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(C) $NAND$ gate has two inputs,say $A$ and $B$,and its output is given by $Y = \overline{A \cdot B}$.If the two inputs are shorted,then $A = B$. Let t

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an $NOT$ gate

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(C) $NAND$ gate has two inputs,say $A$ and $B$,and its output is given by $Y = \overline{A \cdot B}$.If the two inputs are shorted,then $A = B$. Let this common input be $A$.Substituting this into the $NAND$ expression,we get $Y = \overline{A \cdot A}$.Since $A \cdot A = A$ in Boolean algebra,the expression becomes $Y = \overline{A}$.The expression $Y = \overline{A}$ represents the operation of a $NOT$ gate.Therefore,when the inputs of a $NAND$ gate are shorted,it functions as a $NOT$ gate.

If two inputs of a $NAND$ gate are shorted,the resulting gate is

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