Question

Twelve wires of each of resistance $6$ ohms are connected to form a cube as shown in the figure. The current enters at a corner $A$ and leaves at the diagonally opposite corner $G$. The joint resistance across the corners $A$ and $G$ are

• A. 12 ohms
• B. 6 ohms
• C. 3 ohms
• D. 5 ohms

Answer 1

Answer: (D)

Let $A B C D E F G H$ be the skeleton cube formed by joining twelve equal wires each of resistance $r$. Let the current enters the cube at corner $A$ and after passing through all twelve wires, let the current leaves at $G$, a corner diagonally opposite to corner $A$. For the sake of convenience, let us suppose that the total current is $6 i$. At $A$, this current is divided into three equal parts each $(2 i)$ along
$A E, A B$ and $A D$ 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 Kirchhoff's first law. The current leaving the cube at $G$ is again $6 i$.
Applying Kirchhoff's second law to the closed circuit $A D C G A$, we get
$2 i r+i r+2 i r=E $
or $5 i r=E...$(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 Ohm's law, we have
$6 i \times R=E ..... $(ii)
From Eqs. (i) and (ii), we have $6 i R=5 i r$
or $R=\frac{5}{6} r$
Here, $r=6 \Omega$
$\therefore R=\frac{5}{6} \times 6$
Or $R=5 \Omega$