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Kirchoffs Voltage Law Understanding

  1. Nov 21, 2015 #1
    << Mentor Note -- Thread moved as requested >>

    I am trying to complete a mock paper for an upcoming assessment. The mock comes with the answers supplied in bold under the question yet im having trouble understanding how the answers are obtained. I know that the law states that the sum of all emfs equals all the sums of the voltage drops and i have looked at several online articles but they seem to be very complex for what I need to understand. Im hoping someone can explain using the example I have. Im not sure how the arrows play a part in this either

    EDIT: After much banging my head i have come up with the following.Can someone comment if this is correct and if there is a more excepted way of writing it out:

    ΣV=0
    B-E = -2V
    E-D = +10V
    D-C = -5V

    BE+ED+DC = +3
    ∴ C-B = 3V

    And

    ΣV=0

    B-E = -2
    E-F = +8
    A-B = 0

    BE+EF+AB = +6
    ∴ F-A = 6V


    IMG_6241.jpg
     
    Last edited: Nov 21, 2015
  2. jcsd
  3. Nov 21, 2015 #2

    gneill

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    Staff: Mentor

    The arrows depict a change in potential, increasing towards the arrow head. So for example, the arrow on the battery in the left bottom corner depicts an 8 V rise in potential across the battery going from node E to node F. Similarly, on the resistor connected from E to B there's a 2 V potential increase.

    Kirchhoff's voltage law states that the sum of the potential changes around a closed path is zero.
     
  4. Nov 21, 2015 #3
    Ive updated what i think is correct! If anyone can confirm
     
  5. Nov 21, 2015 #4

    gneill

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    Staff: Mentor

    Looks okay.
     
  6. Nov 22, 2015 #5

    CWatters

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    Science Advisor
    Homework Helper

    gneill has already mentioned this but it's far better to think...

    and start by writing equations that sum to zero. For example you wrote..

    BE+ED+DC = +3

    I find I make fewer mistakes with the signs if I specify where I'm starting and write my equation to sum to zero like this..

    Going anticlockwise from B..
    BE + ED + DC + CB = 0

    and then rearrange it to solve for the unknown CB.
     
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