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 Electrical Cube 3 (Posted on 2016-04-15)
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 No Rating 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.

 Subject Author Date solution Charlie 2016-04-15 19:29:21

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