help on an exercise

Discussion in 'Homework Help' started by yiannistamv, Nov 9, 2014.

  1. yiannistamv

    Thread Starter Member

    Jul 23, 2014
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    can please someone help me (see attached file)
     
    • 2.pdf
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  2. MrChips

    Moderator

    Oct 2, 2009
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    There is no equation for v1 and v2.
    You need to write down equations based on Kirchhoff current and voltage laws.
     
  3. yiannistamv

    Thread Starter Member

    Jul 23, 2014
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    ok the equation in the pdf are for corrents v/r=i ok but at the first is v1-v2 and at the second equation is v2-v1 why that i am trying to solve such exersices and i always see that part in a wrong way can you point me my mistake because i would have writen and in the second equation v1-v2
     
    Last edited: Nov 9, 2014
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    When we write the nodal equations for a given node we assume that this node has the highest voltage in the circuit. Or we assume that all current leaving from the node. And this means the highest voltage in the circuit. So the end result is the same.
    So for V1 node all currents leaving the node so:
    KCL 2.4A + V1/125 + (V1 - V2)/25 = 0 (zero currents entering the node)
    And for node V2 we do the same thing (V2 node has the highest voltage in the circuit)
    V2/250 + V2/375 + (V2 - V1)/25 = 3.2A
    Or
    V2/250 + V2/375 + (V2 - V1)/25 - 3.2A = 0
    We give minus 3.2A because 3.2A current source entering the node.
    http://www.wolframalpha.com/input/?i=2.4 + V1/125 + (V1 - V2)/25 = 0,V2/250 + V2/375 + (V2 - V1)/25 - 3.2 = 0
     
    Last edited: Nov 9, 2014
  5. yiannistamv

    Thread Starter Member

    Jul 23, 2014
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    thank you for your time but as i understand is the current discribed as vi-v2/25 is a leaving from the first node when it is entering in the second node is -(v1-v2)/25 which is equal to v2-v1/25 taking the minus on the enumarator is that right? (and sorry but i did not understand your link)
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    We treat each node independently, so when we write KCL for V2 node we don't care about V1 node current directions.
    We simply assume that for a given node all currents will leaves the node (Vx node has the highest voltage in the hole circuit).
    So because of this V2 is first because the current flow from high potential to low potential.
    But when we do KCL for V1 we assume that V1 has the highest voltage in the hole circuit.
     
  7. yiannistamv

    Thread Starter Member

    Jul 23, 2014
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    you mean that every time we write the equation for the node (any node in the circuit) we ALWAYS assume that all currents are leaving the specific node?
     
  8. upopads2

    Member

    Sep 20, 2014
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    You define your conditions. If you consider leaving to be positive then it is, but you must be consistent.
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes exactly, this is what I was trying to say
     
    yiannistamv likes this.
  10. yiannistamv

    Thread Starter Member

    Jul 23, 2014
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    thanks you are really a help jony130
     
  11. MrChips

    Moderator

    Oct 2, 2009
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    The correct answer is no.

    You can draw the current in either direction. After you have solved for the current, if the answer is negative then your initial assumption on the direction is the reverse direction.
     
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