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I'm pretty sure I did this right but it seemed abnormally easy so I wanted to check.

A 10-m length of wire consists of 5.0 m copper followed by 5.0 aluminum, both of diameter 1.0 mm. A voltage difference of 85 mV is placed across the wire. What is the total resistnace (sum) of the two wires? What is the current through the wire? What are the voltages across the alumnium part and across the copper part?

FYI Resistivity of copper = 1.68E-8, resistivity of aluminum = 2.45E-8

So for the first question I calculated the two resistances and added them using R = p (L/A), where that p is really a rho (resistivity)

I got Rcopper part = .11 = 1.68E-8 x 5.0 (pi x (5E-4)^2)

and Raluminum = .17 (calculated the same way, except with the rho of aluminum

Added together = .28

Then, for current I used V=IR so I = .3

And finally, I calculated I = .39 for copper and .25 for aluminum using V = IR again.

Is this really this simple?

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# Homework Help: What is the current through the wire?

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