UGC NET CS 2016 July- paper-3
November 5, 2023UGC NET CS 2012 Dec-Paper-2
November 5, 2023UGC NET CS 2012 Dec-Paper-2
| Question 1 |
Consider the circuit shown below. In a certain steady state, Y is at logical ‘1’. What are possible values of A, B, C ?
| A = 0, B = 0, C = 1 | |
| A = 0, B = C = 1 | |
| A = 1, B = C = 0 | |
| A = B = 1, C = 1 | |
| A and D correct |
Question 1 Explanation:
The output f of the circuit is given as feedback input which should always be 1.
From the given circuit
f= (A+B’)C (Assuming that feedback input is 1)
So, the output remains 1 iff C=1 and (A+B’)=1.
A=1, B=0, C=1
A=1, B=1, C=1
A=0, B=0, C=1
Hence option A and D are correct.
From the given circuit
f= (A+B’)C (Assuming that feedback input is 1)
So, the output remains 1 iff C=1 and (A+B’)=1.
A=1, B=0, C=1
A=1, B=1, C=1
A=0, B=0, C=1
Hence option A and D are correct.
Correct Answer: E
Question 1 Explanation:
The output f of the circuit is given as feedback input which should always be 1.
From the given circuit
f= (A+B’)C (Assuming that feedback input is 1)
So, the output remains 1 iff C=1 and (A+B’)=1.
A=1, B=0, C=1
A=1, B=1, C=1
A=0, B=0, C=1
Hence option A and D are correct.
From the given circuit
f= (A+B’)C (Assuming that feedback input is 1)
So, the output remains 1 iff C=1 and (A+B’)=1.
A=1, B=0, C=1
A=1, B=1, C=1
A=0, B=0, C=1
Hence option A and D are correct.