# Resistance and Resistivity ENGAA Question

So there's this question on the ENGAA 2018 Section 2. Q16, where it aims to find the resistance of the whole cable when it is made up of different materials. I don't really understand why they use the Area of the entire wire A=M/(9dL) when finding the resistivity of part of the wire instead of the area of that part or why you can add 6dAL + 3dAL to make 9dAL even though the areas are not the same. Thanks for any help

Paper and Solutions
The notation used is a bit confusing - $A$ here refers to the area of an individual wire (not whole cable), but each of the wires can have a different resistance and density, so they have to be combined correctly to get the right answer.

The expression $6dAL + 3dAL = 9dAL = M$ is simply a statement that the total mass of cable is equal to the sum of the masses of the component wires. When this is rearranged for $A$, it can then be used to substitute for the area of individual Al and Cu wires, remembering that there are 6 Al wires for every Cu wire (hence the factor of $6A$ in denominator of expression for $R_\mathrm{Al}$).
Original post by lordaxil
The notation used is a bit confusing - $A$ here refers to the area of an individual wire (not whole cable), but each of the wires can have a different resistance and density, so they have to be combined correctly to get the right answer.

The expression $6dAL + 3dAL = 9dAL = M$ is simply a statement that the total mass of cable is equal to the sum of the masses of the component wires. When this is rearranged for $A$, it can then be used to substitute for the area of individual Al and Cu wires, remembering that there are 6 Al wires for every Cu wire (hence the factor of $6A$ in denominator of expression for $R_\mathrm{Al}$).

Oh ok, thank you very much