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Resistors in Parallel - why the 1/R?

Zs
Hello, Genuinely, I hope this finds you well.


The electrical lessons which I am working on from home are going down well, despite my dodgy diagrams and rather plain English.  They have actually been rolled out to more people which I'm pleased about.  Anyway,  and after each one I get some comments or questions.


in the last one, I covered calculations for resistors.  You know the one for resistors in Parallel which is 1/R = 1/ value of resistor1 + 1/ value of resistor 2 +... and so on...?


Well, one of our ladies has suggested that it would be nice to know the derivation of the 1/R aspect, why it is divided into 1. I expect she already knows because she is a most interesting physicist and I have no doubt her maths is superb.  OMS will have met her.  Good question though. 


Truth is, I'm not sure and I have always rather taken it for granted as something I learned. Care to enlighten me? ...all I can think of is Kirchoff but I would love your advice if you have some time.


Thank you, stay well eh?  

Zs


  • Former Community Member
    0 Former Community Member
    Its maths and you are dealing with what is called a 'reciprocal'.


    Lets take the number Rt = R/1 and dividing by 1 does not change the value so this is a true statement.


    So the 'reciprocal' of this equation is to take the numerator and denominator and flipping them upside down so the denominator becomes the numerator and the numerator becomes the denominator, hence the 'reciprocal' of Rt = 1/R.


    Rob





  • Former Community Member
    0 Former Community Member
    Does it help at all to think of it as adding the resistances together when they are in series but the conductances when they are in parallel? The final 1/ just converting the answer back to resistance. Conductance just being the inverse of resistance, more conductance being more conductive.
  • Former Community Member
    0 Former Community Member
    Here is an extra quote for you and I hope you impress your physicist friend,

    The equivalent resistance of a number of resistors in parallel can be found using the reciprocal of resistance, 1/R. The reciprocal of the equivalent resistance is equal to the sum of the reciprocals of each resistance. The unit of resistance is the Ohm (Ω), which is equal to a Volt per Ampere (1 Ω = 1 V/A).


    Rob
  • 0
    Well, one of our ladies has suggested that it would be nice to know the derivation of the 1/R aspect, why it is divided into 1

    If it helps at all, I think of resistors in parallel means that the currents through each of the resistors simply add up to the total. Current through each individual resistor is inversely proportional it's resistance - as R increases I decreases (1/R).


    Or if you want to think of it all mathematically, the current for the whole lot Itotal = the sum of the currents through each of the resistors I1 + I2 etc.


    Remembering that I=V/R, that's the same as saying:


    Vtotal/Rtotal = V1/R1 + V2/R2 etc


    But as all the resistors have their connections in common, all the voltages are the same, so we can cancel them out, which leaves us with:


    1/Rtotal = 1/R1 + 1/R2 etc.


    With resistors in series the resistance simply add up and it's the voltages across them that are proportional to their individual resistances.


    Does that help at all?


       -  Andy.
  • 0
    Perhaps curiously, for I am an electronics guy at heart and do a lot with Rs and Cs in parallel and series, I have never considered there to be a special formula, though now I look around, indeed there it is.

    For things in parallel they all see the same voltage, so we just add the currents to get a total.

    For things in series, they all pass the same current, so we add the voltages.

    The when converting back to an equivalent single value of Z, as we only need a ratio of V to I,  the unknown voltage or current just cancels.

  • 0
    Thank you,  You have shed some light and I'll try to put that into an example. Love the quote and I shall walk around the empty house saying that until I am word perfect ;) 

    If you don't mind, I will pass the new explanation in front of you once I've got it down.


    Lady Physicist has come back with a photo of her notes on the subject.  I am sure it's perfect but I can't read her writing!


    Mapj - because we have to calculate all the parallels before treating everything as series, I too have never paid much attention to this.  To be honest, this was put away some time ago in my mental archive.  In taking the readers of these pages of mine back to basics I am doing the same to myself and rather enjoying it.


    BBC radio 4 tonight at 9.30   In Our Time is on the subject of Gauss.  I caught it this morning - fantastic.  I recommend. 


    Must go - today it's an easier one - Watts and kilowatt hours etc. 


    Thank you, Zs


    Struggling to post this reply - you might get several.
  • 0
    Zs 


    This is how I look at it.

    By Kirchoffs, in a parallel the total current is the sum of the currents in the individual resistors.

    I = IR1 + IR2 etc                                             (1)
    By ohms law  the current is
    I = V/R                                                                        (2)
    So transposing  I/V = 1/R                               (3)

    The voltage across all the resistors is the same, therefore from (1) and (2) the current in a parallel circuit is

    I = V/R1+ V/R2 +  etc.
    As all the V’s are the same, transposing
    I/V = 1/R1 + 1/R2 + 1/R2 + etc
    So from (3)
    1/RT = 1/R1 + 1/R2 + etc


    David
  • 0
    Although not relevant to the kernel of your question, I often found that those with less mathematical prowess preferred Rt = product/sum. It only works for two resistors but any network can be reduced to pair equivalence.
  • 0
    And never forget that in a circuit with parallel resistors, the total circuit resistance will always be less than the lowest individual resistance in the circuit. This is because the circuit current can travel trough the lowest resistor value AND the others in the parallel group.


    Z.
  • 0
    lyledunn:

    Although not relevant to the kernel of your question, I often found that those with less mathematical prowess preferred Rt = product/sum. It only works for two resistors but any network can be reduced to pair equivalence.


    Thank you for that wee nugget. It makes the mental arithmetic much easier. ?


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