JEE Main 2021 · 31 August · morning · Single correct · +4 / −1
In the following logic circuit the sequence of the inputs A, B are (0, 0), (0, 1), (1, 0) and (1, 1). The output Y for this sequence will be :
- (a) 1, 0, 1, 0
- (b) 0, 1, 0, 1
- (c) 1, 1, 1, 0
- (d) 0, 0, 1, 1
Answer and solution
Answer (c) 1, 1, 1, 0
Solution
Physics sir · 3 steps
Name the two middle wires P and Q, fill a four-row truth table, and the output reads itself off.
- Middle wires. The top gate is an AND gate, so P = A · B. The bottom gate is an OR gate, so Q = A + B. The final gate is a NAND gate, so Y = P · Q̅.
Y = (A · B)·(A + B)̅ = A · B̅
- Why it simplifies. Whenever A · B = 1, both inputs are 1, so A + B = 1 too. So P · Q is just A · B, and the whole circuit behaves like a single NAND gate.
A, B P = A · B Q = A + B Y = P · Q̅ 0, 0 0 0 1 0, 1 0 1 1 1, 0 0 1 1 1, 1 1 1 0 - So. The output sequence is 1, 1, 1, 0, which is option (c).
The idea to keep. In a combination of gates, label every intermediate wire and build the truth table column by column. Often the Boolean expression collapses to one basic gate.
Answer from the NTA final answer key. Solution by PrepWiser.