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.



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