JEE Main202330 Jan 2023Evening ShiftPhysicsSemiconductorsActual
The output Y for the inputs A and B of circuit is given by Truth table of the shown circuit is :
Options
- AA B Y 0 0 1 0 1 1 1 0 1 1 1 0
- BA B Y 0 0 1 0 1 0 1 0 0 1 1 1
- CA B Y 0 0 0 0 1 1 1 0 1 1 1 1
- DA B Y 0 0 0 0 1 1 1 0 1 1 1 0
Correct answer
D. A B Y 0 0 0 0 1 1 1 0 1 1 1 0
Step-by-step solution
Output using DeMorgan's theorem can be written as, Y = A · A · B · A · B · B = A · A · B + A · B · B = A · A · B + A · B · B = A · A + B + A + B · B = A · A + A · B + A · B + B · B = A + B · A + B Which represents XOR gate. Therefore, required truth table will be A B Y 0 0 0 0 1 1 1 0 1 1 1 0