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Resistors in Parallel - why the 1/R?

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


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  • 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





Reply
  • 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





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