JEE MainPhysicsSemiconductors
A digital logic circuit has three inputs A , B , and C . It is constructed using four logic gates as follows: Gate 1 is a NOR gate with inputs A and B . Gate 2 is an AND gate whose inputs are the output of Gate 1 and the input C . Gate 3 is a NAND gate with inputs B and C . Gate 4 is an OR gate that takes the outputs of Gate 2 and Gate 3 to produce the final output Y . Which of the following conditions will result in
Options
- AB = 1 and C = 1 , regardless of the value of A
- BA = 0 , B = 1 and C = 1 only
- CA = 1 , B = 0 and C = 1
- DA = 0 and B = 0 , for any value of C
Correct answer
A. B = 1 and C = 1 , regardless of the value of A
Step-by-step solution
Let us write the Boolean expression for the output of each gate. Output of Gate 1 (NOR): Y₁ = A + B Output of Gate 2 (AND): Y₂ = Y₁ C = ( A + B ) C Output of Gate 3 (NAND): Y₃ = B C Final output Y from Gate 4 (OR): Y = Y₂ + Y₃ = ( A + B ) C + B C Now, simplify the Boolean expression for Y using De Morgan's laws: Y = ( A B ) C + B + C Y = A B C + B + C Using the distributive law, we can factor out B from the first two terms: Y = B ( A C + 1) + C Since 1 + anything = 1 , the expression simplifies to: Y = B (1) + C =