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(A) tetrahedron has $4$ vertices and $6$ edges. If each edge contains a capacitor of capacitance $C$, we want to find the equivalent capacitance betwe

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$2$

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(A) tetrahedron has $4$ vertices and $6$ edges. If each edge contains a capacitor of capacitance $C$, we want to find the equivalent capacitance between any two vertices (junctions).Let the vertices be $A, B, C, D$. We want the equivalent capacitance between $A$ and $B$.By symmetry, the potential at $C$ and $D$ will be equal if a potential difference is applied across $A$ and $B$.Since the potentials at $C$ and $D$ are equal, no charge flows through the capacitor connected between $C$ and $D$. Thus, this capacitor can be removed.The circuit then simplifies to two parallel branches connected between $A$ and $B$:$1$. $A$ branch with two capacitors in series ($C/2$), and the path $A-D-B$ has two capacitors in series ($C/2$). These two paths are in parallel with the direct capacitor $A-B$ ($C$).Therefore, $C_{eq} = C + \frac{C}{2} + \frac{C}{2} = 2C$.

If the capacitance of each capacitor in the tetrahedron shown is $C$, then the effective capacitance of the network across any two junctions is ......$C$.

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