lower mid is I3, and right box is I4. There is a supermesh at I2 and I3. I go clockwise

I1 = -4 (A)

KCL at SuperMesh
3(I3-I1)-2+5(I2)+6(I2-I4)+8(I3-I4)

KCL at I4 is
8(Iy-I3) + 6(Iy-I2)+3Vx

KVL for the dependent current source is
Iy=(I2-I3)/7
These are not equations, they are expressions. Equations involve this thing called an equals sign.

Also, you need to specify which direction your mesh currents are circulating. Otherwise there is no way to know if your value for I1 is correct or not.