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GATE | GATE CS 1999 | Question 33

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Consider the circuit shown below . In a certain steady state, the line Y is at ‘ 1 ‘. What are the possible values of A, B and C in this state?

q33

a). A=0, B=0, C=1

b). A=0, B=1, C=1

c). A=1, B=0, C=1

d). A=1, B=1, C=1
(A) b only
(B) Both a and b
(C) Both a and c
(D) a, c and d


Answer: (D)

Explanation: Since f = 1,

2nd level AND gate must give 1 as a result for which C has to be 1 along with the lower first level NAND gate.

Thus, C = 1 is compulsory

Now NAND is 1 when any of the inputs are zero.

Let G be the NAND gate with input A and H be the NAND with input B.

Since, C =1 and H = 1 is needed so clearly we have following possibilities :

B = 0 , G = 1  ………(1)

B = 0 , G = 0  ………(2)

B = 1,  G = 0  ……….(3)

Since G is also a NAND gate and f = 1

G = 1 means A = 0

and G = 0 means A = 1

using these in (1),(2) and (3) :

A = 0 , B = 0 , C = 1

A = 1 , B = 0 , C = 1

A = 1,  B = 1 , C = 1

Option (D) is correct.

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Last Updated : 19 Oct, 2020
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