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Question
Twelve wires, each having equal resistance of 2 Ω, are joined to form a cube. What is the equivalent resistance between diagonally opposite points?
Twelve wires, each having equal resistance of 2 Ω, are joined to form a cube. What is the equivalent resistance between diagonally opposite points?
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Solution
The correct option is B 1Ω
I can very clearly see that if I exchange the point A and B, it changes nothing.
So this circuit has point symmetry for sure. So the way current enters has to be the same way it leaves.
Now if you look at the paths, the way all three currents flow in the branches is exactly the same. One cannot differentiate these paths from each other. So we have path symmetries as well, which implies that i1=i2=i3.
i1=i2=i3=2i, let’s say. Soon you’ll know why we are taking it like that. But you can try to guess though, nobody can stop you to wonder!
As discussed in earlier part of this answer that there is a path symmetry in all the three paths diverging from point A. So at junctions D, H and F, current will again distribute itself symmetrically. So now you can see how the current distribution looks like as shown below:
Let’s attach a battery of ‘V’ volts to find the equivalent resistance between A and B, as shown:
Kirchhoff’s Loop rule: Let’s choose the loop ADCBJA
Note: Every edge of the cube or every wire has a resistance of 2 Ω.
−2i×2−i×2−2i×2+V=0
6i=V
V6i=1 Ω
I can very clearly see that if I exchange the point A and B, it changes nothing.
So this circuit has point symmetry for sure. So the way current enters has to be the same way it leaves.
Now if you look at the paths, the way all three currents flow in the branches is exactly the same. One cannot differentiate these paths from each other. So we have path symmetries as well, which implies that i1=i2=i3.
i1=i2=i3=2i, let’s say. Soon you’ll know why we are taking it like that. But you can try to guess though, nobody can stop you to wonder!
As discussed in earlier part of this answer that there is a path symmetry in all the three paths diverging from point A. So at junctions D, H and F, current will again distribute itself symmetrically. So now you can see how the current distribution looks like as shown below:
Let’s attach a battery of ‘V’ volts to find the equivalent resistance between A and B, as shown:
Kirchhoff’s Loop rule: Let’s choose the loop ADCBJA
Note: Every edge of the cube or every wire has a resistance of 2 Ω.
−2i×2−i×2−2i×2+V=0
6i=V
V6i=1 Ω
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