JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Semiconductors

The truth table corresponding to the circuit given below is

1. (A)A=0, B=0, C=0; A=1, B=0, C=0; A=0, B=1, C=0; A=1, B=1, C=1
2. (B)A=0, B=0, C=0; A=0, B=1, C=0; A=1, B=0, C=1; A=1, B=1, C=1✓ Correct
3. (C)A=0, B=0, C=1; A=1, B=0, C=0; A=0, B=1, C=0; A=1, B=1, C=0
4. (D)A=0, B=0, C=1; A=0, B=1, C=0; A=1, B=0, C=0; A=1, B=1, C=0

Answer: (B)

Solution

The two inputs first pass through an OR gate to give $A+B$, and this output is fed to an AND gate together with $A$ itself, so $C=A\cdot (A+B)$. By absorption this reduces to $C=A$, independent of $B$. Checking the four combinations, $(0,0)$ and $(0,1)$ give $C=0$, while $(1,0)$ and $(1,1)$ give $C=1$. The output therefore follows input $A$: $C=0,0,1,1$ for $(A,B)=(0,0),(0,1),(1,0),(1,1)$.

JEE Main · 2025 · 03 Apr 2025 · Shift 2 · Single correct option