Find the voltage Vm (mesh / nodal analysis)

[Color_of_Cyan]

Homework Statement

http://img541.imageshack.us/img541/5039/homeworkprobsg27.jpg


Find the value of the voltage Vm

Homework Equations

V = IR,
KVL,
KCL,
mesh equations,
nodal equations

The Attempt at a Solution

http://imageshack.us/a/img507/5556/homeworkprobsg27edit.jpg

Doing it this way, and doing mesh, I came up with:

I1 + 5ia = ia;

I1 = -4ia


then for the left loop:

24V - 32Ω*I1 - Vx = 0

Vx = 24V - 32Ω*I1


then for the right loop:

Vx - 3Ω*5ia - Vm = 0

Vx = - 3Ω*5ia - Vm

So my equations are:

1.) I1 = -4ia
2.) Vx = 24V - 32Ω*I1
3.) Vx = - 3Ω*5ia - Vm

combining the last two above:

- 3Ω*5ia - Vm = 24V - 32Ω*I1

- 3Ω*5ia - Vm = 24V - 32Ω*(-4ia)

Then

24 V + 32Ω*(4ia) + 3Ω*(5ia) = Vm

24V + 128Ωia + 15Ωia = Vm

then

24V + 143Ωia = Vm but not sure what to do now, or if I approached this the wrong way.

Any suggestions? Thank you

 

[The Electrician]

http://imageshack.us/a/img507/5556/homeworkprobsg27edit.jpg

This isn't going to work because the current Ia is passing downward through a wire, which has a resistance of zero ohms. There can't be any voltage across this wire; Vx is zero in other words.

Superposition seems like the appropriate method here.

The 24 volt source causes 3/4 amp downward through the wire. The 3 ohm resistor has no effect on the current the dependent current source delivers to Ia.

So, we have Ia = the sum of 3/4 amp plus whatever the dependent source delivers. Can you write an expression for that? Then you can solve for Ia.

Once you have Ia, you also have 5*Ia, which is the current through the 3 ohm resistor. Since the left end of the 3 ohm resistor is grounded, the voltage across the dependent current source will be the same as the voltage across the 3 ohm resistor. Just be sure you get the sign of the voltage across the dependent source right.

 

[NascentOxygen]

2.) Vx = 24V - 32Ω*I1

That branch shown carrying ia is just a wire, it's not a current source.

 

[Color_of_Cyan]

So far as I understand the superposition method, it goes finding each current though (in this case) ia with each voltage source short-circuited and each current source open circuited,and then summing those currents through it to get ia.. ? :

5ia becomes open circuited leaving only the 32 ohm resistor with current flowing through and with V = IR i got 0.75A like you said, Electrician.

So

say ia = i (current through)

this i(current through) = I' + I''

so I' = 0.75A

Then I'' would be from short circuiting the 24V

But then I made a short circuit out of the 24V getting this

http://img600.imageshack.us/img600/2826/homeworkprobsg27edit2.jpg stuck again; I know I would have to solve for I'' in the new diagram now though, right?

 

[gneill]

If you have a dependent current source in your circuit then you need at least one independent source active in order to analyze the circuit. Otherwise there's nothing for the dependent source to 'sample' and react to to get started.

In this circuit you've got one dependent source and one independent source, so you'll need to leave both active to analyze. Why not just write KCL at the top node and solve for the current of interest?

 

[The Electrician]

Then I'' would be from short circuiting the 24V

But then I made a short circuit out of the 24V getting this

http://img600.imageshack.us/img600/2826/homeworkprobsg27edit2.jpg


stuck again; I know I would have to solve for I'' in the new diagram now though, right?

The 32 ohm resistor is shorted by the wire containing I''.

It looks to me like I'' = 5Ia

Proceed from there.

 

[Color_of_Cyan]

let me check that,

so ia = 0.75A + 5ia

-4ia = 0.75A

ia = -0.1875

5ia = -0.9375Aso Vm = (0.9375A)(3 ohm) = -2.81V, but not sure, and no answer given.Trying what gneill said:With this:

http://imageshack.us/a/img507/5556/homeworkprobsg27edit.jpg

then 5ia + I1 = ia

I1 = -4ia

I think I1 = 24V / 32ohm

I1 = 0.75A

-4ia = 0.75A

ia = -0.1875 A

5ia = -0.9375 A

Vm = (-0.9375 A)(3 ohm)

Vm = -2.81V again, surprising if it's really just KVL too.

Would this be correct?

 

[gneill]

Looks fine.

 

[The Electrician]

It seems we can't do a nodal analysis because the place where we would want to have a node (the junction of the 32 ohm and the 3 ohm resistors) is shorted to ground.

A technique I've used in cases like this is to replace the wire carrying Ia with a resistor R. Then perform the nodal analysis and take the limit as R goes to zero.

Let the aforementioned junction be V1 and the right end of the 3 ohm resistor be V2 (also Vm). Then we can form two nodal equations and solve them with matrix algebra:

Vm is -2.8125 volts.