Resistance of a Conductor Problem

[Jayce]

Problem: The resistance of the wire used for the telephone is 35 Ω per kilometer when the weight of the wire is 5 kg per kilometer. If the specific resistance of the material is 1.95 x 10^-8 Ω-m, (A) what is the cross-sectional area of the wire? (B) What will be the resistance of a loop to a subscriber 8 km from the exchange if wire of the same material but weighing 20 kg per kilometer is used?

From the book, the solution is:

A.)

R = 35 Ω
l = 1 km = 1000 m
p (resistivity) = 1.95 x 10^-8 Ω-m.

So from R = p(l/A) ; A = pl / A = (1.95x10^-8)(1000)/35 = 55.7 x 10^-8 sq.m.

B.)

A = (20/5)(55.7 x 10^-8) = 222.8 x 10^-8 sq. m.

Length of wire = 2 x 8 = 16 km = 16000 m

R = pl / A = (1.95 x 10^-8)(16000) / 222.8 x 10^-8 = 140.1 Ω

My question:

On part (A), even though the weight of a material does not affect the resistance (based from the formula), can we really just ignore it on the problem?

On part (B), why is 20/5 multiplied to the acquired area on the first part? Yeah, the weight from the first part is 5 kg and on the 2nd case is 20 kg, but why should we multiply it? If it is to cancel the kg unit, then shouldn't the answer on the first part have the unit of sq.m per 5 kg/km or "x-answer" at weight of 5 kg / km?

Also, on the new length of wire why did we multiply the 8 km to 2? Where did the 2 come from?

I am so confused.. any help would be appreciated. Thank you.

 

[FvM]

A. Cross section can be either calculated from weight or resistance. Both calculations should give similar results, which is the case. I get 0.558 mm² based on the weight (copper density is 8.96 g/cm³)

B. total loop length is twice the distance

 

[smijesh]

On part (A), even though the weight of a material does not affect the resistance (based from the formula), can we really just ignore it on the problem?

First we have to consider 1Km length(l) wire (its resistance(R) is 35 ohm)
If you consider 2Km then resistance will become 70 ohm
Cross section area remains same but Weight of the wire will vary with length (5kg /Km)



On part (B), why is 20/5 multiplied to the acquired area on the first part? Yeah, the weight from the first part is 5 kg and on the 2nd case is 20 kg, but why should we multiply it? If it is to cancel the kg unit, then shouldn't the answer on the first part have the unit of sq.m per 5 kg/km or "x-answer" at weight of 5 kg / km?


You consider 1Km wire Density = mass/volume = kg/m^3

Density is constant (same material)

Here mass increased to 4 times (5 to 20kg)

Definitely volume reduce by for times (volume = area x length ) Length =1KM

So we can directly multiply 4 with area = 20/5 x Old wire area = New wire area

Hope it will help you

 

[Jayce]

A. Cross section can be either calculated from weight or resistance. Both calculations should give similar results, which is the case. I get 0.558 mm² based on the weight (copper density is 8.96 g/cm³)

B. total loop length is twice the distance

Oh, so it's an extra given wherein I get to choose what formula to use. So, weight = (A)(length)(density).

Is the total loop length always twice the distance in a conductor?

On part (B), why is 20/5 multiplied to the acquired area on the first part? Yeah, the weight from the first part is 5 kg and on the 2nd case is 20 kg, but why should we multiply it? If it is to cancel the kg unit, then shouldn't the answer on the first part have the unit of sq.m per 5 kg/km or "x-answer" at weight of 5 kg / km?


You consider 1Km wire Density = mass/volume = kg/m^3

Density is constant (same material)

Here mass increased to 4 times (5 to 20kg)

Definitely volume reduce by for times (volume = area x length ) Length =1KM

So we can directly multiply 4 with area = 20/5 x Old wire area = New wire area

Hope it will help you

I don't get it, I'm so sorry. From what I understand from the above, since case 1 and 2 uses the same material, density is constant but knowing that, I don't know where to go next with that analysis or how can I connect the two cases.

You mentioned Density = mass / volume, so how does the 1 km connect on to it? is the 1 km wire density (kg/m^3) and not length? Then V = (A)(l) and the length is 1km, then New Area = Old Area x (new weight / old weight). I can't seem to visualize how it works. Sorry for being an idiot.

 

[Ratch]

Oh, so it's an extra given wherein I get to choose what formula to use. So, weight = (A)(length)(density).

Is the total loop length always twice the distance in a conductor?



I don't get it, I'm so sorry. From what I understand from the above, since case 1 and 2 uses the same material, density is constant but knowing that, I don't know where to go next with that analysis or how can I connect the two cases.

You mentioned Density = mass / volume, so how does the 1 km connect on to it? is the 1 km wire density (kg/m^3) and not length? Then V = (A)(l) and the length is 1km, then New Area = Old Area x (new weight / old weight). I can't seem to visualize how it works. Sorry for being an idiot.

Don't be confused, be cognizant of what is going on. First of all, the problem says weight in kg. Kg is a unit of mass, a newton is a unit of force. If they wanted to say a kilogram of force, they should have written kgf, which is the weight of 1 kg of mass in Earth gravity.

Resistance of an object depends on two things, shape and resistivity (composition). The mass of a substance or its density has no relation to its resistivity. A long slender object such as a wire is an easy shape to evaluate. Other shapes can be a challenge to calculate.

You do not need the answer from part A to find the answer to part B. You need 16 km of wire for a loop of 8 km distance away. The old wire is 35 ohms./km. The new wire has 4 times the mass/km, so its cross section area is 4 times greater and its resistance is one-fourth of what the old wire is. We simply multiply (35 ohms/4)*16= 140 ohms. Any questions?

Ratch