Q149 / 150 · MHT CET 2026 (18 April Shift 2) · Physics
The output Y of the given logic circuit is
- A$\overline{(A \cdot B)} + \overline{(C \cdot B)}$
- B$\overline{(A + B)} \cdot \overline{(B \cdot C)}$
- Correct answerC$\overline{(A \cdot B)} + \overline{(B + C)}$
- D$(A + B) + (B + C)$
Solution
The given logic circuit consists of three gates.
The upper gate is a NAND gate with inputs $A$ and $B$. Its output is $\overline{A \cdot B}$.
The lower gate is a NOR gate with inputs $B$ and $C$. Its output is $\overline{B + C}$.
The final gate is an OR gate which takes the outputs of the NAND and NOR gates as its inputs.
Therefore, the final output $Y$ is given by the logical OR operation of these two intermediate outputs:
$Y = \overline{A \cdot B} + \overline{B + C}$
Answer: $\overline{(A \cdot B)} + \overline{(B + C)}$
The upper gate is a NAND gate with inputs $A$ and $B$. Its output is $\overline{A \cdot B}$.
The lower gate is a NOR gate with inputs $B$ and $C$. Its output is $\overline{B + C}$.
The final gate is an OR gate which takes the outputs of the NAND and NOR gates as its inputs.
Therefore, the final output $Y$ is given by the logical OR operation of these two intermediate outputs:
$Y = \overline{A \cdot B} + \overline{B + C}$
Answer: $\overline{(A \cdot B)} + \overline{(B + C)}$
Hence, the correct answer is option: 'C'.