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 $1 \Omega$ (Fig. 3.16). Determine the equivalent resistance of the network and the current along each edge of the cube.
Example-(3.5)
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The network is not reducible to a simple series and parallel combinations of resistors. There is, however, a clear symmetry in the problem which we can exploit to obtain the equivalent resistance of the network.
The paths $AA ^{\prime}, AD$ and $AB$ are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, $I$. Further, at the corners $A ^{\prime}, B$ and $D$, the incoming current $I$ must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of $I$, using Kirchhoff's first rule and the symmetry in the problem.
Next take a closed loop, say, ABCC'EA, and apply Kirchhoff's second rule:
$
-I R-(1 / 2) I R-I R+\varepsilon=0
$
where $R$ is the resistance of each edge and $\varepsilon$ the emf of battery. Thus,
$
\varepsilon=\frac{5}{2} I R
$
The equivalent resistance $R_{e q}$ of the network is
$
R_{e q}=\frac{\varepsilon}{3 I}=\frac{5}{6} R
$
For $R=1 \Omega, R_{e q}=(5 / 6) \Omega$ and for $\varepsilon=10 V$, the total current ( $=3 I$ ) in the network is
$
3 I=10 V /(5 / 6) \Omega=12 \text { A, i.e., } I=4 A
$
The current flowing in each edge can now be read off from the Fig. 3.16.
The paths $AA ^{\prime}, AD$ and $AB$ are obviously symmetrically placed in the network. Thus, the current in each must be the same, say, $I$. Further, at the corners $A ^{\prime}, B$ and $D$, the incoming current $I$ must split equally into the two outgoing branches. In this manner, the current in all the 12 edges of the cube are easily written down in terms of $I$, using Kirchhoff's first rule and the symmetry in the problem.
Next take a closed loop, say, ABCC'EA, and apply Kirchhoff's second rule:
$
-I R-(1 / 2) I R-I R+\varepsilon=0
$
where $R$ is the resistance of each edge and $\varepsilon$ the emf of battery. Thus,
$
\varepsilon=\frac{5}{2} I R
$
The equivalent resistance $R_{e q}$ of the network is
$
R_{e q}=\frac{\varepsilon}{3 I}=\frac{5}{6} R
$
For $R=1 \Omega, R_{e q}=(5 / 6) \Omega$ and for $\varepsilon=10 V$, the total current ( $=3 I$ ) in the network is
$
3 I=10 V /(5 / 6) \Omega=12 \text { A, i.e., } I=4 A
$
The current flowing in each edge can now be read off from the Fig. 3.16.
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