JEE Main 2026 · 22 January · morning · Single correct · +4 / −1
Find the correct combination of A,B,C and D inputs which can cause the LED to glow.
- (a) 1000
- (b) 0011
- (c) 0100
- (d) 1101
Answer and solution
Answer (d) 1101
Solution
Physics sir · 4 steps
The LED sits between two gate outputs, so it glows only when its top end is 1 and its bottom end is 0.
- Top branch. A NOR gate with both inputs joined is a NOT gate, so the NAND receives Ā and B̄. Its output is Y₁ = Ā·B̄̅ = A + B, which is an OR of A and B.
- Bottom branch. The bottom gate is a plain NOR, so Y₂ = C + D̅. This end connects to the cathode of the LED.
- Forward bias. The anode connects to Y₁ through the resistor and the cathode to Y₂. Current flows only for Y₁ = 1 and Y₂ = 0. That means A + B = 1 and C + D = 1: at least one of A, B is 1, and at least one of C, D is 1.
ABCD Y₁ = A + B Y₂ = C + D̅ LED (a) 1000 1 1 off (both ends high) (b) 0011 0 0 off (both ends low) (c) 0100 1 1 off (both ends high) (d) 1101 1 0 glows - So. Only 1101 puts the anode at 1 and the cathode at 0, so the answer is (d).
The idea to keep. When a diode or LED is connected between two gate outputs, a 1 at one end is not enough. You need 1 at the anode and 0 at the cathode, so set up both conditions together.
Answer from the NTA final answer key. Solution by PrepWiser.