The combinational logic circuit shown in the given figure has an output Q which is \n\n[Figure needs to be uploaded later]
- A. ABC
- B. A+B+C
- C. ABC (bar)
- D. A.B+C
Answer: B. A+B+C
Explanation: [Figure needs to be uploaded later]\n$Q = \overline{A} \cdot \overline{B} \cdot C + \overline{A}B \cdot 1 + A\overline{B} \cdot 1 + AB \cdot 1$ \n$= \overline{A} \cdot \overline{B}C + \overline{A}B + A(\overline{B} + B)$ \n$= \overline{A} \cdot \overline{B}C + \overline{A}B + A \cdot 1$ \n$= \overline{A} \cdot \overline{B}C + (A + B)$ \n$= \overline{A} \cdot \overline{B}C + A + B$ \n$= (\overline{A} + A)(A + \overline{B}C) + B = 1 \cdot (A + \overline{B}C) + B$ \n$= A + (B + \overline{B})(B + C) = A + 1(B + C)$ \n$= A + B + C$