NEET · Physics · STD 12 - 14. Semicondutor electronics
The Truth table for the given logic circuit is:
- A
\(A\) \(B\) \(C\) \(0\) \(1\) \(0\) \(0\) \(1\) \(0\) \(1\) \(0\) \(0\) \(1\) \(1\) \(1\) - B
\(A\) \(B\) \(C\) \(0\) \(0\) \(1\) \(0\) \(1\) \(0\) \(1\) \(0\) \(1\) \(1\) \(1\) \(0\) - C
\(A\) \(B\) \(C\) \(0\) \(0\) \(0\) \(0\) \(1\) \(0\) \(1\) \(0\) \(0\) \(1\) \(1\) \(1\) - D
\(A\) \(B\) \(C\) \(0\) \(0\) \(0\) \(0\) \(1\) \(1\) \(1\) \(0\) \(1\) \(1\) \(1\) \(0\)
Answer & Solution
Correct Answer
(B) \(A\) \(B\) \(C\) \(0\) \(0\) \(1\) \(0\) \(1\) \(0\) \(1\) \(0\) \(1\) \(1\) \(1\) \(0\)
Step-by-step Solution
Detailed explanation
\(C=\overline{A \cdot B} \cdot \bar{A} \cdot B\)
using De-Morgan Theorem
\(C=\overline{A \cdot B+\bar{A} \cdot B}\)
\(C=\overline{B(A+\bar{A})}=\bar{B}\)
Therefore
using De-Morgan Theorem
\(C=\overline{A \cdot B+\bar{A} \cdot B}\)
\(C=\overline{B(A+\bar{A})}=\bar{B}\)
Therefore
| \(A\) | \(B\) | \(C\) |
| \(0\) | \(0\) | \(1\) |
| \(0\) | \(1\) | \(0\) |
| \(1\) | \(0\) | \(1\) |
| \(1\) | \(1\) | \(0\) |
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