$\begin {array} {1 1} (A)\;5A & \quad (B)\;\large\frac{10}{3}A \\ (C)\;\large\frac{20}{3}A & \quad (D)\;\large\frac{5}{3}A \end {array}$

This question can be solved without much calculation. If A and B are connected,

then by the symmetry of the circuit, one can say that the current going through the $12\: ohm $

resistor on the left is the same as the current going out through the $12\: ohm$ resistor on the right.

Similarly is the case for the $6\: ohm$ resistor. Hence, there exist two current channels,

one of $24\: ohms$ and the other of $12\: ohms$. Hence the current is divided in $1:2$ ratio.

Effective resistance with A and B shorted is $ 8\: ohms$. Total current is $10A$.

Hence the current through the shorting wire is $ \large\frac{10}{3}$$ A$.

Ans : (B)

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