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(C) The given circuit consists of a $NOR$ gate,a $NOT$ gate (implemented by a $NAND$ gate with shorted inputs),and an $AND$ gate.$1$. The output of th

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$B$

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(C) The given circuit consists of a $NOR$ gate,a $NOT$ gate (implemented by a $NAND$ gate with shorted inputs),and an $AND$ gate.$1$. The output of the $NOR$ gate with inputs $A$ and $B$ is $\overline{A+B}$.$2$. This output is fed into a $NAND$ gate whose inputs are shorted,acting as a $NOT$ gate. Thus,the output becomes $\overline{\overline{A+B}} = A+B$.$3$. This signal $(A+B)$ and the input $B$ are fed into an $AND$ gate.$4$. Therefore,the final output $Y = (A+B) \cdot B$.$5$. Using Boolean algebra: $Y = A \cdot B + B \cdot B = A \cdot B + B = B(A+1) = B \cdot 1 = B$.

The output $Y$ of the given network is :-

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