KVL question about signs

Discussion in 'Homework Help' started by hacker804, Mar 6, 2015.

  1. hacker804

    Thread Starter New Member

    Mar 3, 2015
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    Hi there I am having a little confusion about the circuit below.It is included as an example in my textbook:
    [​IMG]
    We need to solve for the current I.Assume a clockwise current and apply KVL around the loop:
    -Vx-12+8I+7I+4Vx=0
    -12+15I+3Vx=0
    Now the confusion here is that in the book it is written that Vx=-30I.I need to ask why the negative sign is included?Shouldn't it just simply be Vx=IR=30I??

    please help clear my confusion.The book makes no mention of why the -ve sign is there.
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Does the book state the nominal direction of the current I? This is an important point and perhaps the key to resolving your confusion.
     
  3. hacker804

    Thread Starter New Member

    Mar 3, 2015
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    The book does mention using the Passive Sign Convention.
    Can you please clarify how it applies here?
     
  4. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Conventional wisdom requires one to choose the direction in which the current is flowing. The choice can be either clockwise or counter-clockwise. For the purposes of analysis it doesn't matter which direction you choose but your equations arising out of that choice must be consistent with that choice.
    I suggest you firstly indicate on the schematic the current direction you have assumed as the basis for your equations. Then mark the nominal polarities on the passive elements for the current direction chosen.
    Then ask how the indicated polarity of Vx on the original schematic relates to the marked polarities you have added to the schematic. Also ask how the same voltage polarity would effect the dependent source.

    Returning again to your choice made for the direction of current flow. What might you conclude if you make a choice that results in a negative value for the current in the completed analysis?
     
    Last edited: Mar 7, 2015
  5. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    tnk,
    you missed this in the problem statement: Assume a clockwise current and apply KVL around the loop:

    A clock-wise current I enters the bottom of the 30Ω resistor.

    What is the voltage developed across the 30Ω resistor? Ohm says E=IR = 30I, but the polarity of the voltage drop caused by I is opposite to the labeled Vx, so Vx = -30I.
     
    Last edited: Mar 7, 2015
  6. WBahn

    Moderator

    Mar 31, 2012
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    Always annotate your diagrams so that all of your variables and terms are clearly defined, which include their polarity.

    KVL4.png
    This will not only communicate your thoughts and intentions to others, but will serve as a reference for YOU when doing the work.

    We want Vx and we know that Ohm's Law relates the voltage across the resistor to the current through the resistor but that the directions are not random -- the current flows from the more positive terminal to the less positive terminal. Vx is defined in the original problem as the voltage on the top terminal minus the voltage on the bottom terminal. Thus the current that Ohm's Law will yield is the current in the resistor flowing from the top terminal to the bottom terminal, which is equal and opposite to your current I. Hence the minus sign.
     
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