# Supermesh analysis w/ dependent sources

• rms5643
In summary, the conversation was about using mesh analysis to find the power delivered by a current-controlled voltage source in a circuit. The conversation included discussions on assigning loop variables and writing equations for the super mesh, source, and regular loops. There was also a clarification on the direction of current in the 32Ω resistor and the correct assignment of W in the equations. Finally, the solution to the system of equations was given.
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!

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.

Last edited by a moderator:
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?

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 = -Vx/8 but messed up when trying to use this. You have the right idea, but are making careless mistakes.)

Last edited:
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

rms5643 said:
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.

Please explain why W = Y*6

NascentOxygen said:
Please explain why W = Y*6

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

NascentOxygen said:
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)?

rms5643 said:
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.

## What is Supermesh analysis?

Supermesh analysis is a method used in circuit analysis to simplify the process of solving a circuit with multiple dependent current or voltage sources. It involves combining meshes or loops in a circuit that share a dependent source to create a "supermesh" and reduce the number of equations that need to be solved.

## When is Supermesh analysis used?

Supermesh analysis is typically used in circuits with dependent sources, such as transistors or operational amplifiers. It is also useful when analyzing circuits with multiple current or voltage sources that are connected in series or parallel.

## How is Supermesh analysis different from regular mesh analysis?

The main difference between Supermesh analysis and regular mesh analysis is that Supermesh analysis allows for the combination of meshes that share a dependent source, whereas regular mesh analysis treats each mesh as a separate entity. This makes Supermesh analysis more efficient and less time-consuming.

## What are the steps involved in Supermesh analysis?

The steps involved in Supermesh analysis are as follows:
1. Identify and label all the meshes in the circuit.
2. Identify any meshes that share a dependent source and combine them to create a supermesh.
3. Write equations for the supermesh and any remaining meshes.
4. Solve the equations simultaneously to find the unknown currents or voltages.
5. Check the solutions for accuracy.

## What are the advantages of using Supermesh analysis?

Supermesh analysis offers several advantages, including:
- Reducing the number of equations that need to be solved, making the analysis process more efficient.
- Allowing for the inclusion of dependent sources in circuit analysis.
- Simplifying the analysis of circuits with multiple current or voltage sources.
- Offering a more systematic approach to solving complex circuits.

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