MHT CET · Physics · Semiconductors
The Boolean expression for 'XOR' gate \(C=(A \oplus B)\) is equal to
- A \((\mathrm{A} \cdot \mathrm{B})+(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}})\)
- B \(\mathrm{A}+(\overline{\mathrm{A}} \cdot \overline{\mathrm{B}})\)
- C \((\mathrm{A} \cdot \mathrm{B})+\overline{\mathrm{B}}\)
- D \((\overline{\mathrm{A}} \cdot \mathrm{B})+(\mathrm{A} \cdot \overline{\mathrm{B}})\)
Answer & Solution
Correct Answer
(D) \((\overline{\mathrm{A}} \cdot \mathrm{B})+(\mathrm{A} \cdot \overline{\mathrm{B}})\)
Step-by-step Solution
Detailed explanation
The Boolean expression of \(X O R\) gate is
\(y=A \oplus B=\bar{A} \cdot B+A \cdot \bar{B}\)
\(y=A \oplus B=\bar{A} \cdot B+A \cdot \bar{B}\)
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