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

https://www.physicsandmathstutor.com/admissions/engaa/solutions-2018/

Paper and Solutions

https://www.physicsandmathstutor.com/admissions/engaa/solutions-2018/

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 $R_\mathrm{Al}$).

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 $R_\mathrm{Al}$).

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

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