# A question about resistivity ( high school physics ) [duplicate]

Resistivity = (Resistance times area ) divided by length

$$\rho = (RA)/l$$

now my question is that in order to calculate the resistivity for some metal we have to take into account its length and area then why is it said that the resistivity of a metal is independent of its area or length . i.e. Why does every piece of copper have the same resistivity regardless of its area or length?

## marked as duplicate by sammy gerbil, John Rennie, Qmechanic♦Mar 1 '18 at 18:20

One way to think about it is as an empirical law. If you have a bunch of copper wires of varying lengths $l$, all with the same cross-sectional area, and you measure their resistances, you will find that the resistance is directly proportional to the length: $$R \propto l \qquad \text{(constant A)}.$$ Similarly, if you have a bunch of wires with different cross-sectional areas $A$, all with the same length, you will find that their resistance is inversely proportional to their area: $$R \propto \frac{1}{A} \qquad \text{(constant l)}.$$ Putting these two equations together, we have that $R \propto l/A$, or $$R = \rho \frac{l}{A}.$$ for some proportionality constant $\rho$. This number is what we define to be the resistivity of copper.
If you turn it around, and calculate $\rho = RA/l$, you're right that you'll get different numbers for $R$, $A$, and $l$ depending on what wire you're talking about. But because of the above proportionalities, these differences will all cancel each other out. For example, if you had two copper wires with the same $A$, but one was twice as long as the other, then the resistance $R$ of the longer wire would be twice as high as well. The factors of two would cancel out, and you'd get the same value of $\rho$.
Because $R$ also depends on the dimensions of wire. Recall that: $$R=\dfrac{m_e l} {n e^2 A \tau}$$ So, the resistivity is \begin{align}\rho &=R\left(\dfrac {A}{l}\right)\\&=\left(\dfrac{m_e l}{n e^2 A \tau}\right )\left(\dfrac Al\right)\\&= \dfrac{m_e}{n e^2 \tau}\end{align} This is independent of $A$ and $l$.