A bettery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors each of resistance `1Omega` Determine the equivalent resistance of the network and the current along each edge of the cube.

Explanation

Above cubical network ls shown as per the statement, where each of 12 identical sides, is a resistor of resistance R = 1 `Omega`. We have to find equivalent resistance of this network between its diagonally opposite points, say between points A and G in above figure. For this purpose, connect a battery across these two points.
Now, current I starting from positive terminal of battery reaches point A, after which it is offered three branches AB, AD and AE and so it is divided into three parts. Here, minimum resistance offered to each of these three parts to reach negative terminal of battery is same and so all of these three parts are equal. Hence, each part wouJd be I/3. Thus, current arriving at points E, D and B are all I/3. These all currents are further divided into two parts as all of them are further offered two branches after points B, D and E. Here, minimum resistance offered to each of these two parts to reach negative terminal of battery is same and so each 1/3 amount of current gets divided into two equal parts after coming to points B, D and E. Hence, currents passing through BC and BF, DC and DH, EF and EH are all
`((I)/(3))/(2) = (1)/(6)` .
Now, at each of the junctions C, F and H, two cu.rrents each of amount 1/6 add up to give 1/3 amount of current, which flow in the branches CG, FG and HG respectively. These three cu.rrents each of amount 1/3 associate at exit point G to give total current I which is equal to current at the entry point A of given cubical network..
Now, applying KVL (Kirchhoff.s Voltage Law) in the direction of current along the closed path AEHGJKA we get,
- `((I)/(3)) (R) - ((I)/(6)) (R) - ((I)/(3)) (R) = - epsilon `
`therefore epsilon = (IR)/(3) + (IR)/(6) + (IR)/(3) `
`therefore epsilon = (5 IR)/(6) " " `... (1)
Now, if equivalent resistance of given cubical network between its diagonally opposite end points is `R_(eq)`, then
`R_(eq) = (epsilon)/(I) = (5R)/(6)` [ From equation (1) ] ... (2)
Here R = 1 `Omega` and so : `R_(eq) = (5)/(6) Omega`
From equation (1) ,
`10 = (5I xx 1)/(6)`
`therefore I = 12 A `
Current through branches AB, AD, AE, CG, FG, HG is , `(I)/(3) = (12)/(3) = 4` A
Current through branches BC, BF, DC, DH, EF, EH is , `(I)/(6) = (12)/(6) = 2 ` A