# Circuit with Two dependent voltages

#### RrD555

Joined Sep 8, 2011
8
I can get dependent source questions fine, but I think Va is throwing me off for this one.

. I tried using KVL which got me
26 + 2Va -2I - 4Vx - 9I = 0 and since Vx = 9I
26 + 2Va - 47I = 0
Then I thought Va = 2Va - 2I which means Va = 2I
Substituting in gives I = 0.60465A and Vo = 2.42V.

Of course, it's wrong. Either I messed up the signs or I completely screwed up with Va...

Any help is appreciated, thanks!

#### PRS

Joined Aug 24, 2008
989
Your loop equation is not consistent with respect to the polarities.

#### t_n_k

Joined Mar 6, 2009
5,455
You've probably assumed I as flowing anti-clockwise which would make Vx=-9I rather than just 9I.

Try that and see how it goes.

#### RrD555

Joined Sep 8, 2011
8
You've probably assumed I as flowing anti-clockwise which would make Vx=-9I rather than just 9I.

Try that and see how it goes.
Sorry for not mentioning, yes, i did draw the arrow in anti-clockwise direction.
As for overall sign, my prof said that as long as it's consistent within the equation/question, it doesn't matter.

I was thinking about that after seeing PRS' post...does that mean that the Va is -2I instead of 2I as well?

Edit: wait...it should be 2I? Completely thrown off now.

#### t_n_k

Joined Mar 6, 2009
5,455
Your original assertion that Va=2I was correct. Only your Vx relationship was incorrect - and then only in terms of the sign.

#### RrD555

Joined Sep 8, 2011
8
Your original assertion that Va=2I was correct. Only your Vx relationship was incorrect - and then only in terms of the sign.
Ok, tried again with Vx = -9I, got 3.59 as the solution. I originally thought it was +3.59V, but the solution was -3.59V...(messed up with sign again)

Is there a really good way to memorize what's + and what's -? I always think too much about which pole is which and current flowing from which direction gives what voltage sign and end up guessing...

#### t_n_k

Joined Mar 6, 2009
5,455
Did you find I was negative?

I found I=-26/29 A.

Hence Va=2I=2*(-26/29)=-3.59V

Note:

That negative sign for the current I just tells me that the choice of anti-clockwise conventional current flow was incorrect. The positive conventional current flow in the circuit is actually clockwise.

#### RrD555

Joined Sep 8, 2011
8
Did you find I was negative?

I found I=-26/29 A.

Hence Va=2I=2*(-26/29)=-3.59V

Note:

That negative sign for the current I just tells me that the choice of anti-clockwise conventional current flow was incorrect. The positive conventional current flow in the circuit is actually clockwise.
Yep I did get that...but I always look at the sign at the end because I could've assigned the sign the other way around at the beginning.
This was how the prof taught it, but maybe this is not a good practice?

Basically I changed the I clockwise, saw that positive current was going into 2Va at positive terminal. I always get confused if that's positive or negative here...

#### justtrying

Joined Mar 9, 2011
439
Yep I did get that...but I always look at the sign at the end because I could've assigned the sign the other way around at the beginning.
This was how the prof taught it, but maybe this is not a good practice?
That is what is taught and is good practice. I find it best to assign direction based on the power source(s) polarity. The key is consistency. What is great about this method is that the solution will tell you if your current direction assignment was correct or not (as was seen in the problem). This comes in handy when doing things like Mesh analysis where loops are many.