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Twelve equal wires each of resistance r are joined to form a skeleton cube. Find the equivalent resistance between two corners on the same edge of the cube. |
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Answer» Solution :Method-I CONNECT a source between points 1 and 2. Let current `i` enter through point 1 into the network . The network is symmetrical about dotted lines. The currents above and below dotted line are symmetrically distributed as shown in fig. By junction RULE at 1, we have `i= x+ 2y` `therefore R_(12) = V/i= V/(x+2y) ""....(i)`  In close loop 1-2-9-10-1 , we have `-rx + V=0 ` or `x= (V)/r ""....(ii)` In close loop 1-4-3-2-1 `-rz-rz-ry+rx=0` (or) `x-2y - z =0""....(iii)` In close loop `4-8-7-3-4 ` `-r(y-z) - r xx 2(y-z) + rz=0` or `-4(y-z) + z =0` or `4Y+ 5z=0""....(iv)` From equations (iii) and (iv) , we GET `-4y + 5(x-2y) = 0 ` or `5x=14y` Since `x = V/r therefore y= 5/(14) xx V/r` Now `R_(12) = V/(x+2y) = (V)/(V/r + 2 xx(5V)/(14r)) =(7r)/(12` Method II: Out previous knowledge reveals that points 3 and 6 must be at the same potential . So must be 4 and 5 . If points of equal potential are joint by a wire, the currents in the circuit do not change. The given network of resistors can be reduced successively as shown in figure.  `R_(12) =(rxx(7r)/(5))/(r+(7r)/(5)) = 7/(12) r` |
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