Current due to Potential Difference

Click For Summary
SUMMARY

The discussion focuses on calculating the current flow through a rectangular block of metal when a potential difference V is applied across different faces. The initial current Iab is observed when V is applied across the A x B faces, while the current Ibc is calculated for the B x C faces. The derived formula for Ibc is Iab(C^2/A^2), demonstrating that the resistivity and dimensions of the block significantly influence the current flow. The analysis confirms that the relationship between the currents is dependent on the geometry of the block and the applied potential difference.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Knowledge of resistivity (rho) and its role in electrical resistance
  • Familiarity with the concept of cross-sectional area in electrical circuits
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of resistivity and its impact on current flow in different geometries
  • Learn about the relationship between voltage, current, and resistance in three-dimensional conductors
  • Explore advanced topics in electrical engineering, such as circuit analysis and network theorems
  • Investigate the effects of varying dimensions on electrical resistance in materials
USEFUL FOR

This discussion is beneficial for physics students, electrical engineering students, and professionals involved in circuit design and analysis, particularly those focusing on current flow in conductive materials.

ihearyourecho
Messages
61
Reaction score
0

Homework Statement



Consider a rectangular block of metal of height A, width B, and length C , as shown in the figure. If a potential difference V is maintained between the two A X B faces of the block, a current Iab is observed to flow.

Find the current that flows if the same potential difference V is applied between the two B X C faces of the block

http://session.masteringphysics.com/problemAsset/1126202/2/Walker4e.ch21.Pr023.jpg


Homework Equations



V=IR, R=rho(L/A)


The Attempt at a Solution



V=IR, R=rho(L/A)

V=I*rho*(L/A)

Iab=(V)/(rho*(C/AB))

Ibc=(V)/(rho*(A/BC))

I thought this last equation was right, but apparently it isn't. When I entered it, it said that B, V, and rho had no impact upon the answer. I'm not really sure what to do now...
 
Physics news on Phys.org
By the way, it wants the answer in "terms of some or all of the variables A, B, C, and Iab."
 
the answer is Iab(C^2/A^2) due to the fact the voltages had to be set equal
IabR =IbcR
Iab(rhoL/A) =Ibc(RhoL/A)
Iab(RhoC/AB) = Ibc(RhoA/BC)
Rho's and B's cancel
IabC/A = IbcA/C
cross multiply
IabC^2=IbcA^2
Ibc =Iab(C^2/A^2)
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 49 ·
2
Replies
49
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
1
Views
2K
Replies
6
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
7
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K