In the following circuit, the output 
for all possible inputs 
and 
is expressed by the truth table: [2007]
A | B | Y |
0 | 0 | 0 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 1 |
A | B | Y |
0 | 0 | 1 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 0 |
A | B | Y |
0 | 0 | 1 |
0 | 1 | 0 |
1 | 0 | 0 |
1 | 1 | 0 |
A | R | Y |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 1 |
Text Solution
Verified by ExpertsThe correct answer is:
D
Gates‐ 
and 
are NOR gates. We can simplify the gate circuit as
Here, gates 
and 
are NOR gates. The output 
of gate 
will be appeared as input of gate II. The final output is
This is the Boolean expression of OR gate whose truth table is given below
A | B | Y |
0 | 0 | 0 |
0 | 1 | 1 |
1 | 0 | 1 |
1 | 1 | 1 |
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