Solve 5 Wire Junction Current w/ Conservation of Current

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The discussion revolves around calculating the current at the fifth wire using the conservation of current principle. Participants emphasize the importance of converting measurements to meters for accurate calculations, although some found success without conversion due to consistent units. The formula J = I/(πr^2) is mentioned as a key equation for determining current density. Ultimately, the calculated current for the fifth wire is approximately -9.9 A, confirming the importance of unit consistency in calculations. This exchange highlights the collaborative effort in solving a complex electrical engineering problem.
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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

 
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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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