Can any one point me in the right direction. #1. You are given an \"empty\" cube

Question

Can any one point me in the right direction.
#1. You are given an "empty" cube (that is having only its edgesbut no faces) made of 12 identical wires, each wire edge (each wireforming an edge) having same electric resitance R=1 Ohm. Find thetotal electric resistance in DC of this cube between one vertex andits other, most opposite vertex. That is, if you connect anOhm-meter to one of its vertices and the other terminal of theohmmeter to the opposite vertex, what will it show?
Can any one point me in the right direction.
#1. You are given an "empty" cube (that is having only its edgesbut no faces) made of 12 identical wires, each wire edge (each wireforming an edge) having same electric resitance R=1 Ohm. Find thetotal electric resistance in DC of this cube between one vertex andits other, most opposite vertex. That is, if you connect anOhm-meter to one of its vertices and the other terminal of theohmmeter to the opposite vertex, what will it show?

Explanation / Answer

The idea is that there is some total voltage V applied acrossthe corners and some total current I runs through the wires. . By symmetry, the current must split three ways when it leavesthe first corner because there are three identical wires. So eachof the first three resistors will have current I/3 . Then the current splits again when it reaches the nextcorner... two ways. So each of the next two resistors willget current I/6. . Then the currents recombine... and the next resistor getscurrent I/3. . And these three currents meet at the opposite corner. . To get from one corner to the other you must gothrough:   one wire with current I/3;  one wire with current I/6;   and one wire with currentI/3. . The voltages across each of these are:    (I/3) R       (I/6)R     (I/3)R           . because for each, voltage = current * resistance . So the total voltage from one corner to the other is: .      total voltage =      (I/3)R       (I/6)R     (I/3) R     . But total voltage is also      Itotal * R total. . So...           I total * R total =    (1/3 + 1/6 + 1/3) I R   .     But I total is the same as"I"   and R = 1 ohm so... .     R total = (5/6) * 1ohm   =    5/6 ohm   = 0.8333 ohms

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