Calculating Current in a Wire: Applying PD and Cross-Sectional Area

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Homework Help Overview

The discussion revolves around calculating the current in a wire when a potential difference (pd) is applied, considering the wire's cross-sectional area and length. The subject area includes concepts from electricity and circuit theory, particularly Ohm's law and resistivity.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between current, cross-sectional area, and resistance, questioning how changes in geometry affect current flow. Some participants raise concerns about the assumptions made regarding the velocity of charge carriers and the application of formulas.

Discussion Status

There is an ongoing exploration of the effects of changing the cross-sectional area and length on resistance and current. Some participants have provided insights into the relationship between resistance, length, and area, while others have noted potential misunderstandings in the application of the relevant equations. The discussion is productive, with various interpretations being examined.

Contextual Notes

Participants are navigating assumptions about the physical setup of the problem, including the orientation of the applied potential difference and the implications for resistance. There is a recognition of the need to consider both length and area when analyzing the problem.

ravsterphysics
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Homework Statement


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Homework Equations

The Attempt at a Solution



I know I = nqvA

When the pd is applied the surface that is 8cm long, the cross sectional area is 32cm (8x4) but when the pd is applied across the 4cm side, the cross sectional area is now 16cm (4x4) so I has decreased by a factor of 2 so the new I is I/2 thus answer A? But the correct answer is D.
 
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When comparing I=nqvA for each case, you assumed that only A changed. q stays the saem OK - but is v the same?
HInt: Ohms' law and resistivity.
 
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You've got the order of the faces backwards. The initial scenario has the PD applied across the ends of the rectangle (the 4 x 4 ends are separated by 8 cm).

Also, remember that resistance is proportional to path length as well as being inversely proportional to the cross sectional area. Look up the definition of resistivity.
 
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ravsterphysics said:

Homework Statement


View attachment 111643

Homework Equations

The Attempt at a Solution



I know I = nqvA

When the pd is applied the surface that is 8cm long, the cross sectional area is 32cm (8x4) but when the pd is applied across the 4cm side, the cross sectional area is now 16cm (4x4) so I has decreased by a factor of 2 so the new I is I/2 thus answer A? But the correct answer is D.

I think that the relevant equation is: R \propto \frac{l}{A} where l is the length of the resistor and A is the cross-sectional area. Switching from one face to the other changes both A and l.
 
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People are always very quick to use formulas ... much easier just to look at it ...the distance the current has to travel has halved so this alone will make the resistance halve ... but in addition the area the current moves through has doubled .. this will also halve the resistance ... so in total the resistance has dropped to a quarter of what it was ...
 
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Simon Bridge said:
When comparing I=nqvA for each case, you assumed that only A changed. q stays the saem OK - but is v the same?
HInt: Ohms' law and resistivity.

gneill said:
You've got the order of the faces backwards. The initial scenario has the PD applied across the ends of the rectangle (the 4 x 4 ends are separated by 8 cm).

Also, remember that resistance is proportional to path length as well as being inversely proportional to the cross sectional area. Look up the definition of resistivity.

stevendaryl said:
I think that the relevant equation is: R \propto \frac{l}{A} where l is the length of the resistor and A is the cross-sectional area. Switching from one face to the other changes both A and l.

oz93666 said:
People are always very quick to use formulas ... much easier just to look at it ...the distance the current has to travel has halved so this alone will make the resistance halve ... but in addition the area the current moves through has doubled .. this will also halve the resistance ... so in total the resistance has dropped to a quarter of what it was ...

Yep, I totally disregarded the effects on the wire's resistance. It's clear now why current goes up by a factor of 4. thanks for the help.
 
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