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Electrical Cube 3 (Posted on 2016-04-15) Difficulty: 3 of 5
A third wire cube with 1 ohm edges is made, just like the cubes in Electrical Cube and Electrical Cube 2.

This time one of the 1 ohm edges is removed entirely and replaced with a 1 volt battery. How much current is in the wire connecting the battery to the cube structure this time?

  Submitted by Brian Smith    
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Solution: (Hide)
Charlie provides a solution in the comments, my solution is below.

Place the battery on segment AB with the remaining 11 segments 1 ohm resistors.
 A----VV----B
 |\        /|
 | \      / |
 |  C----D  |
 |  |    |  |
 |  |    |  |
 |  E----F  |
 | /      \ |
 |/        \|
 G----------H
One way to calculate the currents is to use a system of mesh currents. Let those currents be:
I1: A-B-D-C-A
I2: A-C-E-G-A
I3: C-D-F-E-C
I4: B-H-F-D-B
I5: E-F-H-G-E
Note that the current flowing through the battery is I1.

The five equations from currents are:
1V = 1*(I1-I4) + 1*(I1-I3) + 1*(I1-I2)
0V = 1*(I2-I1) + 1*(I2-I3) + 1*(I2-I5) + 1*I2
0V = 1*(I2-I1) + 1*(I3-I4) + 1*(I3-I5) + 1*(I3-I2)
0V = 1*I4 + 1*(I4-I5) + 1*(I4-I3) + 1*(I4-I1)
0V = 1*(I5-I3) + 1*(I5-I4) + 1*I5 + 1*(I5-I2)

This system solves to yield:
I1 = 5/7
I2 = 5/14
I3 = 2/7
I4 = 5/14
I5 = 3/7

I1=5/7 amp is the current through the battery.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionsolutionCharlie2016-04-15 19:29:21
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