What is the output Y in the following circuit | vedclass

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(C) The given circuit consists of two $NAND$ gates. Let the output of the first $NAND$ gate be $X$. The inputs to this gate are $A$ and $B$. Thus, $X

Vedclass · Accepted Answer

$1, 0$

Vedclass · Answer

(C) The given circuit consists of two $NAND$ gates. Let the output of the first $NAND$ gate be $X$. The inputs to this gate are $A$ and $B$. Thus, $X = \overline{A \cdot B}$.The second $NAND$ gate takes $X$ and $C$ as inputs. Thus, the final output $Y = \overline{X \cdot C} = \overline{(\overline{A \cdot B}) \cdot C}$.Case $1$: When $A = 0, B = 0, C = 0$:$X = \overline{0 \cdot 0} = \overline{0} = 1$.$Y = \overline{1 \cdot 0} = \overline{0} = 1$.Case $2$: When $A = 1, B = 1, C = 1$:$X = \overline{1 \cdot 1} = \overline{1} = 0$.$Y = \overline{0 \cdot 1} = \overline{0} = 1$.Wait, re-evaluating the circuit diagram: The first gate is a $NAND$ gate, and the second gate is also a $NAND$ gate. Let's re-check the logic.For $A=0, B=0, C=0$: $X = \overline{0 \cdot 0} = 1$. Then $Y = \overline{1 \cdot 0} = 1$.For $A=1, B=1, C=1$: $X = \overline{1 \cdot 1} = 0$. Then $Y = \overline{0 \cdot 1} = 1$.Looking at the provided solution image, the first gate is actually an $AND$ gate. Let's re-read the diagram. The first gate is an $AND$ gate, and the second is a $NAND$ gate.If first is $AND$: $X = A \cdot B$. Then $Y = \overline{X \cdot C} = \overline{(A \cdot B) \cdot C}$.Case $1$: $A=0, B=0, C=0 \implies X = 0 \cdot 0 = 0 \implies Y = \overline{0 \cdot 0} = 1$.Case $2$: $A=1, B=1, C=1 \implies X = 1 \cdot 1 = 1 \implies Y = \overline{1 \cdot 1} = 0$.Thus, the outputs are $1, 0$. The correct option is $C$.

What is the output $Y$ in the following circuit, when all the three inputs $A, B, C$ are first '$0$' and then '$1$'?

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