Question:
The truth table for the circuit given in the fig. is:
The truth table for the circuit given in the fig. is:
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The output of the OR gate (inputs A and B) feeds one input of the NAND gate whose other input is A. Find Y for each (A, B).
Updated On: Nov 5, 2026
A B Y 0 0 1 0 1 1 1 0 1 1 1 1 A B Y 0 0 1 0 1 0 1 0 0 1 1 0 A B Y 0 0 1 0 1 1 1 0 0 1 1 0 A B Y 0 0 0 0 1 0 1 0 1 1 1 1
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The Correct Option is C
Solution and Explanation
The OR gate gives \(X = A + B\). The NAND gate has inputs A and X, so \(Y = \overline{A\cdot X} = \overline{A(A+B)}\).
By absorption, \(A(A+B) = A\), hence \(Y = \overline{A}\).
Row by row: (A,B) = (0,0): X=0, Y = NAND(0,0) = 1.
(0,1): X=1, Y = NAND(0,1) = 1.
(1,0): X=1, Y = NAND(1,1) = 0.
(1,1): X=1, Y = NAND(1,1) = 0.
The table is Y = 1, 1, 0, 0, which is option (3).
Answer: (3) the table with Y = 1, 1, 0, 0
By absorption, \(A(A+B) = A\), hence \(Y = \overline{A}\).
Row by row: (A,B) = (0,0): X=0, Y = NAND(0,0) = 1.
(0,1): X=1, Y = NAND(0,1) = 1.
(1,0): X=1, Y = NAND(1,1) = 0.
(1,1): X=1, Y = NAND(1,1) = 0.
The table is Y = 1, 1, 0, 0, which is option (3).
Answer: (3) the table with Y = 1, 1, 0, 0
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