ac power analysis, IrwinExt9.5

Discussion in 'Homework Help' started by PG1995, Jan 15, 2012.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
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    5
    Hi

    Please have a look on the attachment. Please help me with the query. Thank you.

    Regards
    PG
     
  2. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Help, please!
     
  3. Zazoo

    Member

    Jul 27, 2011
    114
    43
    Since one side of the branch is at ground (zero), the voltage at node A is the same as the voltage across the branch (19<72°V)

    i.e. V_{AG} = V_{A} - V_{G} = V_{A} - 0 = V_{A}
     
  4. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Thank you.

    But suppose you don't know the voltage across the branch AG. In the attachment in my previous I stated how I found the voltage across AG by starting at G and then proceeding toward A. I'm just interested to know if it is possible to find the voltage across the branch AG by starting at A and proceeding toward G, the way I did in the attachment. Do you get me? Please let me know if it's possible. Thank you.

    Best regards
    PG
     
  5. Zazoo

    Member

    Jul 27, 2011
    114
    43
    When you "travel" from G to A, you're basically completing a KVL loop that includes the unknown branch voltage (see the attached image):

    So KVL for this loop is: -25.4<45^{o} +12<0^{o} +V_{AG} = 0

    Applying the KVL loop in the opposite direction (A-to-G, clockwise) gives:
    -12<0^{o} +25.4<45^{o} -V_{AG} = 0

    The branch voltage is the same either way. You don't need to know the node voltage A (or even G)
     
    • KVL.jpg
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  6. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Thanks a lot, Zazoo.

    Best wishes
    PG
     
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