A battery of 10V and negligible internal resistance, is connected across the diagonally opposite corners of a cubical network consisting of 12 resistors, each of resistance 1 $Ω$ . The equivalent resistance of the network and the current along each edge of the cube is

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5/6 ohms, 2 amperes

5/6 ohms, 4 amperes

6/5 ohms, 2 amperes

6/5 ohms, 4 amperes

answer is A.

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Detailed Solution

Applying kirchoff’s voltage law in closed mesh BCGHB
-Ri/3 - Ri / 6 - Ri / 3 + V = 0
$\begin{array}{l} V = \frac{Ri}{3} + \frac{Ri}{6} + \frac{Ri}{3}; V = \frac{5 Ri}{6} \\ {iR}_{eff} = \frac{5 Ri}{6} \Rightarrow R_{eff} = \frac{5 R}{6} \end{array}$
∴ effective resistance, $R_{eff} = \frac{5 R}{6} = \frac{5}{6} \times 1 = \frac{5}{6} Ω$
From the figure, it is clear that current in each edge of the cube is i/3
$∴ \frac{i}{3} = \frac{\frac{V}{R}}{3} = \frac{\frac{10}{5 / 6}}{3} = \frac{12}{3} = 4 A$