Example 1:

A length of wire is cut into five equal pieces. The five pieces are then connected in parallel

with the resulting resistance being 2.009. What was the resistance of the original length of wire

before it was cut up?

50 Ohms

Explanation:

Given that

the original resistance of the wire is 2 ohms, then

it is known that the equivalent resistance in a parallel circuit is

1/R (eq) = 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5 ... + 1/Rn

Giving that there are 5 pieces, then we stop at number 5.

Since R (eq) is 2, we substitute for that, and then we have

1/2 = 1/R + 1/R + 1/R + 1/R + 1/R

1/2 = 5/R

R = 5 * 2

R = 10 ohms

Since each R is 10 ohms, and the wire was cut into 5 pieces, we can then say that the original resistance of the wire is

R + R + R + R + R

10 + 10 + 10 + 10 + 10

total original resistance is 50 Ohms