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Circuit analysis problem

  1. Dec 1, 2011 #1
    why the author is using negative sign there?

    r27omo.jpg

    And here the problem is different the author is using different current flow signs and one answer is positive and other is negative why is this so?

    See this?
    http://i42.tinypic.com/2rqdj5w.jpg
     
    Last edited: Dec 1, 2011
  2. jcsd
  3. Dec 1, 2011 #2

    Simon Bridge

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    Means that the current calculated points in the opposite direction to the arrow drawn on the diagram. Look at the current source - which direction is the current going to flow?
     
  4. Dec 1, 2011 #3

    berkeman

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    Because the direction of IL is defined in the diagram to be down, and the solution finds that the actual current is flowing up. So you indicate that with IL = -0.25mA.

    EDIT -- Beaten out by Simon! :smile:
     
  5. Dec 1, 2011 #4
    please see the second pic and reply please! it is contrary with the ist picture!
     
  6. Dec 1, 2011 #5

    berkeman

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    Same reason. Look at how the arrows are defined for those two currents, and look at how the actual current flows.
     
  7. Dec 1, 2011 #6
    direction of current in both pics is downward but in one pic there is negative sign and in other there is positive ?
     
  8. Dec 1, 2011 #7

    berkeman

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    Look more closely in the 2nd pic. I1 is up and I2 is down as drawn on the figure.
     
  9. Dec 1, 2011 #8
    understanding that.
    sir do you have any book or video that will clear my concept about current flow ? because it's very puzzling to me!
     
  10. Dec 1, 2011 #9

    berkeman

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  11. Dec 1, 2011 #10

    Simon Bridge

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    To me this is one of the beauties of the physics - you can draw the arrows on your circuits how you like and if you get it wrong, the math gives you a negative sign.

    I think there are two sources of confusion here;
    1. people get used to currents flowing clockwise around circuits so they end up going top-to-bottom through resistors - just turn the page around.
    2. the minus sign was just inserted apropos of nothing in the working ... probably from considering the current source to be negative. If the circuit was a single loop, the reason for the choice would have been trivially obvious. Of course it would also work if you actually compute the equivalent circuit, considering "up" to be positive for current sources.
     
  12. Dec 2, 2011 #11
    In the second diagram we can assume any direction of load current, and if the load comes positive than the assumed direction is correct otherwise it is opposite.
     
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