Resistance of 2 materials, different resistivity, combined

[zachem62]

Homework Statement

A 60-cm long gold wire is soldered to a silver wire of the same length as shown in the diagram. If each wire is 2.00 mm in diameter, determine the equivalent resistance of the combination between A and B.
Diagram is attached.

Homework Equations

R = ρL/A

The Attempt at a Solution

My approach was to calculate resistance for the 3 separate parts and add them together. So 10cm of gold, 10cm of silver, and 40cm of the combined part in the middle. But for the middle combined part, my approach was to combine the areas and resitivity of the 2 wires, but is this really the right way?

 

[scottdave]

The problem statement says that each wire is 60 cm, but in your attempt, you have 10cm + 40cm.
It looks like the section where they are running alongside each other should be treated as 2 resistors in parallel. Then that parallel combination is in series with the strand of gold and that is in series with the strand of silver.

But make sure you get it to add up to the correct lengths.

 

[zachem62]

The problem statement says that each wire is 60 cm, but in your attempt, you have 10cm + 40cm.
It looks like the section where they are running alongside each other should be treated as 2 resistors in parallel. Then that parallel combination is in series with the strand of gold and that is in series with the strand of silver.

But make sure you get it to add up to the correct lengths.

just to clarify, I didn't say it was 10cm + 40cm, I said it was 10 + 40 + 10, with 40 being the combined part in the middle. How can you tell if those 2 resistors should be treated as parallel? don't you calculate the resistance with R = ρL/A formula and combining their resistivity values somehow?

 

[scottdave]

You have a voltage gradient, between one end of the wire and the other end (in the soldered section). Whether they are separated, and only soldered at two points, or are touching each other all the way along, the gradient will be the same. If they were separated, and only joined at the tips, you could go along, a certain amount of length, and measure voltage for the gold wire and the silver. The voltage would be the same.

I think of it similar to how a Wheatstone Bridge works. If the resistance is proportional, then the galvometer will read zero across the bridge, and no current will flow. If no current is flowing across, it doesn't matter if they are touching or separate - they operate the same.

Here are a couple of links on it. https://www.grc.nasa.gov/WWW/k-12/airplane/tunwheat.html
http://www.electronicshub.org/wheatstone-bridge/

 

[scottdave]

just to clarify, I didn't say it was 10cm + 40cm, I said it was 10 + 40 + 10, with 40 being the combined part in the middle. How can you tell if those 2 resistors should be treated as parallel? don't you calculate the resistance with R = ρL/A formula and combining their resistivity values somehow?

The problem statement says the gold wire is 60 cm, and the silver is the same length. The part where they are together is 40 cm, according to your picture. So if 40 cm of the gold wire shares with the silver, then there is 20 cm of gold wire by itself. Similarly, the silver will have 20 cm left over by itself. The entire combined wires should be 20cm + 40cm + 20cm

 

[zachem62]

The problem statement says the gold wire is 60 cm, and the silver is the same length. The part where they are together is 40 cm, according to your picture. So if 40 cm of the gold wire shares with the silver, then there is 20 cm of gold wire by itself. Similarly, the silver will have 20 cm left over by itself. The entire combined wires should be 20cm + 40cm + 20cm

Oh I see mistake now. Thanks for pointing that out.

 

[zachem62]

You have a voltage gradient, between one end of the wire and the other end (in the soldered section). Whether they are separated, and only soldered at two points, or are touching each other all the way along, the gradient will be the same. If they were separated, and only joined at the tips, you could go along, a certain amount of length, and measure voltage for the gold wire and the silver. The voltage would be the same.

I think of it similar to how a Wheatstone Bridge works. If the resistance is proportional, then the galvometer will read zero across the bridge, and no current will flow. If no current is flowing across, it doesn't matter if they are touching or separate - they operate the same.

Here are a couple of links on it. https://www.grc.nasa.gov/WWW/k-12/airplane/tunwheat.html
http://www.electronicshub.org/wheatstone-bridge/

So in this case, you're saying I would treat it like a circuit and calculate the resistances with the formula R = ρL/A and add them all up as if they were series and parallel accordingly?

 

[scottdave]

So in this case, you're saying I would treat it like a circuit and calculate the resistances with the formula R = ρL/A and add them all up as if they were series and parallel accordingly?

Yes, that is how I would approach this problem.