JEE Mains · Physics · STD 12 - Semicondutor electronics
- JEE Mains 2026
- Easy
- MCQ
If \(X\) and \(Y\) are the inputs, the given circuit works as _______.

Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) NOR gate
Step-by-step Solution
Detailed explanation
The circuit consists of four NAND gates.
The first two NAND gates have their inputs tied together, acting as NOT gates. Their outputs are \(\overline{X}\) and \(\overline{Y}\).
These outputs are fed into the third NAND gate. The output of this gate is:
\(y' = \overline{\overline{X} \cdot \overline{Y}}\)
Using De Morgan's theorem:
\(y' = \overline{\overline{X}} + \overline{\overline{Y}} = X + Y\)
This output is fed into the fourth NAND gate, which also has its inputs tied together, acting as a NOT gate. The final output is:
Output \(= \overline{X + Y}\)
This represents the Boolean expression for a NOR gate.
Answer: NOR gate
The first two NAND gates have their inputs tied together, acting as NOT gates. Their outputs are \(\overline{X}\) and \(\overline{Y}\).
These outputs are fed into the third NAND gate. The output of this gate is:
\(y' = \overline{\overline{X} \cdot \overline{Y}}\)
Using De Morgan's theorem:
\(y' = \overline{\overline{X}} + \overline{\overline{Y}} = X + Y\)
This output is fed into the fourth NAND gate, which also has its inputs tied together, acting as a NOT gate. The final output is:
Output \(= \overline{X + Y}\)
This represents the Boolean expression for a NOR gate.
Answer: NOR gate
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