JEE Mains · Physics · STD 12 - 14. Semicondutor electronics
The truth table for this given circuit is _______.
- A
\(A\) \(B\) \(Y\) \(0\) \(0\) \(1\) \(0\) \(1\) \(1\) \(1\) \(0\) \(1\) \(1\) \(1\) \(0\) - B
\(A\) \(B\) \(Y\) \(0\) \(0\) \(0\) \(0\) \(1\) \(1\) \(1\) \(0\) \(0\) \(1\) \(1\) \(1\) - C
\(A\) \(B\) \(Y\) \(0\) \(0\) \(0\) \(0\) \(1\) \(0\) \(1\) \(0\) \(0\) \(1\) \(1\) \(1\) - D
\(A\) \(B\) \(Y\) \(0\) \(0\) \(1\) \(0\) \(1\) \(0\) \(1\) \(0\) \(1\) \(1\) \(1\) \(0\)
Answer & Solution
Correct Answer
(B) \(A\) \(B\) \(Y\) \(0\) \(0\) \(0\) \(0\) \(1\) \(1\) \(1\) \(0\) \(0\) \(1\) \(1\) \(1\)
Step-by-step Solution
Detailed explanation
\(\mathrm{Y} =\mathrm{A} \cdot \mathrm{B}+\overline{\mathrm{A}} \cdot \mathrm{B} \) \( =(\mathrm{A}+\overline{\mathrm{A}}) \cdot \mathrm{B} \) \(\mathrm{Y} =1 \cdot \mathrm{B} \) \(\mathrm{Y} =\mathrm{B}\)
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