Nichrome and Copper Circuit w/ Electron Mobility

  • #26
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Yup. Standard circuit analysis.

I absolutely hate circuits :) lol

So the first part, the copper wire on top. V=IR, so the voltage at the end of the first resistor is V=531.42A*(1.13E-5) and that voltage continues until we hit the next resistor, yeah? 1.13E-5 is the R of copper
 
  • #27
gneill
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I absolutely hate circuits :) lol

So the first part, the copper wire on top. V=IR, so the voltage at the end of the first resistor is V=531.42A*(1.13E-5) and that voltage continues until we hit the next resistor, yeah? 1.13E-5 is the R of copper
I'm seeing a slightly different value for the resistance of the copper wire sections based on the given values. Did you use a table value for resistivity or calculate the resistance from the given values?

Either way, potential is dropped across a resistance. It does not "continue" until another resistance. What remains the same in a series circuit is the current. It's the same value for all the components. Each component creates an end-to-end drop in potential. The sum of all these potential drops must equal the original applied potential (the voltage source in this case). This is Kirchhoff's voltage law. So what you want to do is calculate the individual potential drops across the resistors. These drops should sum to 1.5 V.
 
  • #28
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I'm seeing a slightly different value for the resistance of the copper wire sections based on the given values. Did you use a table value for resistivity or calculate the resistance from the given values?

Either way, potential is dropped across a resistance. It does not "continue" until another resistance. What remains the same in a series circuit is the current. It's the same value for all the components. Each component creates an end-to-end drop in potential. The sum of all these potential drops must equal the original applied potential (the voltage source in this case). This is Kirchhoff's voltage law. So what you want to do is calculate the individual potential drops across the resistors. These drops should sum to 1.5 V.

I calculated the conductivity by using the information given in the initial problem. The reciprocal gives me the resistivity. I used that value in the resistance equation and that gave me R.

Okay, so I calculated that both copper wires have voltage drops of 0.006V and the Nichrome has a voltage drop of 1.48. Which makes sense because 2(0.006)+1.48~1.5V.

And now I want the electric field in the copper wire and in the nichrome wire. And the electric field is just the change in voltage, right?
 
  • #29
Mister T
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It really is best to get in the habit of referring to the voltage "across" a circuit element, and the current "through" a circuit element.
 
  • #30
gneill
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I calculated the conductivity by using the information given in the initial problem. The reciprocal gives me the resistivity. I used that value in the resistance equation and that gave me R.

Okay, so I calculated that both copper wires have voltage drops of 0.006V and the Nichrome has a voltage drop of 1.48. Which makes sense because 2(0.006)+1.48~1.5V.

And now I want the electric field in the copper wire and in the nichrome wire. And the electric field is just the change in voltage, right?
What are the units associated with an electric field? It's not just Volts, which is just the unit of electric potential.

I'll give you a hint. One of the common ways of expressing the units of an electric field is N/C, that is, the force exerted per coulomb of charge. But another entirely equivalent way is V/m (you can massage the units to prove they're equivalent!). That's the change in potential over distance. You have calculated the change in potential that occurs over each wire segment, and you have the lengths of those wire segments....
 
  • #31
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What are the units associated with an electric field? It's not just Volts, which is just the unit of electric potential.

I'll give you a hint. One of the common ways of expressing the units of an electric field is N/C, that is, the force exerted per coulomb of charge. But another entirely equivalent way is V/m (you can massage the units to prove they're equivalent!). That's the change in potential over distance. You have calculated the change in potential that occurs over each wire segment, and you have the lengths of those wire segments....

Ah that makes sense!! Thank you very much for your time and assistance!
 
  • #32
gneill
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Ah that makes sense!! Thank you very much for your time and assistance!
You're very welcome.
 

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