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: (d). A NOR gate with joined inputs is a NOT gate: outputs A and B. NAND of these: AB = A + B (this is the LED anode side). Lower NOR gives C + D (the LED cathode side). LED glows when anode = 1 and cathode = 0: A + B = 1 and C + D = 1. Only 1101 has at least one of A, B and at least one of C, D equal to 1. So the answer is (d) 1101.

Find the correct combination of A, B, C and D inputs which can cause… | JEE Main 2026 (22 Jan, Shift 1)

Q22JEE Main 2026 · 22 JAN · SHIFT 1Medium

Logic Gates

Find the correct combination of $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D inputs which can cause the LED to glow.

Show answer and solution

Answer: (d)

A NOR gate with joined inputs is a NOT gate: outputs $\bar{A}$ and $\bar{B}$ .
NAND of these: $\overline{\bar{A}\bar{B}} = A + B$ (this is the LED anode side).
Lower NOR gives $\overline{C + D}$ (the LED cathode side).
LED glows when anode $= 1$ and cathode $= 0$ : $A + B = 1$ and $C + D = 1$ .
Only $1101$ has at least one of A, B and at least one of C, D equal to 1. So the answer is (d) 1101.

Report a mistake

Logic Gates question from JEE Main 2026 (22 Jan, Shift 1)

More questions on Logic Gates