KVL orientation question

Discussion in 'General Electronics Chat' started by Transatlantic, Apr 21, 2016.

  1. Transatlantic

    Thread Starter Member

    Feb 6, 2014
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    upload_2016-4-21_13-31-31.png

    I was following an example that was using KVL to calculate the Voltage across A/B.

    The way they did it was to assume the current was flowing like this :

    upload_2016-4-21_13-32-55.png

    Vr1 = 4.5v (using the R1/R2 divider)
    Vr3 = 13.5v (using the R3/R4 divider)

    And then calculate the top right loop using KVL, going clockwise :

    Vr3 + Vab - Vr1 = 0
    Vab = -Vr3 + Vr1
    Vab = -13.5 + 4.5
    Vab = -9v

    I've been reading though, that if you don't know the current direction, you can just assume a direction, and that as long as you stay consistant, the value will still come out correct. You'll just have to remember the direction is actually flipped if you get a negative current.

    So I thought I'd try a different current direction.

    upload_2016-4-21_13-43-32.png

    But I get a different result using KVL :

    Vr3 + Vab + Vr1 = 0
    Vab = Vr3 + Vr1
    Vab = 13.5 + 4.5
    Vab = 18v

    Have a misunderstood something here? .. I appreciate that in the above example the currenly is clearly going the other way, but I wanted to just try a different direction as I understood the direction could be arbitrary?
     
  2. ericgibbs

    Senior Member

    Jan 29, 2010
    2,499
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    hi T,
    Why is that upper arrow in your circuits at that location.?
    Move the arrow to the left of node 'X', is the assumed direction of current flow consistent for all the circuit.????

    E
     
  3. Picbuster

    Member

    Dec 2, 2013
    373
    50
    Calculating this type of circuits;
    always calculate from one pole of battery normally the minus pole.
    Volt A from -batt = R2 * 18/(r2+r1)
    Volt B from -batt = R4 * 18/(r4+r3)
    The voltage between A and B= |(Volt A-volt B) I took the absolute value reason volt A is, in this case, higher then Volt B but don't know where A and B are connected to so I defined the potential difference.

    Picbuster
     
  4. Rodney Phillips

    New Member

    Jun 12, 2015
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    Vab would be equal to the sum of Vr2 and Vr4, not Vr1 and Vr3.
     
  5. Dodgydave

    Distinguished Member

    Jun 22, 2012
    4,969
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    6/8 x 18v = A

    3/12 x 18v = B,

    subtract the difference, 9V,,, simple ohms law.
     
  6. Transatlantic

    Thread Starter Member

    Feb 6, 2014
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    I think you missed the point. I was trying to calculate it using KVL, which lead to the current direction question.
     
  7. Dodgydave

    Distinguished Member

    Jun 22, 2012
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    I think you missed the point, A is more positive than B,.
     
  8. dl324

    Distinguished Member

    Mar 30, 2015
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    Where in KVL does it state you can assume arbitrary current directions?
     
  9. Transatlantic

    Thread Starter Member

    Feb 6, 2014
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    For example here http://hades.mech.northwestern.edu/index.php/Kirchhoff's_Current_and_Voltage_Laws

    "The directions for the currents were chosen arbitrarily. Keep in mind that there is no "wrong" direction when choosing the direction of the arrow, as long as we are consistent with each current's direction."

    It is my understanding that when using KVL to sum the voltages around a loop, you need to know whether to add or subtract the voltage at a certain segment in the loop.

    "It doesn't matter if you go clockwise or counter-clockwise. If you go from a "+" sign to a "-" sign when going around the loop, subtract the voltage across that element. Otherwise, add it. Once you've completed the loop, set the summation equal to zero".

    I thought you used the current direction to figure out if it's a + or - you are going through.
     
  10. dl324

    Distinguished Member

    Mar 30, 2015
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    The equations should be the same regardless of direction. The second loop equation is correct. The first has a minus sign that you can't assume.

    upload_2016-4-21_10-16-0.png
     
  11. dl324

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    Mar 30, 2015
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  12. Transatlantic

    Thread Starter Member

    Feb 6, 2014
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    But how have you decided on the orientation of the '+' '-' at each point in your example? Thats the part I don't understand.

    I thought you work them out from the current direction?
     
  13. dl324

    Distinguished Member

    Mar 30, 2015
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    I used the voltage divider equation to calculate the voltage across R2 and R4. That established the voltages at A and B.
     
  14. Transatlantic

    Thread Starter Member

    Feb 6, 2014
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    [​IMG]
    But take an example like that. Why have they chosen the + to be on the left of R2? Why isn't it on the right and the - on the left? How is it determined?
     
  15. dl324

    Distinguished Member

    Mar 30, 2015
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    They made arbitrary current direction assignments and that resulted in the polarity of the drop across R2 being what it is.
     
  16. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    The current directions determines the voltage polarity across R2 resistor (across all resistors).
     
  17. Transatlantic

    Thread Starter Member

    Feb 6, 2014
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    Which is what I thought I did? .. I made an arbitrary current direction.

    [​IMG]

    Vr3 + Vab + Vr1 = 0
    Vab = Vr3 + Vr1
    Vab = 13.5 + 4.5
    Vab = 18v

    ..and get the wrong answer. This is my confusion.
     
  18. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    But how did you solve for VR1 and VR2 ??
     
  19. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Also did your solution for VR1 is consistent with your assumed current direction ? The answer is NO.
    Because if the current is flow from point A toward upper R1 terminal VR1 is negative (-4.5V) therefore
    Vab = -Vr3 - Vr1 = 13.5V - (-4.5V) = -9V
     
    Last edited: Apr 21, 2016
  20. dl324

    Distinguished Member

    Mar 30, 2015
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    If you determine the currents at the top node, you'll find that that gives a negative current for the direction you assumed through R1 and that will reverse the polarity of the voltage drop.

    And you still need to show how you determined the voltage drops.

    To do it with KVL and KCL, you need to setup equations for KCL and solve the simultaneous equations.
     
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