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Kirchhoff's Law

  1. Aug 26, 2013 #1
    When you start going around a loop to try and determine whether to add or subtract each voltage in that loop, how do you decide which?

    For example see the diagram just above Q7. http://www.facstaff.bucknell.edu/mastascu/elessonsHTML/Basic/Basic5Kv.html

    Starting from any point and going around, how do we find which terminals to assign + and which to assign - for, for each component?
     
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  3. Aug 26, 2013 #2

    Simon Bridge

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    1st step - put current arrows on the wires around the loops.
    2nd step: put voltage arrows on the components: the rule is, the arrow points the opposite way to the current arrow that goes through it. (Except for voltage sources - they point in the direction of the current.)

    When you apply the loop rule - run your finger around the loop whichever way you want - if your finger goes in the directon of a voltage arrow, you add it, if it goes opposite the direction you subtract it.

    It does not matter which direction you pick for the current arrows in step 1 - though it helps to make sure they don't all point into or out of a node. If you got the direction wrong, you'll just get a negative number.
     
  4. Aug 26, 2013 #3
    You have to make a decision for each element in the circuit and stick with it. It doesn't matter which way you choose (except for the battery) as long as you are consistent.
     
  5. Aug 26, 2013 #4

    Simon Bridge

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    eg. this example:

    Basic4Ki1Ckt.gif

    Label nodes:
    between R1 and R2 = n1
    between R2 and gnd = n2

    current arrows:
    I1 up from the battery and down into R1 and down into n1.
    I2 to the right out of n1, through R3 and down through R4, right into n2.
    I3 down through R2 and down into n2.

    voltage arrows:
    V: up through the battery
    V1: up through R1 (because current goes up and it's a battery)
    V2: up through R2 (because current goes down)
    V3: right through R3 (because current goes right)
    V4: up through R4 (because current goes down)

    So the left hand loop, starting with n1 and going clockwise ...
    -V2+V-V1=0 see?

    If you go anticlockwise, you get:
    V1-V+V2=0 ... which is the same thing.

    For the second loop - starting at n1 and going clockwise:
    -V3-V4+V2=0 ... because you traverse the loop against the arrows of R3 and R4, but along the arrow in R2.
     
  6. Aug 28, 2013 #5
    Thanks, what you wrote makes sense. But it is in disagreement with the equation listed on the website, which is -V2 - V3 + V4 = 0. The sign that appears to be taken on V3 doesn't make sense. That's why I'm wondering if there's a problem with the understanding you just suggested.
     
  7. Aug 28, 2013 #6

    sophiecentaur

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    Whatever sign you choose to assign to the PDs, initially, the actual correct values (with sign) will come out of the calculation. i.e. if you choose your 'Voltage Arrow' to be one way then the calculation will give you a positive or negative result and make up for what you may regard as a 'mistake' in your original choice. This is true for all vector calculations, in fact.
     
  8. Aug 28, 2013 #7
    -V3-V4+V2=0 and -V2 - V3 + V4 = 0 are fundamentally different equations, they cannot both be correct as far as I can see. that's a mathematical observation not a physical one.
     
  9. Aug 28, 2013 #8

    sophiecentaur

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    It only depends upon how you define your V1 V2 V3 V4. i.e. how you connect your virtual meter across each resistance in the circuit. (The V's in each of the equations you have written, above, are not the same - they can have different signs). Once you have decided on that, the sums will tell you whether the volts you will measure will be + or -, on each meter. If it works for one resistor, it will work for a circuit with a hundred resistors in it.
    Let's face it- we are looking for a reason why you have got it wrong and not a reason to find Kirchoff to be wrong. Approach this with that thought in mind. :wink:
     
  10. Aug 28, 2013 #9

    Simon Bridge

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    If you have a look at my notes about the current going in and out of n1 again - you'll see that I made an arbitrary choice in the directions of two of the current arrows. Change that choice and you'll get the model answers.

    Kirkiffs loop law does not work all by itself - it is part of constructing a set of simultaneous equations.
    What you have discovered is that there is more than one set of simultaneous equations that will give you the right answers.
     
  11. Aug 29, 2013 #10
    Ok, so let's say I label the end of a component where the current goes in as - for a battery and the component where current comes out as + for a battery, and the opposite for everything else. I then define a certain loop and pick an arbitrary point on it and begin to go around the loop in a certain direction. Any component where I first cross the end marked + is counted as positive (let's say); any component where I first encounter the end marked - is counted as negative. Let's say all voltages are given to me as magnitudes. Then I sum up the magnitudes from the ends marked + and subtract from them the magnitudes from the end marked -, for that loop. Is this right? The problem with that definition would seem to be component (3) in SimonBridge's example, because if the current enters through the left end of it, then the left end should be marked + and the right end marked -, whereas in the diagram on the webpage it is the opposite.
     
  12. Aug 29, 2013 #11

    sophiecentaur

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    The point is that you can't necessarily know which way the current is going to flow, just by looking at the circuit. You Have to choose an arbitrary 'sense' to your voltage variable in order to enter it into an equation. When the equation has been solved, you will find the V may be positive or negative. (BTW it will work for a single battery and a single resistor!!)
    You are assuming too much in what you say. However you mark the voltages, initially. Kirchoff will give you the same actual answer and will tell you which direction the current will flow. If you don't believe it then you just have to try it. there are plenty of examples of Kirchoff exercises around.
    As I said before - you are wrong and you need to prove that to yourself as you cannot believe what we are telling you. Put your money where your proverbial mouth is and try to prove it "won't" work. Forget about your intuition.
     
  13. Aug 29, 2013 #12

    analogdesign

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    Big Daddy,

    When I was learning this stuff, one way I really made it stick was to take a few problems and arbitrarily choose different nodes as "Ground". Even though the equations sometimes would change a good bit, the final answers were always consistent. I would follow sophiecentaur's advice and work some problems.

    Learning electronics is kind of like joining the Marines. You have to let go of your pre-conceived intuitions and start acquiring new ones that are empirically based. Just like in basic where your drill instructor has to tear you down before you can be built up again.
     
  14. Aug 29, 2013 #13
    The only inconsistency is resistor (3). Or are you saying that if I use -V3 -V4 +V2 = 0 instead of -V2 - V3 + V4 = 0, along with the other equations, I would get the same magnitudes of voltage?

    It's not that I don't believe you. I am just trying to check I understand what you're saying correctly.
     
  15. Aug 29, 2013 #14

    Simon Bridge

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    That's pretty much it - put some arbitrary values in the diagram and work out the voltages and currents by different schemes. Then you'll see for yourself and it is not a matter of belief.

    You are going to have V3=I2.R3 notice - so it is the current direction that counts. (you may want to change the labels so I3 goes through R3?) The voltage is positive if you traverse the components against the current flow because that direction is "uphill".

    Perhaps I should mark-up that diagram to show you how it can go?
     
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