How to Calculate the Current Flowing Through a Gold Wire?

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Homework Help Overview

The discussion revolves around calculating the current flowing through a gold wire connected to a 0.70 V battery. The wire's dimensions and resistivity are provided, and participants are attempting to apply Ohm's law and resistance formulas to find the current.

Discussion Character

  • Mathematical reasoning, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the calculation of resistance using the resistivity of gold and the dimensions of the wire. There are questions about the accuracy of the resistivity value used and its impact on the final current calculation. Some participants express confusion over why their calculated current does not match expected results.

Discussion Status

Several participants have shared their calculations and expressed uncertainty about the resistivity value. There is a suggestion that the discrepancy may relate to significant figures or the specific values expected by the MasteringPhysics platform. Some participants are exploring the possibility that the question may have issues.

Contextual Notes

Participants note the importance of significant figures in their calculations and question whether the resistivity value provided aligns with what is expected in the homework context. There is also mention of a potential reference table for constants that may not be available to them.

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Homework Statement


The terminals of a 0.70 Vwatch battery are connected by a 70.0-m-long gold wire with a diameter of 0.100 mm.
What is the current in the wire?

Homework Equations


V = IR
R = pL/a

The Attempt at a Solution



I try to find resistance first, resistivity of gold is 2.2 * 10 ^-8

Area of the wire is a = pi * r ^ 2
r = ( 0.1 * 10^-3 ) /2

So I found a = 7.85 * 10^-9

Then,
R = 2.2 * 10^-8 * 70 / (7.85 * 10^-9) = 196.1

I plug this into V/R = I, .70/196.1 = .00357 = 3.57 mA, but this answer ends up being wrong. Any help on where I went wrong would be appreciated, thanks.
 
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david12445 said:

Homework Statement


The terminals of a 0.70 Vwatch battery are connected by a 70.0-m-long gold wire with a diameter of 0.100 mm.
What is the current in the wire?

Homework Equations


V = IR
R = pL/a

The Attempt at a Solution



I try to find resistance first, resistivity of gold is 2.2 * 10 ^-8

Area of the wire is a = pi * r ^ 2
r = ( 0.1 * 10^-3 ) /2

So I found a = 7.85 * 10^-9

Then,
R = 2.2 * 10^-8 * 70 / (7.85 * 10^-9) = 196.1

I plug this into V/R = I, .70/196.1 = .00357 = 3.57 mA, but this answer ends up being wrong. Any help on where I went wrong would be appreciated, thanks.
Where does that value for resistiviuty of gold come from?
 
I searched online for the value of the resistivity of gold wire
 
Does Mastering Physics provide a reference table of constants to use? Perhaps they have a value for the the resistivity of gold that they expect you to use. The value that you found has just two significant figures and you're told to supply a result with three. This could make a difference in your least significant figure.
 
gneill said:
Does Mastering Physics provide a reference table of constants to use? Perhaps they have a value for the the resistivity of gold that they expect you to use. The value that you found has just two significant figures and you're told to supply a result with three. This could make a difference in your least significant figure.

No, I don't think there is one. I tried a value to three significant figures, 2.25 * 10^-8 but it still wasn't right. MasteringPhysics is usually pretty good about saying it's right as long as you are close, so that must mean my answer is a good bit off. Are there any numbers I put in wrong?
 
The value should be in the neighborhood of 3.5 mA. Using various values I've seen for the resistivity of gold I get a range from about 3.51 mA to 3.55 mA.

Could be the question is "broken" in Mastering Physics.
 
Mastering Physics is a stickler for significant figures. Your textbook has a table of resistivities.
 

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