Passive Sign Convention (Specific Issue)

Discussion in 'Homework Help' started by Jake1234, Jan 27, 2010.

  1. Jake1234

    Thread Starter Member

    Oct 14, 2007
    19
    0
    I know there are plenty of posts on passive sign convention (PSC) in the forums and I've looked at them, but this deals with a specific on PSC that's really bothering me.

    I searched PSC on Google and the first hit is this wikiuniversity link that describes it and gives a few exercises. The link is here: http://en.wikiversity.org/wiki/Passive_sign_convention

    In particular, I was observing Example 1.3 where they give the current as entering the negative terminal and ask to calculate the power with both the voltage and current negative. Now, I know they say it's a resistor and all resistors dissipate power (i.e. power is positive) however, that is beside the point I want to make here.

    In the example they show the diagram as following active sign convention, but they don't add the sign to the power equation when they calculate power and they end up calculating it as it should be (with current in the opposite direction). They explain the logic and I understand it, but shouldn't you make all calculations based on what the diagram says? So, sense the diagram doesn't follow PSC, shouldn't the power be negative i.e. p = -v*i?

    I have a program called circuit tutor written for a circuits class (here's the site for validity reasons: http://ewh.ieee.org/soc/es/Aug1996/002/cd/ms.htm) it's an old software program that basically generates circuit problems. However, I've been using the following mental process (Follows PSC, don't add sign, p=vi OR Doesn't follow PSC, add sign, p=-vi) and I've been getting the problems right. A completed problem is attached.

    I just need someone to verify this confusion. Because I've always been confused by PSC because every time I try and figure it out, I get in a confusing situation like this where theory isn't meshing with application.

    Thank you, I appreciate your help if given.
     
  2. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
    16
    Passive Sign convention states that "A positive current flows from + to - across a resistor, forming a voltage drop across that resistor." Typically the procedure is to have the current direction flowing the opposite direction of the voltage polarity. In other words, the arrow of the current flows into the resitor on the + side. However, it doesn't always have to be the case. We just "assume" it is in that direction. See this link:

    http://www.ent.ohiou.edu/~manhire/basic_ee/chapter02/sld018.htm

    In your attachement, let's solve for V1 in your first loop. In element two, they have -30V, and in element 3 they have 20V. Notice how element 1 and 2 has the polarity from + to -, thus adding them together nicely? We always want a positive end to go to the negative end. Otherwise, if I had two 1.5V batteries and the negative or positive ternimals were connected, I would get 0V. But now you'll also see that V2 isn't nicely configured in the polarity we would like it to have; it's opposite. So that means we'll have -V2. Thus forming the equation: V1 + V2 - V3 = 0, according to KVL. So now we can implement our true voltages:

    V1 + 20 - (-30) = 0 which simplifies to V1 + 50 = 0 and finally V1 = -50V.

    If you have any further questions, I'd be glad to assist you.

    Austin
     
    Last edited: Jan 31, 2010
  3. ELECTRONERD

    Senior Member

    May 26, 2009
    1,146
    16
    For power, if power is dissipated that means it is negative. So if we have a power source that supplies 50W and it goes into a resistor which dissipates 20W, that means -20W is being dissipated. So 50 - 20W = 30W left of power.

    The key terms you'll have to learn is "supplied" and "received".

    Austin
     
  4. Jake1234

    Thread Starter Member

    Oct 14, 2007
    19
    0
    I thought resistors were suppose to dissipate power though. If power dissipation is negative, then how do you explain the power calculations for a resistor being I²R and V²/R? These, mathematically speaking, will never be negative due to the square.

    Sourced power is suppose to be negative because of active sign convention, i.e. current is moving into the negative terminal. Any component that follows ASC is a source or supplies power to the circuit. Anything that follows PSC, like resistors, are suppose to absorb or dissipate power. Hence the power is positive. But I guess it doesn't really matter if you're consistent.

    However, I was mostly confused on example 1.3 in the wiki link I posted http://en.wikiversity.org/wiki/Passive_sign_convention

    I mean, clearly the current is entering in the negative terminal, meaning that the power equation should be p=-(v * i) right? However, they fail to add the additional sign and that's what is confusing me the most.

    Thank you for your reply.
     
  5. Jony130

    AAC Fanatic!

    Feb 17, 2009
    3,962
    1,097
    Power in the resistor is always positive, resistor always dissipate power and never supply the power.
     
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