Question

A 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 $1\Omega$ in figure. Determine the equivalent resistance of the network and the current along each edge of the cube.
Refer to point no. 16[Important Terms, Definitions and Formulae]

Answer 1

Sol. Let us number the corners as $1,2,3,\dots$ and so on as shown

$a(1-8-7-6),b(1-2-7-6),c(1-8-5-6),d(1-2$

$-3-6),e(1-4-5-6),f(1-4-3-6)$

Now for each of these paths, we have identical resistances.

Let the current entering at point $A$ be $I.$ The current will be

distributed equally among all the three parts. Therefore, we can

say that the current is

$I_{12}=I_{14}=I_{18}=I/3$

Now look at all paths from 2 to 6 . We have the following paths

$g(2-7-6)$ and $h(2-3-6)$ .

since they also have the same resistances, we can assume that

current in $2-3$ and current in $2-7$ must be same. Therefore,

$I_{27}=I_{23}=\frac{1}{2}\left(\frac{I}{3}\right)=\frac{I}{6}$

we can say that $I_{43}=I_{45}=I_{85}=I_{87}=I/6$

Therefore, to calculate the potential difference across $1-6$

(see Fig. 5.34 ) the path $A_{12\%},$ we have

$V_1-\frac{I}{3}R-\frac{I}{6}R-\frac{I}{3}R=V_6\Rightarrow V_1-V_6=I[\frac{R}{3}+\frac{R}{6}+\frac{R}{3}]=\frac{5}{6}IR$

Hence, $\frac{(V_1-V_6)}{I}=\frac{5}{6}R$

Therefore, the equivalent resistance is $\frac{5}{6}R$ .

$I=\frac{V}{\frac{5}{6}R}=\frac{10}{5/6}=12A$

Hence, equivalent resistance is $\frac{5}{6}\Omega$ , $I_{12}=I_{14}=I_{18}=4A$ and$I_{43}=I_{45}=I_{85}=I_{87}=2A$