JEE Main 2025 · 29 January · evening · Single correct · +4 / −1
The truth table for the circuit given below is:
- (a) A | B | Y; 0 | 0 | 0; 1 | 1 | 1; 1 | 0 | 1; 0 | 1 | 1
A B Y 0 0 0 1 1 1 1 0 1 0 1 1 - (b) A | B | Y; 0 | 0 | 0; 1 | 0 | 1; 0 | 1 | 0; 1 | 1 | 0
A B Y 0 0 0 1 0 1 0 1 0 1 1 0 - (c) A | B | Y; 0 | 0 | 0; 1 | 0 | 0; 1 | 1 | 0; 0 | 1 | 1
A B Y 0 0 0 1 0 0 1 1 0 0 1 1 - (d) A | B | Y; 0 | 0 | 0; 0 | 1 | 1; 1 | 0 | 1; 1 | 1 | 0
A B Y 0 0 0 0 1 1 1 0 1 1 1 0
Answer and solution
Answer (d) A | B | Y; 0 | 0 | 0; 0 | 1 | 1; 1 | 0 | 1; 1 | 1 | 0
Solution
Physics sir · 3 steps
Read each AND gate's two inputs first; the OR then just adds the two branches, and you get XOR.
- Top branch. The upper AND gate gets A directly and B through a NOT gate. Its output is AB̄, which is 1 only when A = 1 and B = 0.
- Bottom branch. The lower AND gate gets A through a NOT gate and B directly. Its output is ĀB, which is 1 only when A = 0 and B = 1.
Y = AB̄ + ĀB
A, B AB̄ ĀB Y 0, 0 0 0 0 0, 1 0 1 1 1, 0 1 0 1 1, 1 0 0 0 - So. Y is 1 only when the inputs differ. This is the XOR gate, and its truth table is option (d).
The idea to keep. Whenever you see AB̄ + ĀB, it is XOR: the output is 1 only when the inputs are different. Each AND gate catches one of the 'different' cases.
Answer from the NTA final answer key. Solution by PrepWiser.