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Kirchoffs Current Law

  1. Dec 19, 2007 #1
    Is Kirchoffs current law basically saying the current through any point in a circuit is exactly the same and will remain the same as long as the voltage remains the same?

    Lets say I have a steady current running through a circuit will that current be exactly the same on one end or the current as it is on the other? If I was to put a resistor on one end of the circuit will the current still be the same on the resistors side as it is on the other side without the resistor?
  2. jcsd
  3. Dec 19, 2007 #2
    It's basically a conservation law.
    If a wire 'forks' into 2, the current going into this point is equal to the sum of the 2 currents leaving. (cars entering an intersection = cars exiting an intersection, regardless which street they take)

    "If I was to put a resistor on one end of the circuit will the current still be the same on the resistors side as it is on the other side without the resistor?"
    Assuming there are no losses along the wire, for a circuit with a (for simplicity dc) voltage source and a resistor in series, the current will be the same no matter where you measure it ('in front' of the resistor, 'behind' the resistor).

    If you draw a nice square box the circuit diagram, you can put the resistor on the opposite end of the voltage source, close to it, etc.
    If it has just this voltage source and a resistor you will measure the current the same at every point in the circuit
  4. Dec 19, 2007 #3


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    Current is continuous, at least for DC. When you talk about transmission lines, where the wavelength of the AC excitation is on the order of the length of the line or smaller, then you will obviously see variations in the current (and voltage) along the length of the transmission line. The answer to your question just depends on the size of the component with respect to the wavelength of the excitation.

    And KCL states that the sum of the currents leaving (or entering if you prefer) a node is zero. There is no accumulation of charge at a node.
  5. Dec 19, 2007 #4
    That explains it Curious. I read alot about series and parallel resistors but I can't get it to stick in my mind yet.

    So on a series circuit no matter how many resistors you put throughout the circuit the current will be exactly the same at whatever point you measure it. The voltage on a series circuit though will change depending on whether you measure it before or after a resistor. Lets say the power source initially supplies 10V to the circuit. After it hits a resistor the voltage will decrease a few volts so lets say theres 8V between resistor 1 and 2. Then there would be about 6V between resistor 2 and 3 and so on.

    I'll try and make a ****ty diagram of what Im thinking about.

    +____(10V, 2Amps)____R1____8V, 2Amps____R2_____6V, 2Amps_____-

    Is this correct?

    Then on a parallel circuit the current will be divided between each parallel resistor it runs through and the voltage will be the same throughout the whole circuit?

    One thing I was wondering though is when you have a circuit with 10 amps running through it you add 2 shorts to make a parallel circuit. Will the current on the main wire still be 10 amps while the two parallel wires in the middle have 5 amps or would the whole circuit become 5 amps?

    I didn't think about AC but I suppose you could measure the average of the fluctuations and the average would be the same throughout the circuit wouldn't it?
    Last edited: Dec 19, 2007
  6. Dec 20, 2007 #5
    The statement voltage between resistors doesn't make any sense to me.(i.e. the 8V)

    About the second part of your question. To me the current in the main line will continue to remain 10A.
    When you short two points what you are basically trying to do is get them to the same potential. But in the above given circuit there is a voltage source (due to which there is current) which will maintain a potential difference in the main wire.
    This is my view and am not sure about the same.
  7. Dec 20, 2007 #6
    I think you have a good understanding of the dc series resistive circuits, I've attached an (also crappy) picture to help a bit.
    That symbol the the left/bottom indicates where I've (arbitrarily) selecting my 'reference' voltage of 0 volts to be.

    If you provide an alternate path for the current, with a very small resistance (basically just that of the wire), almost all of the current will flow through this (less resistive) path. (of course this is relative to the resistance of the load that the 10Amps is flowing through, but assuming this is much higher than the wire)

    If the battery is happy to provide such power, the current through the short circuits will be very high (V=IR, if we short circuit the battery, R is very low (just a wire) and the current I will be high in an attempt to uphold the difference in potential across the battery).
    Such large currents can damage your battery or cause explosions (or so is commonly taught)

    I've attached a picture of what I believe you are trying to say here, where the two red wires are the two shorts you mention.

    AC circuits are a bit more complicated with things like capacitors and inductors (which are modelled very nicely under DC loads :))

    Attached Files:

    Last edited: Dec 20, 2007
  8. Dec 21, 2007 #7


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    The statement of Kirchoff's Current Law is just a statement of local conservation of charge. Charges can't just disappear. All of the charge entering a junction has to come out...it has to take one of the paths available to it.

    The statement that the current in a DC series circuit (i.e. a circuit consisting of a single closed loop) is the same at every point in the loop (i.e. it is a steady current) is, in my opinion, a slightly different idea. Think about it this way. Let's say for the sake of argument that the current is different at two different points in the loop. I.e., the current 'here' is fast, but the current up ahead is slow. What happens? Charges start to pile up! And what happens when charges start to pile up? Well, like charges repel. So the charges coming in hot from behind give the slowpokes up ahead a kick in their rear ends...and everything starts moving again. Due to the "self-correcting" nature of this system, the current tends to remain steady.

    Now as for your specific questions...

    To be absolutely clear, the current measured at any point in a series circuit will be the same as the current measured at every other point in that same circuit, regardless of the nature of the circuit (i.e. how many elements it has in it).

    I know that this is what you meant to say, but the way you worded it almost made it sound like you were saying that a series circuit with 3 resistors would have the same current as one with 10 resistors (but I know that's not what you meant).


    Umm...no? How could the voltage be the same throughout the circuit? There are still resistive elements in it causing voltage drops.

    I don't understand this...two shorts??? Anyway, if you add more branches, the current is being divided up more ways, so it stands to reason that the current in any given branch is lower..

    Yes, and the rules would hold true at any given instant as well (instantaneous current would be steady). Keep in mind that AC specifically means a sinusoidal voltage, which means that the average over one cycle would be zero. Instead, we measure (with voltmeters etc) the root mean square (RMS) voltage.
    Last edited: Dec 21, 2007
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