Semiconducting Devices Question 71
Question: The combination of NAND gates shown here under (figure) are equivalent to
[Haryana CEET 1996]
Options:
A) An OR gate and an AND gate respectively
B) An AND gate and a NOT gate respectively
C) An AND gate and an OR gate respectively
D) An OR gate and a NOT gate respectively.
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Answer:
Correct Answer: A
Solution:
$ C=\overline{\bar{A}.\bar{B}}=\overline{\bar{A}+\bar{B}}=A+B $ (De morgan’s theorem)
Hence output C is equivalent to OR gate.
$ C=\overline{\overline{AB}.\overline{AB}}=\overline{\overline{AB}+\overline{AB}}=AB+AB=AB $
In this case output C is equivalent to AND gate.
