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Kirchoff's Loop Rule ∑ΔV=0 question

  1. Apr 22, 2015 #1
    Here is a portion of the explanation my book gives

    "Kirchoff's second rule follows the law of conservation of energy. Let's imagine moving a charge around a closed loop of a circuit. When the charge returns to the starting point, the charge-circuit system must have the same total energy as it had before the charge was moved. The sum of the increases in energy as the charge passes through some circuit elements must equal the sum of the decreases in energy as it passes through other elements."

    Why is the sum always zero in his second rule (∑ΔV=0)? Why can't the battery supply a potential difference greater than elements that cause a decrease (∑ΔV>0) ? Or why can't the elements cause a decrease in the potential difference that stops current pass a certain point (∑ΔV<0)?
  2. jcsd
  3. Apr 22, 2015 #2


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    If you stand on a hillside and walk up hill 10 m, then walk down hill 20 m, and finally walk back to where you started, what is your elevation? When you arrive back at the starting point, is it possible for you to be any higher (or lower) than you were when you were previously at that exact same point?

    There is your answer.
  4. Apr 22, 2015 #3


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    This is not a correct quotation of Kirchhoffs voltage law. Kirchhoff says (at least in Denmark):

    The sum of voltage changes through a closed circulationpath (loop) in a circuit = 0: (It is not through some circuit.)

    Of course that's true: Say you have a voltmeter (with two probes). You place one probe at a starting point, and move the other probe through the closed path, ending up with having the other probe at the starting point where the two probes are short-circuited, and the voltmeter will display 0V.

    Say that ∑ΔV>0 and the path is composed of resistors in series, then the surplus of ΔV would increase the loop-current until ∑ΔV=0.

    Kirchhoff had no intention of making something ingenious. He just systematized Ohms law, (V=R*I), by adding signs to the values/directions, so that voltages and currents could be calculated logically with signs. (computers are very happy for that).
    Last edited: Apr 22, 2015
  5. Apr 22, 2015 #4
    If the total voltage supplied was greater than the total voltage drop around a loop, then the total power supplied would be greater than the total power absorbed, and then you will be destroying energy, and thus you have defied the laws of physics.
  6. Apr 22, 2015 #5
    I always felt the Kirchhoff rules bordered the blatantly obvious. I mean, obviously the current rule must hold (unless you are able to create or destroy electrons!), and the voltage one, well, if you could trace around a full circle and end up with a different potential at the end point than where you started, clearly that would be a contradiction.
  7. Apr 22, 2015 #6


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    It should be noted, that KVL typically is used to calculate currents, and that KCL is used to calculate voltages.

    As such the laws becomes a little more non-obvious.
  8. Apr 22, 2015 #7
    I've really only used them in combination, to establish a set of equations that you can then solve.
  9. Apr 22, 2015 #8


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    Most people uses combinations and ends up with for example 5 equations. But using only KVL or KCL you can typically reduce this number to 2 or 3 equations, solving two loop currents by KVL: L1 and L2. To find the physical currents is a matter of some mental arithmetic in the form:

    I1 = L1, I2 = L2, I3= L2-L1, and so on, which can be seen in the drawing of the circuit. It's much faster and simple.

    How much time do you use to enter 5 equations on your calculator?
    Last edited: Apr 22, 2015
  10. Apr 22, 2015 #9
    My college was on the very theoretical side of things, so the important part was deemed having a fully determined set of equations :smile:
  11. Apr 22, 2015 #10


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    It's true, that computers typically creates a fully determined set, but that's because they cannot do mental arithmetic (idiots).
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