# HW question!!

#### Earless

Joined Aug 23, 2007
11 [/IMG]

The answer in the book V=14.4v, they got that answer cuz they used ix=-V/2000, i used ix=+V/2000.My question is that i thought i had to label the terminals so that the current would enter the + side of the circuit element so it could follow the Passive sign convention.It seems to me that the book labeled the 2KOhm resistor opposite to mine..why??, arent they supposed to label to satisfy to the PSC?

#### Distort10n

Joined Dec 25, 2006
429
The dependent current source, independent current source, and ix are entering the upper node. I would label these as positive, while i6 is leaving the upper node and is negative. The bottom node is the reference node.
However, ix is negative to begin with, so it would be adding a negative current. So the equation works out fine.
The current is negative because of the polarity as so defined across the 2k resistor.

#### Earless

Joined Aug 23, 2007
11
I thought the the current ix would be positive since its entering the 2K resistor thru the positive terminal (i labeled it that way), isnt that how the psc works, a current entering the + terminal gives a +V and a current entering a - terminal give a -V.

#### Distort10n

Joined Dec 25, 2006
429
You're being inconsistant with your polarities and/or direction of current flow. If you choose the bottom node to be the reference node, then how can you say that the current through the 6k resistor is positive (it is entering the bottom node), and the current through the 2k resistor is also positive when it is leaving the bottom node? KCL allows for great flexibility, but you are not remaining consistant in your analysis.

#### Earless

Joined Aug 23, 2007
11
Ok so the polarity across the 2K resistor is negative because its polarity is opposite to the polarity of V right?.Say i wrote the polarity of the 2K resistor like the V polarity, then v across the 2K resistor would be positive but ix would be negative and therefore i would still get a negative v is that correct?

#### Distort10n

Joined Dec 25, 2006
429
Basically yes in order to conform to KVL, but it is rather redundant since you would have to add a negative current. It is probably not the best practice to define voltage across the 2k resistor opposite in polarity since the bottom node is the reference node as it is for V.

If you defined the polarity across the 2k resistor as + to - rather than - to + then you can think of the voltage across the 2k resistor as a voltage rise rather than a voltage drop.

Ignoring the actual direction of current flow, KCL allows you to define currents to be positive or negative, entering or leaving. You can certainly say that ix is positive and entering the upper node, while i6 is negative and leaving the upper node as well.

As long as you are consistent, then the equation will work out.