Another wire cube with 1 ohm edges, just like the cube is
Electrical Cube, is made. But this time the 1 volt battery is connected to two points diagonally opposed on one face of the cube. How much current is in the wire connecting the battery to the cube structure this time?
Let's refer to the face to which the two wires are attached as a "diamond" to emphasize the diagonal nature of the effective passage of the current, and the independence from non-topological measurements.
This diamond, taken by itself, acts as a 1-ohm resistor as each path around it acts as a 2-ohm resistor, and the two paths are in parallel.
The opposite face acts as another 1-ohm resistor. Two of the literal 1-ohm resistors connect this opposite face to the original face at the points at which the the 1-volt potential difference exists. The other two literal 1-ohm resistors connect points with equal potential (by symmetry with either end), 1/2 volt different from either input end, as each is halfway in the two parallel pieces. No current will flow in those two resistors.
Ultimately, then, we have a 1-ohm resistor in parallel with a 3-ohm resistor for a total conductivity of 4/3 mho, or resistance of 3/4 ohm.
The current in the wires connecting the battery terminals with the two points on the face diagonal of the cube will be 4/3 amp. 1/2 amp goes through each of the two routes in the face of the cube to which the leads are attached. 1/3 amp goes through each of the two current-carrying resistors that lead to the opposite face (and no current, as mentioned, in the other two such resistors). The opposite face has 1/6 amp going through each of its edges.
1/2 v
|
1/2 A / \ 1/2 amp
/ \
/ \
/ \
4/3 amp 0 v / \ 1 v 4/3 amp
-------------------- ------------------
|\ /|
| \ / |
| \ 1/2 / |
| \ V / |
| \ / |
| 1/2 A |
| |
| |
1/3 amp | 1/2 V | 1/3 amp
| | |
| / \ |
| / \ |
| / 1/6 \ |
| / A \ |
|/ \|
| |
1/3 V \ / 2/3 V
\ /
\ /
\ /
\ / 1/6 amp
|
1/2 V
Two resistors not carrying current
are not shown.
|
Posted by Charlie
on 2016-03-25 10:22:17 |