Question

The output Y of the given logic circuit is

• A. \(\overline{(A\cdot B)}+\overline{(C\cdot B)}\)
• B. \(\overline{(A+B)}\cdot \overline{(B\cdot C)}\)
• C. \(\overline{(A\cdot B)}+\overline{(B+C)}\)
• D. \((A+B)+(B+C)\)

Hint:

Read the figure gate by gate: a NAND of A and B, a NOR of B and C, and an OR of the two outputs.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The figure shows three gates. The upper gate is an AND gate with a small circle on its output, so it is a NAND gate with inputs A and B. The lower gate is an OR gate with a circle on its output, so it is a NOR gate with inputs B and C. The right-hand gate is an OR gate that takes both outputs.

Step 2: Key Formula or Approach:
NAND: \(\overline{A\cdot B}\). NOR: \(\overline{B + C}\). OR: sum of its inputs.

Step 3: Detailed Explanation:
Output of the NAND gate: \(\overline{A\cdot B}\).
Output of the NOR gate: \(\overline{B + C}\).
The final OR gate combines them:
\[ Y = \overline{A\cdot B} + \overline{B + C} \]
Option (A) has a NAND of C and B in place of the NOR. Option (B) multiplies the two outputs, which would need an AND gate at the end. Option (D) has no inversions at all.

Final Answer:
The output is \(Y = \overline{A\cdot B} + \overline{B + C}\), option (C). \[ \boxed{\overline{A\cdot B}+\overline{B+C} \text{ (C)}} \]