In the following logic circuit,the sequence o | vedclass

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(A) The given circuit consists of an $AND$ gate,an $OR$ gate,and a $NAND$ gate.The inputs to the $AND$ gate are $A$ and $B$,so its output is $A \cdot

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$1,1,1,0$

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(A) The given circuit consists of an $AND$ gate,an $OR$ gate,and a $NAND$ gate.The inputs to the $AND$ gate are $A$ and $B$,so its output is $A \cdot B$.The inputs to the $OR$ gate are $A$ and $B$,so its output is $A + B$.These two outputs are fed into a $NAND$ gate,so the final output $Y$ is given by:$Y = \overline{(A \cdot B) \cdot (A + B)}$Now,we calculate $Y$ for each input sequence $(A, B)$:$1$. For $(0,0): Y = \overline{(0 \cdot 0) \cdot (0 + 0)} = \overline{0 \cdot 0} = \overline{0} = 1$$2$. For $(0,1): Y = \overline{(0 \cdot 1) \cdot (0 + 1)} = \overline{0 \cdot 1} = \overline{0} = 1$$3$. For $(1,0): Y = \overline{(1 \cdot 0) \cdot (1 + 0)} = \overline{0 \cdot 1} = \overline{0} = 1$$4$. For $(1,1): Y = \overline{(1 \cdot 1) \cdot (1 + 1)} = \overline{1 \cdot 1} = \overline{1} = 0$Thus,the sequence of outputs $Y$ is $1, 1, 1, 0$.

In the following logic circuit,the sequence of the inputs $A, B$ are $(0,0), (0,1), (1,0)$ and $(1,1)$. The output $Y$ for this sequence will be:

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