NEET · Physics · STD 12 - 14. Semicondutor electronics
For the following logic circuit, the truth table is:
- A
\(A\) \(B\) \(Y\) \(0\) \(0\) \(0\) \(0\) \(1\) \(0\) \(1\) \(0\) \(0\) \(1\) \(1\) \(1\) - B
\(A\) \(B\) \(Y\) \(0\) \(0\) \(1\) \(0\) \(1\) \(1\) \(1\) \(0\) \(1\) \(1\) \(1\) \(0\) - C
\(A\) \(B\) \(Y\) \(0\) \(0\) \(0\) \(0\) \(1\) \(1\) \(1\) \(0\) \(1\) \(1\) \(1\) \(1\) - D
\(A\) \(B\) \(Y\) \(0\) \(0\) \(1\) \(0\) \(1\) \(0\) \(1\) \(0\) \(1\) \(1\) \(1\) \(0\)
Answer & Solution
Correct Answer
(C)
\(A\) \(B\) \(Y\) \(0\) \(0\) \(0\) \(0\) \(1\) \(1\) \(1\) \(0\) \(1\) \(1\) \(1\) \(1\)
Step-by-step Solution
Detailed explanation
\(Y=\overline{\bar{A} \cdot \bar{B}}=A+B\) It is \(OR\) gate.
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