Supermesh analysis w/ dependent sources

[rms5643]

Homework Statement

Use mesh analysis to find the power delivered by the current-controlled voltage source in the circuit in the Figure: http://i.imgur.com/LUXYFO5.jpg

Homework Equations

1 super mesh equation, 1 source equation, 1 mesh equation, KVL

The Attempt at a Solution

To start off, I made all of my currents flow counter-clockwise and I labeled them, starting from the bottom left loop, then bottom right, then upper left, then upper right loop as: W, X, Y & Z, respectively.

Next, I drew a super mesh equation around Y and Z (Upper left and upper right). My super mesh equation:

6*Y-X+8*Z=0

Next, I wrote down my source equation:

3=Z-Y

Finally, I did a regular loop around X (Bottom right):

6*X+X+32*(X+W/8)=0 (Note, I added W/8 instead of subtracted because I defined all my currents to go counter-clockwise, except, the current coming from the voltage-controlled current source is going the opposite way, so it would technically be 32*(X-(-W/8)). I'm not sure if this assumption is correct. I think my mistake may be there.

Next, since W is an unknown, I need one more equation. Since W is defined as the voltage drop across the upper left resistor, I wrote the following:

W=6*Y

Now, here I am also unsure whether or not this is correct. Since Y is entering the negative terminal of the resistor, or at least I've defined it to be, would that make W=6*-Y?

So, my two questions are whether W/8 is positive or negative in 6*X+X+32*(X(+/-)W/8) and whether Y should be negative or positive in W=6*(+/-)Y. (As well as my mesh loops, not quite sure I completely have the hang of mesh analysis)

Thanks for all of your help again!

 

[NascentOxygen]

Next, I drew a super mesh equation around Y and Z (Upper left and upper right). My super mesh equation:


6*Y-X+8*Z=0

The 6Y is a voltage
The -X is a current http://physicsforums.bernhardtmediall.netdna-cdn.com/images/icons/icon5.gif
The 8Z is a voltage. Why aren't these three terms all voltages?

The dependent voltage source has a gain of 15 volts/ampere. Therefore its voltage is 15X. Correct?

Next, I wrote down my source equation:

3=Z-Y

Right.

 

[rms5643]

Hey Nascent, that was my error. I don't know why, but I kept ignoring the 15. My equations should be

Supermesh: 6*Y-15*X+8*Z=0
Source: 3=X-Y
Voltage-controlled current source: W=6*7
Bottom right mesh: 6*X+X+32*(X+W/8)=0

Do these look correct?

 

[NascentOxygen]

i) right
ii) no, it's 3=Z-Y
iii) how did you get 6*7
iv) wrong

W is not a voltage drop; W is a current. (EDIT: you noticed W = -V_x/8 but messed up when trying to use this. You have the right idea, but are making careless mistakes.)

 

[rms5643]

Thanks for all your help Oxygen, I just noticed I switched up my loop variables when I redid some of the equations, namely ii), which should be 3=Z-Y
For iii), that should be W=Y*6
For iv) 6*X+15*X+32*(X-W/8)=0

Now, why is it 32*(X-W/8) and not plus? I defined my W current to go the opposite direction of X, but since I see the current source, don't I just make W negative? Or should I always just follow my loop direction designation and assume the math will work out in the end? (Just that W will be a negative value?)

When I solve the system, I get the following: (With 32*(X-W/8))
W=-19.972, X=-1.5079, Y=-3.398, Z=-0.3298

 

[The Electrician]

For iv) 6*X+15*X+32*(X-W/8)=0

I think this should be:

6*X+15*X+32*(X-W)=0

The current in the 32Ω resistor is just X-W.

 

[NascentOxygen]

Please explain why W = Y*6

 

[The Electrician]

Please explain why W = Y*6

I assume this question is actually directed to rms5643 rather than to me.

 

[rms5643]

Please explain why W = Y*6

Using Ohm's law V=IR, the voltage drop across the 6 ohm resistor is the current, Y, times the resistance, 6 ohms. I am attempting to use W as Vx in the figure and account for the 1/8 in my X mesh equation since it does not appear anywhere else.

Would it be more appropriate to assign W=(Y*6/8)?

 

[NascentOxygen]

Using Ohm's law V=IR, the voltage drop across the 6 ohm resistor is the current, Y, times the resistance, 6 ohms. I am attempting to use W as Vx in the figure and account for the 1/8 in my X mesh equation since it does not appear anywhere else.

Would it be more appropriate to assign W=(Y*6/8)?

It would be.