
The circuit is equivalent to
(1) NOR gate
(2) OR gate
(3) AND gate
(4) NAND gate
Answer
506.7k+ views
Hint:The circuit has a NOR gate, a NAND gate according to the question. The input is sent to the NOR gate, whose output becomes the input for the NAND gate, and the NAND gate's output becomes the final output. This will help you in answering this question.
Complete step-by-step solution:
Taking outputs of all the gates as ${{y}_{1}},{{y}_{2}}$,
(1) In a NOR gate, the output is the inverse of the sum of A and B. (NOR, OUTPUT $=\overline{A+B}$ ).
(2) In a NAND gate, the output is the inverse of the multiplication of A and B. (NAND, OUTPUT $=\overline{A \cdot B}$ ).
We know that a NOR gate inverses the sum of its inputs, so output at NOR is $y_{1}=\overline{A+B}$
$y_{1}$ becomes the input for NAND gate, so,
$\overline{y_{1} \cdot y_{1}}=y_{2}=\overline{\overline{A+B} \cdot \overline{A+B}}=\overline{\overline{A+B}}+\overline{\overline{A+B}}=A+B$
This will be the final output for the combination of the gates.
That is,
$Y=A+B$
This is the output we obtain for an OR gate. Therefore the correct option is (2).
Note: The small circle beside each of the gates makes them NOR and NAND gates, so don't get them mixed up with the diagram. It is the polar opposite of OR and AND. The N stands for "negation of," which means the opposite. The NOR gate is a digital logic gate that implements logical NOR; it acts in accordance with the truth table on the right.
Complete step-by-step solution:
Taking outputs of all the gates as ${{y}_{1}},{{y}_{2}}$,
(1) In a NOR gate, the output is the inverse of the sum of A and B. (NOR, OUTPUT $=\overline{A+B}$ ).
(2) In a NAND gate, the output is the inverse of the multiplication of A and B. (NAND, OUTPUT $=\overline{A \cdot B}$ ).
We know that a NOR gate inverses the sum of its inputs, so output at NOR is $y_{1}=\overline{A+B}$
$y_{1}$ becomes the input for NAND gate, so,
$\overline{y_{1} \cdot y_{1}}=y_{2}=\overline{\overline{A+B} \cdot \overline{A+B}}=\overline{\overline{A+B}}+\overline{\overline{A+B}}=A+B$
This will be the final output for the combination of the gates.
That is,
$Y=A+B$
This is the output we obtain for an OR gate. Therefore the correct option is (2).
Note: The small circle beside each of the gates makes them NOR and NAND gates, so don't get them mixed up with the diagram. It is the polar opposite of OR and AND. The N stands for "negation of," which means the opposite. The NOR gate is a digital logic gate that implements logical NOR; it acts in accordance with the truth table on the right.
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