question_answer

The circuit is equivalent to

A) AND gate            

B) NAND gate

C) NOR gate  

D) OR gate

Correct Answer: C

Solution :

                The gate circuit can be shown by giving two inputs A and B. Output of NOR gate, \[{{Y}_{1}}=\overline{A+B}\] Output of NAND gate, \[{{Y}_{2}}=\overline{{{Y}_{1}}.{{Y}_{1}}}\] \[=\overline{\overline{A+B}.\overline{A+B}}\] \[=\overline{\overline{A+B}}+\overline{\overline{A+B}}\] \[=(A+B)+(A+B)\]                            \[=A+B\] Output of NOT gate,       \[=Y=\overline{{{Y}_{2}}}\] \[=\overline{A+B}\] which is the output of NOR gate. Alternative:

	NOR			NAND		NOT
A	B	\[{{Y}_{1}}\]	\[{{Y}_{1}}\]	\[{{Y}_{1}}\]	\[{{Y}_{2}}\]	Y
0	0	1	1	1	0	1
0	1	0	0	0	1	0
1	0	0	0	0	1	0
1	1	0	0	0	1	0

Same as NOR Gate NOR Gate 0              0              1 0              1              0 1              0              0 1              1              0 Note: NOR and NAND gates are said to be universal building blocks.