The truth table of the given circuit diagram is :
The Correct Option is B
Solution and Explanation
The given circuit diagram is equivalent to an XOR gate, which outputs a value of 1 if and only if the inputs are different.
| A | B | Y = A ⊕ B |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
This matches the truth table given in option (2).
Top Questions on Digital Circuits
In the digital circuit shown in the figure, for the given inputs the P and Q values are:
- JEE Main - 2025
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The truth table corresponding to the circuit given below is
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The Boolean expression $\mathrm{Y}=\mathrm{A} \overline{\mathrm{B}} \mathrm{C}+\overline{\mathrm{AC}}$ can be realised with which of the following gate configurations.
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B. One 3-input AND gate, 1 NOT gate, One 2-input NOR gate and one 2-input OR gate
C. 3-input OR gate, 3 NOT gates and one 2-input AND gate
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