Resistor

What is (1/r1) + (1/r2)=? It is 2/r1+r2

by Rahul3382
September 03, 2020
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1 Answer

Answer by vanderghast

Let see...

1/a + 1/ b = ( (a b) / (a b) ) ( 1/a + 1/b)

' by multiplication by 1 which is also ab / (ab)

           =  (1/(a b)) ( ab/a   +  ab/b  )

'by distributivity of multiplication over sum 'here, distribution of numerator (a b) over ( 1/a + 1/b )

           =  (1/(a b)) ( b + a )

' by associativity of multiplication

' and because x/x == 1 for any x <> 0 or infinity

               = ( a + b) / ( a b )

' by commutativity and rearrangement of the terms

So, 1/r1 + 1/r2 == (r1 + r2)/(r1 r2)

If you remember that it implies a sum and a product ( r1 + r2 ) and ( r1 * r2 ), but don't remember which one is the numerator and which one is the denominator, add "units".

Example, if r1 and r2 are seconds, 1/r1, and 1/r2 are "per second" (or Hertz). Since r1+r2 has the units of seconds, and r1 times r2 has units of second_square ( such as feet times feet is square-feet), then r1 r2 / (r1 + r2) units will be second * second / second == second, not matching the wanted "per second". While (r1 + r2)/(r1 r2) has units of second / (second * second) == 1/ second, the loveable unit that we are looking for. So, of the two alternatives, that is the one. The seems an overkill the first time that we have to think about it, but once we know it, it become faster, and a second nature.

+1 vote
by vanderghast
September 04, 2020
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