# Calculate the power consumed by dependent source

1. Jun 13, 2013

### ongxom

1. The problem statement, all variables and given/known data

calculate the power consumed by dependent source

2. Relevant equations

3. The attempt at a solution
The 10A source does not build a supermesh.
Loop 3:
(i3-i2).(1/10)=0 => i3=i2
Loop 2 :
i2=10(A)
i3=10(A)
vx.(10)=10=> vx=1(V) (node voltage)
also for loop 2 :
1/10(i2-i3)-3vx+(1/110)(i2-i1)=0
i1=-320(A)
To calculate the power consumed p=v.i=-3vx.i

I am not sure what current goes through the dependent source, i think it is (i2-i1)

Last edited: Jun 13, 2013
2. Jun 13, 2013

### Staff: Mentor

Is there something missing on the left side? A power supply?

Why should this be zero?
Where does that come from?
Right.

3. Jun 13, 2013

### ongxom

1) Nothing is missing.
2) Maybe no voltage source was on the left side.
3) Fix it, it should be vx.(1/(1/10))=-10 (cause the 10A source is going out the node).
Is the equation for Loop 2 correct ?

4. Jun 13, 2013

### Staff: Mentor

If there is nothing missing on the left then there is no loop 3, and no i3. Furthermore, the 10A source will fix the mesh current i2 at 10A (because the 10A source is in a portion of that mesh that is not shared with any other mesh and so that mesh current is the only current that flows in that branch making the mesh current identical to the current in that branch).

So you've got a value for mesh current i2 by inspection, and only have i1 to solve. Given that you know i2, what's Vx? Can you write KVL for the i1 mesh?

5. Jun 13, 2013

### ongxom

vx=-1 (V)
KVL for mesh 1 :
(1/110).(i1-i2)+3vx+(1/100).i1=0

6. Jun 13, 2013

### Staff: Mentor

Yes, that looks good. So what's i1? How about the net current flowing into the dependent source?

7. Jun 14, 2013

### ongxom

substitute vx=-1, i2=10A we have i1=161.9(A).
p=-3vx.(i2-i1)=-3.(-1).(10-161.9)=-455.7(W)
Is the result correct now ?

So according to the diagram, i3 should be eliminated and the above equation of mine for mesh 2 was incorrect, right ?

8. Jun 14, 2013

### Staff: Mentor

The number 455.7 W is good. One thing to verify is whether or not the source is actually producing or consuming power. Suppose you label the device with the actual potential across it. Is the current entering or leaving via the more positive terminal? A voltage supply that is producing power has current leaving via its + terminal.
Right.