question_answer

Twelve wires of each of resistance 60 are connected to form a cube as shown in the figure. The current enters at a comer A and leaves at the diagonally opposite corner G. The joint resistance across the comers A and G are:

A)  12\[\Omega \]           

B)  60\[\Omega \]

C)  30\[\Omega \]            

D)  50\[\Omega \]

Correct Answer: D

Solution :

Let ABCDEFGH be the skeleton cube formed by joining twelve equal wires each of resistance r. Let the current enters the cube at comer A and after passing through all twelve wires, let the current leaves at G, a comer diagonally opposite to comer A. For the sake of convenience, let us suppose that the total current is 61. At A, this current is divided into three equal parts each (21) along AE, AB and AD as the resistance along these paths are equal and their end points are equidistant from exit point G. At the points E, B and D, each part is further divided into two equal parts each part equal to i. The distribution of current in the various arms of skeleton cube is shown according to Kirchhoffs first law. The current leaving the cube at G is again 6i. Applying Kirchhoffs second law to the closed circuit ADCGA, we get \[2ir+ir+2ir=E\] or \[5ir=E\] ??..(i) where E is the emf of the cell of neglegible internal resistance. If R is the resistance of the cube between the diagonally opposite corners A and G, then according to Ohms law, we have \[6i\times R=E\]               ...(ii) From Eqs. (i) and (ii), we have \[6iR=5ir\] or \[R=\frac{5}{6}r\] Here, \[r=6\Omega \] \[\therefore \] \[R=\frac{5}{6}\times 6\] or \[R=5\Omega \]