In the circuit, the logical value of A = 1 or B = 1 when potential at A or B is 5 V and the logical value of A = 0 or B = 0 when potential at A or B is 0 V. The truth table of the given circuit will be: (a) A B Y 0 0 0 1 0 0 0 1 0 1 1 1 (b) A B Y 0 0 0 1 0 1 0 1 1 1 1 1 (c) A B Y 0 0 0 1 0 0 0 1 0 1 1 0 (d) A B Y 0 0 1 1 0 1 0 1 1 1 1 0

Answer: (a). Diode cathodes face the inputs; Y is pulled up to 5 V through R. If A or B is at 0 V, that diode conducts and pulls Y to about 0 V, so Y = 0. Only when both A and B are at 5 V do both diodes stay off, so Y = 1: AND gate. Table: 00 → 0, 10 → 0, 01 → 0, 11 → 1. So the answer is (a).

In the circuit, the logical value of A = 1 or B = 1 when potential at… | JEE Main 2022 (25 Jul, Shift 1)

Q110JEE Main 2022 · 25 JUL · SHIFT 1Medium

Logic Gates

In the circuit, the logical value of $A = 1$ or $B = 1$ when potential at $A$ or $B$ is $5\text{ V}$ and the logical value of $A = 0$ or $B = 0$ when potential at $A$ or $B$ is $0\text{ V}$ .

The truth table of the given circuit will be:

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Answer: (a)

Diode cathodes face the inputs; Y is pulled up to 5 V through R.
If A or B is at 0 V, that diode conducts and pulls Y to about 0 V, so $Y = 0$ .
Only when both A and B are at 5 V do both diodes stay off, so $Y = 1$ : AND gate.
Table: $00 \to 0$ , $10 \to 0$ , $01 \to 0$ , $11 \to 1$ . So the answer is (a).

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