JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
Four \(NOR\) gates are connected as shown in figure. The truth table for the given figure is :

- A
\(A\) \(B\) \(Y\) \(0\) \(0\) \(1\) \(0\) \(1\) \(0\) \(1\) \(0\) \(1\) \(1\) \(1\) \(0\) - B
\(A\) \(B\) \(Y\) \(0\) \(0\) \(0\) \(0\) \(1\) \(1\) \(1\) \(0\) \(1\) \(1\) \(1\) \(0\) - C
\(A\) \(B\) \(Y\) \(0\) \(0\) \(0\) \(0\) \(1\) \(1\) \(1\) \(0\) \(0\) \(1\) \(1\) \(1\) - D
\(A\) \(B\) \(Y\) \(0\) \(0\) \(1\) \(0\) \(1\) \(0\) \(1\) \(0\) \(0\) \(1\) \(1\) \(1\)
Answer & Solution
Correct Answer
(D)
\(A\) \(B\) \(Y\) \(0\) \(0\) \(1\) \(0\) \(1\) \(0\) \(1\) \(0\) \(0\) \(1\) \(1\) \(1\)
Step-by-step Solution
Detailed explanation
\(y=\overline{(A+\overline{A+B})+(\overline{B+\overline{A+B})}}\) \(y=(A+\overline{A+B}) \cdot(B+\overline{A+B})\) \(A\) \(B\) \(Y\) \(0\) \(0\) \(1\) \(0\) \(1\) \(0\) \(1\) \(0\) \(0\) \(1\) \(1\) \(1\)
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