Reading up on the E-Book. Kirchoff's Law

Discussion in 'General Electronics Chat' started by Yarry, May 13, 2009.

  1. Yarry

    Thread Starter New Member

    May 13, 2009
    2
    0
    Hello :)

    I was reading through Kirchoff's Law in the E-Book just now, and found myself confused.

    The confusion I'm refering to is about Kirchoff's Law:
    http://www.allaboutcircuits.com/vol_1/chpt_6/4.html

    What I don't get it why the current (I) is highest over the lowest resistance (R1) when dealing with a parallel circuit. According to a lecture I attended recently it should be infact just the opposite since in a parallel circuit the smallest resistor has the most impact which would lead the largest amount of the current over the largest resistor, and not the smallest.

    Is this a mistake, and if not how is this possible?

    Thanks for your time,
    Yarry

     
  2. hgmjr

    Moderator

    Jan 28, 2005
    9,030
    214
    An easy way to look at this is that when you have two resistors in parallel, the voltage across each resistor is the same. When you apply Ohms Law to each resistor you will see that the parallel resistor with the lowest resistance value will have the highest current flowing through it.

    hgmjr
     
  3. steinar96

    Active Member

    Apr 18, 2009
    239
    4
    You might be familiar with ohms law which states that V = IR. This applies to a single resistor though but still it says what needs to be said.

    I = V/R. Which means that the higher the resistance is the lower current flows.
     
  4. Yarry

    Thread Starter New Member

    May 13, 2009
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    0
    Yes, but I thought when speaking of a parallel system it was 1/R instead of R, since R Total is calculated differently than in a series circuit.

    Series Circuit: R Total = R1 + R2 + ... Rn
    Parallel Circuit: (1 / R Total) = (1 / R1) + (1 / R2) + ... (1 / Rn)

    This would make the formula I = V / (1 / R), meaning that the lowest resistor in the system would be the one with the smallest flow of current.

    My head is about to explode :confused:

    (Thanks for the fast replies btw. :), Yarry)

     
  5. Cabwood

    Member

    Feb 8, 2009
    20
    0
    That formula for calculating combined resistance should not be used for anything but calculating combined resistance. It does not pretend to help in any way to calculate current through or voltage across any individual resistor.

    You have assumed that the 1 / R references in the formula are somehow related to the current through them. They aren't, at least not as directly as you assumed.

    The formula can be applied to calculate the combined equivalent resistance of a group of resistors, whose value can then be used (with Ohm's law) to calculate total combined current through the entire group of resistors, but it does not help in calculating current through any individual resistor.
     
  6. hobbyist

    Distinguished Member

    Aug 10, 2008
    764
    56
    remember the formula is stated as Rtotal= 1/ (1/R1) + (1/R2)....(1/Rn)

    So if you used only 1 resistor 1/ (1/R1) you still end up with the original value.

    So as stated this formula is used to calculate total resistance that the circuit see's in its entireity as one resistor.

    But each individual resistor will be it's normal value when calculating current through it and voltage across it.
     
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