The following logic gate circuit is equivalen | vedclass

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(A) Let the inputs be $A$ and $B$. The first two $NAND$ gates act as $NOT$ gates because their inputs are shorted. Thus,the outputs of these gates are

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$NOR$ Gate

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(A) Let the inputs be $A$ and $B$. The first two $NAND$ gates act as $NOT$ gates because their inputs are shorted. Thus,the outputs of these gates are $\bar{A}$ and $\bar{B}$ respectively.These are fed into a third $NAND$ gate,which produces the output $Y' = \overline{\bar{A} \cdot \bar{B}} = A + B$ (by De Morgan's Law).This output $Y'$ is then fed into a final $NAND$ gate acting as a $NOT$ gate,resulting in the final output $Y = \overline{Y'} = \overline{A + B}$.The Boolean expression $Y = \overline{A + B}$ corresponds to a $NOR$ gate.Truth table:$A$$B$$Y$$0$$0$$1$$0$$1$$0$$1$$0$$0$$1$$1$$0$

The following logic gate circuit is equivalent to which of the following gates?

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