JEE Main 2026 Apr 6 Shift 2, Physics Q18: Logic Gates

JEE Main2026Apr 6, Shift 2Physics

Refer to the logic circuit given below. For two inputs $(A = 1, B = 1)$ and $(A = 0, B = 1)$ , output $(Y)$ will be ______.

A
1, 0 respectively
B
0, 1 respectively
C
0, 0 respectively
D
1, 1 respectively

Solution

Label intermediate signals. The NOT gate gives $\overset{ˉ}{A}$ . The upper OR gate outputs $\overset{ˉ}{A} + B$ . The lower AND gate outputs $\overset{ˉ}{A} \cdot B$ . The final NAND combines them:

$Y = \overline{(\overset{ˉ}{A} + B) + (\overset{ˉ}{A} B)}$

Applying the absorption identity $X + X Y = X$ with $X = \overset{ˉ}{A} + B$ and $Y = \overset{ˉ}{A} B$ (which is contained in $\overset{ˉ}{A} + B$ ): the inner expression simplifies to $\overset{ˉ}{A} + B$ .

$Y = \overline{\overset{ˉ}{A} + B} = A \cdot \overset{ˉ}{B}$ (De Morgan).

For $A = 1, B = 1$ : $Y = 1 \cdot 0 = 0$ .

For $A = 0, B = 1$ : $Y = 0 \cdot 0 = 0$ .

Both outputs are $0$ .

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JEE Main 2026 Apr 6 Shift 2, Physics Q18: Logic Gates

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