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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 :

  1. A
    \(A\) \(B\) \(Y\)
    \(0\) \(0\) \(1\)
    \(0\) \(1\) \(0\)
    \(1\) \(0\) \(1\)
    \(1\) \(1\) \(0\)
  2. B
    \(A\) \(B\) \(Y\)
    \(0\) \(0\) \(0\)
    \(0\) \(1\) \(1\)
    \(1\) \(0\) \(1\)
    \(1\) \(1\) \(0\)
  3. C
    \(A\) \(B\) \(Y\)
    \(0\) \(0\) \(0\)
    \(0\) \(1\) \(1\)
    \(1\) \(0\) \(0\)
    \(1\) \(1\) \(1\)
  4. D
    \(A\) \(B\) \(Y\)
    \(0\) \(0\) \(1\)
    \(0\) \(1\) \(0\)
    \(1\) \(0\) \(0\)
    \(1\) \(1\) \(1\)
Verified Solution

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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