Resistivity what a concept

In summary, the problem involves two identical cylindrical rods, one copper and one iron, joined end-to-end to form one long rod. A 12-volt battery is connected across the free ends of the copper-iron rod. The resistance of the combined rod is found by creating a ratio of the resistivity of iron to copper and multiplying it by the resistance of the copper rod. The voltage drop across the copper rod is then found using Ohm's Law, and the remaining voltage drop across the iron rod can be calculated by subtracting the voltage drop across the copper rod from the total voltage supplied.
  • #1
Physicsit
So here is the problem

Two cylindrical rods, one copper and the other iron, are identical in lengths and cross-sectional areas. They are joined, end-to-end, to form one long rod. A 12-V battery is connected across the free ends of the copper-iron rod. What is the voltage between the ends of the copper rod?

I know that R = resistivity(L/A)


so what I did was created a ratio to get the resistance of the new rod

Riron/Rcopper = (resistivty of iron * L/A)/(resistivty of copper * L/A) = (resistivty of iron)/(resistivty of copper) = 9.7*10^-8/1.72*10^-8

= a resistance of 5.64

I believe that is the resistance of the copper and iron rod connected to each other however I do not know how it helps if it does at all
 
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  • #2
It's a start.

After all that ********, all you did was find the ratio of the resistance of the iron part to the resistance of the copper part.

So now imagine that you have a circuit consisting of a battery and two resistors in series, connected by wire with 0 resistance. One of the resistors is the copper rod, the other is the iron rod.

Remembering that for any resistor, ΔV = IR, can you write an equation that describes the changes in potential at each step going around the entire circuit?

hint: let Rc = resistance of the copper rod. What is the resistance of the iron rod, in terms of Rc ?
 
  • #3
Not sure I follow

I will have to look at this a little more closley becuase I am not sure I know where to start.
 
  • #4
The problem as stated doesn't makes sense. If you connect a conductor to the two poles of a 12 volt battery, then the voltage (drop) between the two end IS 12 volts. You could then use "V= IR" to determine the current in the two parts, but "12 volts is 12 volts"!
 
  • #5
The problem as stated doesn't makes sense.
?

Of course it makes sense. There is a certain drop across the copper part, and an additional drop across the iron part. The combined drop is 12 volts. The question asks for the drop across the copper part alone.

What's wrong with that?
 
  • #6
Is this correct

I came up with the following

Rcopper/Riron = (resistivty of copper * L/A)/(resistivty of iron * L/A) = (resistivty of copper)/(resistivty of iron) = 9.7*10^-8/1.72*10^-8

= a resistance of .18

Icopper = Iiron
V= IR

so

then we know that

Vcopper = .18*Viron

so

Viron=Vcopper/.18

we also know that

Vcopper + Viron= 25

so with a little subsititution

Vcopper + Vcopper/.18 =25

so

Vcopper = 1.83


This is driving me outta my head please let me know If i am correct
 
  • #7
This is driving me outta my head
Wait till you get to magnetism.

You mixed up the resistivities, and you typed 25 instead of 12 for the voltage, but amazingly you ended up very close to my answer.

I'll take your numbers for the resistivities; my book gives slightly different figures. But it's obvious that the 1.72 must be the copper, and the 9.7 is the iron.

So here's how I set it up:
Let R = resistANCE of the copper rod
therefore IR = voltage drop across the copper rod
since the two rods are identical in length & diameter:
(9.7/1.72)R = 5.64R = resistANCE of the iron rod.
I*5.64R = voltage drop across the iron rod
The sum of the voltage drops across the two rods must equal the total voltage supplied, so
I*R + I*5.64R = 12
IR(1 + 5.64) = 12
IR = 12/6.64 = 1.81 V

as a check:
voltage drop for the iron rod = 5.64*IR = 10.19 V
1.81 + 10.19 = 12
 
  • #8
Thanks again

I am glad that I finally got to the bottom of it.
Wait till you get to magnetism.


Luckily this is only a one semester class and as we are drawing to the end quickly I do not think we will get to magnetism. So I will have to conquer that another day
 

What is resistivity and how is it measured?

Resistivity is a measure of a material's ability to resist the flow of electrical current. It is typically represented by the Greek letter rho (ρ) and is measured in ohm-meters (Ω·m). It can be measured using a device called an ohmmeter, which applies a known voltage to the material and measures the resulting current.

What factors affect the resistivity of a material?

The resistivity of a material is affected by several factors, including the material's composition, temperature, and physical dimensions. Materials with high electrical conductivity tend to have low resistivity, while materials with low electrical conductivity have high resistivity. Additionally, as temperature increases, resistivity generally increases. The geometry and dimensions of a material can also impact its resistivity.

How does resistivity differ from resistance?

Resistivity and resistance are often confused, but they are not the same thing. Resistivity is an intrinsic property of a material, while resistance is a measure of how much a material resists the flow of current at a specific point. Resistivity is dependent on the material itself, while resistance is affected by both the material and the dimensions of the object through which the current is flowing.

What is the relationship between resistivity and conductivity?

Resistivity and conductivity are inversely related. This means that as resistivity increases, conductivity decreases, and vice versa. Conductivity is the measure of a material's ability to conduct electricity, and it is the inverse of resistivity. Materials with high resistivity have low conductivity, and vice versa.

How is resistivity used in practical applications?

Resistivity is an important concept in various fields, including electrical engineering, material science, and geology. It is used to design and optimize electrical circuits, determine the suitability of materials for specific applications, and study the composition of Earth's crust. It is also used in the development of new materials with desired electrical properties.

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