Two 4-bit binary numbers, A = 1101 and B = 10 | vedclass

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(C) The circuit consists of a $NOT$ gate applied to input $A$, followed by a $NAND$ gate that takes the output of the $NOT$ gate and input $B$ as its

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$Y = 0111$

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(C) The circuit consists of a $NOT$ gate applied to input $A$, followed by a $NAND$ gate that takes the output of the $NOT$ gate and input $B$ as its inputs.Let $A = 1101$ and $B = 1010$.The output of the $NOT$ gate is $\overline{A} = 	ext{NOT}(1101) = 0010$.The output $Y$ of the $NAND$ gate is given by $Y = \overline{\overline{A} \cdot B}$.First, calculate the bitwise $AND$ operation: $\overline{A} \cdot B = 0010 \cdot 1010 = 0010$.Then, perform the $NOT$ operation on the result: $Y = \overline{0010} = 1101$.Wait, let's re-evaluate the circuit diagram. The diagram shows an inverter on input $A$ and a $NAND$ gate. If the inputs to the $NAND$ gate are $\overline{A}$ and $B$, then $Y = \overline{\overline{A} \cdot B} = A + \overline{B}$.$A = 1101$, $B = 1010$, so $\overline{B} = 0101$.$Y = 1101 + 0101 = 1101$.If the circuit is a $NAND$ gate with inputs $A$ and $B$, $Y = \overline{A \cdot B} = \overline{1101 \cdot 1010} = \overline{1000} = 0111$.Given the options, if we assume the circuit is a $NAND$ gate with inputs $A$ and $B$, the result is $0111$ (Option $C$).

Two $4-$bit binary numbers, $A = 1101$ and $B = 1010$, are given as inputs to the logic circuit shown in the figure below. The output $(Y)$ will be:

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