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Voltage difference, current flow

  • Thread starter kevnm67
  • Start date
  • #1
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Homework Statement


A voltage difference is applied to a piece of copper wire, a 5.0 mA current flows. The copper wire is replaced with a silver wire with twice the diameter of the copper wire.

A) How much current will flow through the silver wire? The lengths of both wires are the same, and the voltage difference remains unchanged. (The resistivity of copper is 1.68 x 10-8 ohms*meters and the resistivity of silver is 1.59 x 10-8 ohms*meters.)

B) How much charge passes through the copper wire in 10 minutes? (in C)


Homework Equations



V=IR
p=RA/L

The Attempt at a Solution



I am not sure where to start. Theres a difference in diameter so I figured p=RA/L might be useful but I am not sure. Looking for some direction, thanks.
 

Answers and Replies

  • #2
gneill
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Write expressions for the resistance of each type of wire. Use the given information to write an expression for the ratio of the two in terms of known values.
 
  • #3
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Write expressions for the resistance of each type of wire. Use the given information to write an expression for the ratio of the two in terms of known values.
The first time I tried this I made a ratio and said it was 4 times that of Cu but that gives 20mA and the answer is 21mA, not sure if I did this correct: p=RA/L and the diameter is twice that of Ag and since A is pi(r)2 I arbitrarily assigned the values of 2 for the diameter of Cu and 4 Ag
 
  • #4
gneill
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20,793
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I suspect that you forgot to keep the ratio of the resistivities in your expression.
 
Last edited:
  • #5
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So does this sound like I am moving in the right direction....
pAg/pCu = .95
Ag: p=RA/L

.95=2^2
=4.23

So 4.23 x 5mA = 21.13
 
  • #6
gneill
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20,793
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The result looks good!
 

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