Discussion
From the circuit of the following logic gates the basic logic gate obtained is ................
Answer: D
The circuit is a NAND gate.
Follow the signals through. The NOT gates give \(\bar{A}\) and \(\bar{B}\). The NOR gate then gives, by De Morgan's theorem,
\[ \overline{\bar{A} + \bar{B}} = A\cdot B \]
The final NAND gate takes \(AB\) and \(B\):
\[ Y = \overline{(AB)\cdot B} = \overline{AB} \]
since \(AB\cdot B = AB\). So \(Y\) is A NAND B.
The output depends on both inputs, so it is not a NOT gate, and it is the complement of AND, so it is neither AND nor OR: for \(A = B = 0\) it gives 1, which an AND or OR gate would not.
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