In the given circuit P and Q form the inputs. The output $Y$ is
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When analyzing logic circuits, always check for the "bubble" at the output of a gate, which signifies an inversion (like the NAND gate in this diagram).
Concept:
To find the output of a logic circuit, identify each gate and its respective inputs.
• NOT Gate: Inverts the input (\(\overline{A}\)).
• OR Gate: Sums the inputs (\(A + B\)).
• NAND Gate: Multiplies the inputs and inverts the result (\(\overline{A \cdot B}\)).
Step 1: Identify the logic gates in the diagram.
The circuit consists of:
• A NOT gate on the $P$ input line.
• An upper OR gate.
• A lower OR gate.
• A final NAND gate producing Y.
Step 2: Trace the inputs through the gates.
The upper OR gate receives $\overline{P}$ and $Q$:
\[
\text{Output 1} = \overline{P} + Q
\]
The lower OR gate receives $P$ and $Q$:
\[
\text{Output 2} = P + Q
\]
Step 3: Find the final output Y.
Outputs 1 and 2 go into the final NAND gate:
\[
Y = \overline{(\overline{P} + Q) \cdot (P + Q)}
\]
Using De Morgan's Law ($\overline{A \cdot B} = \overline{A} + \overline{B}$):
\[
Y = \overline{(\overline{P} + Q)} + \overline{(P + Q)}
\]
Applying De Morgan's again ($\overline{A + B} = \overline{A} \cdot \overline{B}$):
\[
Y = (P \cdot \overline{Q}) + (\overline{P} \cdot \overline{Q})
\]
Factoring out $\overline{Q}$:
\[
Y = \overline{Q} \cdot (P + \overline{P})
\]
Since \(P + \overline{P} = 1\):
\[
Y = \overline{Q} \cdot 1 = \overline{Q}
\]
Note: Based on the logical trace of the provided image, the output simplifies to $\overline{Q}$, which corresponds to option (D). If your source material lists (B) as correct, please double-check the gate symbols in the original diagram.
