Semiconductors

JEE Main level, single correct, medium

131 of 209

Question 131 of 209

Two 44 bits binary numbers, A=1101A = 1101 and B=1010B = 1010 are given in the inputs of a logic circuit shown in figure below. The output (Y)(Y) will be:

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Answer: (A) Y=1101Y = 1101

From the given logic circuit, the input AA passes through a NOT gate, giving A‾\overline{A}.

This output A‾\overline{A} and the input BB are then fed into a NAND gate.

The Boolean expression for the final output YY is:
Y=A‾⋅B‾Y = \overline{\overline{A} \cdot B}

Given the 44-bit binary inputs:
A=1101A = 1101
B=1010B = 1010

First, find the bitwise NOT of AA:
A‾=1101‾=0010\overline{A} = \overline{1101} = 0010

Next, find the bitwise AND of A‾\overline{A} and BB:
A‾⋅B=0010⋅1010=0010\overline{A} \cdot B = 0010 \cdot 1010 = 0010

Finally, apply the NOT operation for the NAND gate:
Y=0010‾=1101Y = \overline{0010} = 1101

Alternatively, using De Morgan's laws:
Y=A‾⋅B‾=A+B‾Y = \overline{\overline{A} \cdot B} = A + \overline{B}
Y=1101+1010‾=1101+0101=1101Y = 1101 + \overline{1010} = 1101 + 0101 = 1101

Therefore, the output YY is 11011101.

Answer: Y=1101Y = 1101

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