The output of a NAND gate is shown in the tru | vedclass

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(C) $NAND$ gate is a combination of an $AND$ gate followed by a $NOT$ gate.The Boolean expression for a $NAND$ gate is $Y = \overline{A \cdot B}$.Let

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$S$

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(C) $NAND$ gate is a combination of an $AND$ gate followed by a $NOT$ gate.The Boolean expression for a $NAND$ gate is $Y = \overline{A \cdot B}$.Let us evaluate the output $Y$ for all possible input combinations of $A$ and $B$:$1$. If $A = 0, B = 0$, then $A \cdot B = 0$, so $Y = \overline{0} = 1$.$2$. If $A = 0, B = 1$, then $A \cdot B = 0$, so $Y = \overline{0} = 1$.$3$. If $A = 1, B = 0$, then $A \cdot B = 0$, so $Y = \overline{0} = 1$.$4$. If $A = 1, B = 1$, then $A \cdot B = 1$, so $Y = \overline{1} = 0$.Comparing this with the given tables:- Table $(P)$ represents an $OR$ gate.- Table $(Q)$ represents an $XOR$ gate.- Table $(R)$ represents an $AND$ gate.- Table $(S)$ represents a $NAND$ gate.Thus, the correct truth-table is $(S)$.

The output of a $NAND$ gate is shown in the truth-table ($A$ and $B$ are inputs, $Y$ is output). Identify the correct truth-table.

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