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=A¯.B¯=A¯+B¯=A+B (De morgan’s theorem)

Hence output C is equivalent to OR gate.

C=AB.AB=AB+AB=AB+AB=AB

In this case output C is equivalent to AND gate.



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