In: Physics
It is necessary to double the length of a wire, but the resistance of the wire needs to remain APPROXIMATELY the same. If the wire is initially 22 gauge, what gauge of wire should be used to accomplish this?
Resistance, R = (
* L) / A ...(1)
Where
is the resistivity of the wire, L is the length of the wire and A
is the area of the wire.
L becomes 2L but resistance remains the same
R = (
* 2L) / Anew ...(2)
Where Anew is the new area of the wire
(1) / (2) gives
1 = Anew / (2 * A)
2 = Anew / A
2 =
Rnew2 /
R2
Where Rnew and R are the radius of the new and old wire
respectively.
2 = (Rnew/R)2
Taking square root on both the sides,
= Rnew/R
Rnew = 1.414 * R
Dnew = 1.414 * D
Where Dnew is the diameter of the new wire and D is the diameter of
the old wire
According to American wire gauge system, 22 gauge corresponding
to a diameter of 0.644 mm
Diameter of new wire, Dnew = 0.644 * 1.414
= 0.911 mm
This corresponds to 19 gauge wire