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soccerkid12 soccerkid12
wrote...
Posts: 10
Rep: 0 0
11 years ago
Here's the full question:
A length of wire is cut into 5 equal pieces. The 5 pieces are then connected in parallel, with the resulting being 2.0 ohms. What was the resistance of the original length of wire before it was cut up?
What's the answer and how do you do it?
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4 Replies
Replies
wrote...
11 years ago
For parallel resistors:

R(eff) = 1/((1/R1) + (1/R2) + (1/R3) +.....)

so if there are 5 and the R values are all equal,

R(eff) = 1/(5/R) = 2 ohm
R/5 = 2
R = 10 ohms  (for each piece)

So the total resistance of the original wire was 50 ohms
wrote...
11 years ago
Before they were cut, the wires were effectively in series, so their resistances added. Let R be the resistance of the total length of wire, and let r be the resistance of each of the 5 sublengths. We know that R = r+r+r+r+r = 5r. let the parallel resistance of the wires be p. 1/p = 1/r + 1/r + 1/r + 1/r + 1/r = 5/r so r = 5p. R = 5r = 5(5p) = 25p, or R = 50 ohms.
wrote...
11 years ago
Since the wire pieces are in parallel, letting the resistance of each wire piece = R

(for a parallel, sum of reciprocal of resistances = reciprocal of total resistance)

1/R + 1/R + 1/R + 1/R + 1/R = 1/2, because each piece is equal in length and from the same wire, so they should have identical resistance.

5/R = 1/2
5 = R/2
R = 10 ohms

So the resistance of each wire is 10 ohms.
The resistance of the wire is directly proportional to its length, so let's say that each 10 ohm wire piece has a length of 2m, making the original wire 10m, letting resistance of original wire = x:

10/2 = x/10
100 = 2x
x = 50 ohms

So the resistance of the original wire = 50 ohms.
Answer accepted by topic starter
microbiology1microbiology1
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Posts: 44
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11 years ago
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