Does Ohm's Law Equate to the Resistivity Formula R = ρ(L/A)?

[Chromium]

1st question
Ohm's Law is R = V/I; however there is another law that states : R = p (L/A) (the p should be a rho, i know)


so this means that V/I = p (L/A) ?


2nd question
A resistor in a circuit can have a smaller current in it than the actual wire because of a higher resistivity, right? Example if the circuit's wires are copper, but there is a resistor that has a higher resistivity (like aluminum), will the resister have a smaller I than the copper?

 

[mezarashi]

To your second question, no. You can't simply separate circuit components like that. The actual current that flows through the circuit is the battery voltage V divided by the total circuit resistance which is R + the resistance (albeit very small) of the copper wire + the battery's internal resistance.

Imagine a circuit with two resistors of different values in series. Would it make sense that there is more current flowing in one resistor than the other? Where would this "extra" current disappear to between them?

Ohm's Law is an interesting observation relating the voltage and current in a circuit. Not all materials obey Ohm's Law. In fact, resistance of most real materials are to a certain degree voltage and temperature dependent. The other relationship you mention for the calculation of resistivity is more or less empirical. We deduce this relationship.