# Twelve wires each having a resistance of 3 Ω are connected to form a cubical network. A battery of 10 V and negligible internal resistance is connected across the diagonally opposite - Physics

Twelve wires each having a resistance of 3 Ω are connected to form a cubical network. A battery of 10 V and negligible internal resistance is connected across the diagonally opposite corners of this network. Determine its equivalent resistance and the current along each edge of the cube.

#### Solution

A battery of 10 V is connected across the diagonally opposite points a and f of the given network. Each side of the cube represents a 3-ohm resistor. Current entrees through the point a and leaves through the point f. Let I current is drawn from the battery and enters through point a, due to the symmetry of cube between point a and f, the current will divide itself into equal parts at each corner of the cube.
As you can see in the figure given below:

If we consider the current path ABGFA, then we can write, VA-B + VB-G  + VG-F = V

⇒ ("I"/3 xx"R") + ("I"/6 xx "R") + ("I"/3 xx "R") = ("I" xx "R"_"net")

⇒ "R"_"net" = "R"/3 + "R"/6 + "R"/3 = (5"R")/6

∵ R = 3Ω

⇒ "R"_"net" = (5xx3)/6 = 2.5 Ω

Current, "i" = "V"/"R"_"net" = 10/2.5 = 4 "A"

⇒ i = 4 A

As i = 4 A, hence current through each side of the cube can be marked. The below figure shows the current through each resistance of the network. i.e. each side of the cube.

Is there an error in this question or solution?