Mesh Current Analysis with Dependent Voltage Source

[grekin]

Homework Statement

Use the mesh-current method to find the power developed in the dependent voltage source in the circuit in figure (Figure 1) if v = 29V.

Homework Equations

∑R in mesh 1 * i_1 - ∑R in common between meshes 1 and 2 * i_2 = v_source in mesh 1
∑R in mesh 2 * i_2 - ∑R in common between meshes 1 and 2 * i_1 = v_source in mesh 2
P = i*v

The Attempt at a Solution

Putting mesh 1 on the top, mesh 2 on the left, and mesh 3 on the right, I have:

Mesh 1:
(3+5)*i_1 - 3*i_2 -5*i_3 = 53*i_delta

Mesh 2:
(3+20+7)*i_2 - 3*i_1 - 20*i_3 = 29

Mesh 3:
(5+20+2)*i_3 - 5*i_1 - 20*i_2 = 29

Aux:
i_2 = i_delta + i_3

Solving, I got i_1 = 2.09, i_2 = 4.25, i_3 = 4.61, and i_delta = -0.360

V_dependent = 53*i_delta = 110.77

P = i*V = 2.09 * 110.77 = -39.8 W (incorrect)

Any ideas?

EDIT: Solved it, just had a couple signs of the source voltages mixed up. P=43582.5 W

 

[collinsmark]

Homework Statement

Use the mesh-current method to find the power developed in the dependent voltage source in the circuit in figure (Figure 1) if v = 29V.

Homework Equations

∑R in mesh 1 * i_1 - ∑R in common between meshes 1 and 2 * i_2 = v_source in mesh 1
∑R in mesh 2 * i_2 - ∑R in common between meshes 1 and 2 * i_1 = v_source in mesh 2
P = i*v

The Attempt at a Solution

Putting mesh 1 on the top, mesh 2 on the left, and mesh 3 on the right, I have:

Mesh 1:
(3+5)*i_1 - 3*i_2 -5*i_3 = 53*i_delta

Mesh 2:
(3+20+7)*i_2 - 3*i_1 - 20*i_3 = 29

Mesh 3:
(5+20+2)*i_3 - 5*i_1 - 20*i_2 = -29

Aux:
i_2 = i_delta + i_3

Pay a bit more attention to the polarities (i.e., "signs"). I've marked a couple above in red for investigation.

I noticed that for each of your mesh equations, on the left hand side, you have used a positive sign if the current is flowing from a higher to lower potential (from + to -). Which is fine; there's nothing wrong with that.

But that means you must do the opposite for the right hand side of the equation: Use a negative sign if the current flows from higher to lower potential (from + to -), on that side of the equation.

If it helps to understand why, you could rewrite all of your equations such that everything is on left hand side, and everything sums to 0. (i.e. zero is the only thing on the right hand side).

The important thing is that all voltages in a given loop must sum to zero; the potential at the end of a loop must equal the potential at the beginning of the loop since the loop's beginning and end are the same thing. Then use whatever convention you want after that. But in a couple of your mesh equations, you did not stick to a consistent convention.

 

[grekin]

Pay a bit more attention to the polarities (i.e., "signs"). I've marked a couple above in red for investigation.

I noticed that for each of your mesh equations, on the left hand side, you have used a positive sign if the current is flowing from a higher to lower potential (from + to -). Which is fine; there's nothing wrong with that.

But that means you must do the opposite for the right hand side of the equation: Use a negative sign if the current flows from higher to lower potential (from + to -), on that side of the equation.

If it helps to understand why, you could rewrite all of your equations such that everything is on left hand side, and everything sums to 0. (i.e. zero is the only thing on the right hand side).

The important thing is that all voltages in a given loop must sum to zero; the potential at the end of a loop must equal the potential at the beginning of the loop since the loop's beginning and end are the same thing. Then use whatever convention you want after that. But in a couple of your mesh equations, you did not stick to a consistent convention.

Thanks for the response, but it seems you opened the page before I made my edit. I figured it out after looking over my equations for a bit, thanks anyway though. You are indeed correct.

 

[collinsmark]

Thanks for the response, but it seems you opened the page before I made my edit. I figured it out after looking over my equations for a bit, thanks anyway though. You are indeed correct.

Right. I didn't see the edit.

Well, good job then!