A wire with a resistance of 6.0 ohms is drawn out through a die so its new length is 3X its original length?

Question

Find the resisance of the longer wire, assuming that the resistivity and density of the material are unchanged.

Answer 1

Volume is a product of length L and cross sectional area A and will remain constant
V= A l=const.
V1=V2
A1L1= A2L2
A2= A1(L1/L2)
L2=3L1
A2= A1( L1/3L1)= A1/3

R= p L/A where p is specific material resistance
R1= p (L1/A1)
R2= p(L2/A2)
R2=p(3L1/ A1 / 3)
R2= 9 p(L1/A1)
R2= 9 R1 or

R2= 54 ohms

That was a hard one....I need a break Jack.

Answer 2

Resistance depends directly on the length of the wire (the longer the higher the resistance) and inversely on the area of the wire (the wider the pipe, the greater the flow).

Also, since matter is conserved, the volume of the wire is unchanged if its length is changed. Assuming the wire to be a cylinder, we know that cross section area x L=constant, or that

pi*r^2*L=cst. If the length is tripled, then the radius decreases by a factor of sqrt[3].

Using these data in the resistance relationship, we have that:

R=cst L/A, so if L increases by a factor of 3 and the cross sectional area decreases by a factor of 3, the total resistance becomes 9 times greater, so the new resistance is 54 ohms.

Answer 3

I'm not going to bet my life on it, but I'm thinking 54 ohms. Each third of the drawn wire is the same length as the original wire and 1/3 the volume of the original wire, so has 1/3 the cross sectional area. Imagine the thirds of the wire were in parallel. They they'd have the same resistance as the original wire, since the length is the same and the cross sectional area is the same. But in reality, the three sections are in series, so their resistances add up, so it's nine times the original. Each section has 3 times the resistance of the original (if my reasoning above is right), and they're in series, so that makes 9 times. Six nines is 54, so you get 54 ohms.

Answer 4

That's Easy !!


Resistance = Resistivity x Length / Area
6 = pxL/A
6=pL/A

New Resistance .

Mass of wire remain as it is and Density remains the same .

Density = mass / Volume
Therefore Volume remains the same .

Volume = Length x Area
Volume is constant
Length increases by 3
Area decreases by 1/3

Resistance = Resistivity x Length / Area
Resistance = p x 3L / 1/3 A
Resistance = 9pL/A
6=pL/A
Resistance = 9x6 = 54 Ω

Answer 5

It is necessary to know the diameter of the wire, if it is a very thin wire the resistance may be increased do the change in the cross sectional area.

Answer 6

resistance relies upon quickly on the dimensions of a twine and inversely on its bypass sectional section the longer the twine, the greater resistance encountered by employing the present the broader (the better the bypass sectional section) the better the "pipe" by which the present flows; narrowing the "pipe" makes the resistance improve in case you triple the dimensions, that on my own triples the resistance in case you're making the section smaller by employing a element of three, that on my own triples the resistance the two one in each and every of those mutually improve the resistance by employing a element of 9, so the hot resistance is 9x0.7=6.3 ?

Answer 7

the length is 3 times bigger and area is 3 times smaller... you have to do the rest though.