Battery 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 `1 Omega` (Fig. 3.23). Determine the equivalent resistance of the network and the current along each edge of the cube.

Explanation

Let 6I be the current through the cell. At junction A, this current will get divided into three equal parts (due to symmetry). Current through AD will again get divided into two equal parts at junction D. So DC and DE will have current I. Similarly, BC, BG, HE and HG will also have current I flowing through them. The current through DC and BC will meet and add at junction C and thus, current 2I will flow through the path CF. Similarly, current 2I will flow through EF and GF. These three currents will reunite at F again and will add to 6I again. Applying Kirchhoff.s loop rule to AHGFA, we get `-2IR-IR -2IR+ epsi= 0`
`rArr 2IR + IR + 2IR - epsi=0`
`:. 5IR=epsi`
The equivalent resistance of the network is
`rArr R.= ("Total emf")/("Total current") = (epsi)/(6I) = (5IR)/(6I) = (5)/(6)R`
Given, `R= 1Omega`
`:. R. =(5)/(6)Omega`
`:.` Total current in the network is
`6I= (epsi)/(R.) = (10)/(5) =12A`
`rArr I= (12)/(6)=2 Lamda`