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Answer: 4. NOR

Key idea: write the output of each gate in Boolean form and simplify.

The upper gate is a NOR gate with inputs A and B: \(\overline{A+B}\). The lower gate is a NAND gate with inputs B and A: \(\overline{AB}\). Their outputs go to an AND gate:

$$Y=\overline{A+B}\cdot\overline{AB}=(\bar A\bar B)(\bar A+\bar B)=\bar A\bar B$$

(because \(\bar A\bar B\cdot\bar A=\bar A\bar B\) and \(\bar A\bar B\cdot\bar B=\bar A\bar B\)). Since \(\bar A\bar B=\overline{A+B}\), the circuit behaves as a NOR gate.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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