parallel wire ampacity

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Charlie:

I see what you are asking. I think Smart has shown one way to handle different numbers of resistors.

I would have used the basic current divider formula for two resistors in parallel. Then, to add more resistors, I would have used "equivalent circuits" to reduce the circuit to two parallel resistors. Then you are back to a circuit with only two resistors.

Basic Current divider:

IR1 = It * R2/(R1 + R2)
(notice you use R2 in the numerator to find current I1)

Resistance equals some constant (r) times the length L:

IR1 = It * (1/r*L2) /((1/r*L1 + 1/r*L2))
(the r's cancel out)

IR1 = It * (1/L2) /((1/L1 + 1/L2))

If R2 is actually several resistors, we can combine them into an Equivalent resistor, Req, using the formula for parallel resistors:

Req = 1/ ( 1/R2 + 1/R3 + 1/R4 + ...)

Then we substitute lengths:

1/r*Leq = 1/ (1/r*L2 + 1/r*L3....)

Again, the r's cancel out:

Leq = 1/L2 + 1/L3....

Now we just substitute Leq for L2 to solve for the current through R1. For each other current I2, I3, ..., we have to find the new Leq, and substitute it back into the equation with L2, L3, etc.


Steve
 
The formula for the equivalent resistance of parallel resistors is messy. Better to express each resistance as a conductance (in mhos, of course) and then just add, then take the inverse.
 
Steve: That is the set of mathematical manipulations that I was talking about. It does get messier, as you add more resistors in parallel.

Steve and Smart $: The rules of mathematics do not allow, as part of the proof that a formula works for any number of items, the use of an "etcetera" notation - your series of dots that imply that the series continues until you reach the last item on the list. To prove the "it always works" claim, you need to use the Mathematical Induction process.

At this point, I no longer doubt the formula works. All I am saying is that you have not provided satisfactory, formal, mathematical proof. That's OK, however; I don't think it is needed, at least not for the purposes of this forum.
 
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