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- Thread starter Steven7
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- #2

LowlyPion

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What equation would you think to apply to figure this?

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I just don't get the part how the length of the wire affects the cross sectional area/radius.

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LowlyPion

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Don't you figure that there is a conservation of mass?

I just don't get the part how the length of the wire affects the cross sectional area/radius.

What happens to the volume of wire when you stretch it?

Volume = L * A

if the volume is constant, then if L = 1.1*L then A must equal A/1.1

- #5

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I got it. I thought the increase of L decrease the r to r/1.1.

Can I ask another question about resistance?

A square carbon film of thickness 5x10^-7m, rho 4x10^-5 ohm m is formed on an insulator rod of diameter 3mm.What is the length of the rod so that the carbon film on its curve surface has a resistance of 100 ohm.

I tried:

RA/rho = L

100[3.142*(1.5mm+5x10^-7)^2]/4x10^-7 = L

But the answer is 1.18cm which is different from mine. Where did it gone wrong

Can I ask another question about resistance?

A square carbon film of thickness 5x10^-7m, rho 4x10^-5 ohm m is formed on an insulator rod of diameter 3mm.What is the length of the rod so that the carbon film on its curve surface has a resistance of 100 ohm.

I tried:

RA/rho = L

100[3.142*(1.5mm+5x10^-7)^2]/4x10^-7 = L

But the answer is 1.18cm which is different from mine. Where did it gone wrong

Last edited:

- #6

LowlyPion

Homework Helper

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Consider that your cross sectional area is the cross section of a thin shelled tube.I got it. I thought the increase of L decrease the r to r/1.1.

Can I ask another question about resistance?

A square carbon film of thickness 5x10^-7m, rho 4x10^-5 ohm m is formed on an insulator rod of diameter 3mm.What is the length of the rod so that the carbon film on its curve surface has a resistance of 100 ohm.

I tried:

RA/rho = L

100[3.142*(1.5mm+5x10^-7)^2]/4x10^-7 = L

But the answer is 1.18cm which is different from mine. Where did it gone wrong

A = L * T = π * d * 5*10

So:

100Ω = 4*10

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- #8

LowlyPion

Homework Helper

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It's an insulating rod, so only the surface foil conducts.

If you unwrap the foil off the rod its cross sectional area is width (π * d) and its thickness is 5*10

Area then is W*T = (π * d) *( 5*10

Substituting into

R = 100 = p*L/A

yields

100*3.14*.0003*5*10

- #9

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I struggle with A = π r^2 and you cleared those. Thanks a lot.

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