A) OR gate
B) AND gate
C) XOR gate
D) NAND gate
Correct Answer: B
Solution :
For our convenience, the output of first NAND gate is chosen as X as shown, Output of first NAND gate, \[X=\overline{A.B}\] Using De-Morgan's theorem, \[\overline{A.B}=\overline{\overline{A}.\overline{B}}\] So, \[X=\overline{A}+\overline{B}\] Now, output of 2nd NAND gate, \[Y=\overline{X}=\overline{\overline{A}+\overline{B}}\] again \[\overline{\overline{A}+\overline{B}}=\overline{\overline{A}}.\overline{\overline{B}}=A.B\] \[(\because \overline{\overline{A}}=A)\] Hence, \[Y=A.B\] This is the logic function of AND gate.You need to login to perform this action.
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