A) OR
B) NAND
C) AND
D) NOR
Correct Answer: C
Solution :
The truth table for the given circuit isA | B | \[{{\mathbf{y}}_{\mathbf{1}}}\] | \[{{\mathbf{y}}_{\mathbf{2}}}\] | \[\mathbf{y=}\overset{\mathbf{\_\_\_\_\_\_\_\_}}{\mathop{{{\mathbf{y}}_{\mathbf{1}}}\mathbf{+}{{\mathbf{y}}_{\mathbf{2}}}}}\,\] |
0 | 0 | 1 | 1 | 0 |
0 | 1 | 1 | 0 | 0 |
1 | 0 | 0 | 1 | 0 |
d1 | 1 | 0 | 0 | 1 |
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