Solve 5 Wire Junction Current w/ Conservation of Current

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Homework Help Overview

The discussion revolves around a problem involving a five-wire junction where participants are tasked with determining the current in the fifth wire using the conservation of current. The problem includes various known and unknown values related to current density and diameters of the wires.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of Kirchhoff's junction law and the need to convert units for current density and diameter. There are questions about whether to convert measurements to meters for calculations.

Discussion Status

Some participants have provided guidance on unit conversion and shared their calculations, but there is no explicit consensus on the correctness of the answers. Multiple interpretations of the problem setup are being explored.

Contextual Notes

There are uncertainties regarding the conversion of units, as some participants mention leaving values in millimeters while others suggest converting to meters. The problem also includes unknown values that are critical for the calculations.

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Homework Statement



The information about the magnitudes of the current density and the diameters for wires 1, 2, 3, and 4 is given in the table. Some of the values are unknown.
Wire Current density (A/mm^2) Diameter (mm) Total Current (A)
1-------- 1.6 -------- 2.0 -------- ?
2 -------- ? -------- 3.0 -------- 2.0
3 -------- 3.0 -------- 1.1 -------- ?
4 -------- 0.8 -------- ? -------- 4.0

What is the current at the 5th wire?

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Homework Equations



I know you need to use the conservation of current, but I'm just not sure what formula to use.


The Attempt at a Solution

 
Last edited:
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study kirchhoff's junction law...
J = I/pi*r^2 ; pi = 3.14...
Hope this helps... good luck.
 
would I need to convert the current density and diameter to metres before calculations?
 
it I5 is going to be-9.84
 
ravi1611 said:
it I5 is going to be-9.84

Yep, that's the answer I got, just wasnt sure if it was right. Thanks.
 
Last edited:
yes... always go meters for length unless otherwise specified!
and yess the answer is -9.9
 
in this case, I didnt convert; I just left it as it
 
That is because the diameter is in (mm) so is the current density :)
 
usman27 said:
That is because the diameter is in (mm) so is the current density :)

Ah, good to know. I probably would have done this on the test without noticing. You probably saved me a few marks, Thanks.
 

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