The given circuit works as :

### Related Formula De Morgan's Laws: overlineA + overlineB = overlineA cdot B$$\overline{A} + \overline{B} = \overline{A \cdot B}$$ overlineoverlineA = A$$\overline{\overline{A}} = A$$ ### Core Logic Analyzing the circuit diagram:

- Top branch has a NOT gate on A, so P = overlineA$P = \overline{A}$ - Bottom branch has a NOT gate on B, so Q = overlineB$Q = \overline{B}$ - They enter a NOR gate, giving output R = overlineoverlineA + overlineB$R = \overline{\overline{A} + \overline{B}}$ - Finally, R passes through a NOT gate to give S = overlineR$S = \overline{R}$ ### Step 1: Boolean Simplification R = overlineoverlineA + overlineB = overlineoverlineAB = AB$$R = \overline{\overline{A} + \overline{B}} = \overline{\overline{AB}} = AB$$ S = overlineR = overlineAB$$S = \overline{R} = \overline{AB}$$ The expression overlineAB$\overline{AB}$ is the Boolean expression for a NAND gate. ### Pattern Recognition Two NOTs feeding into a NOR equals an AND gate (AB$AB$). Adding a final NOT gate turns the AND into a NAND. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Semiconductor Electronics