JEE Physics 2026 Question with Solution

Given: A logic circuit with inputs $A$ and $B$ is shown.

Find: Which basic logic gate the circuit is equivalent to.

Each input first passes through a NOT operation, so the intermediate outputs are $A'$ and $B'$.

These two signals are then applied to a NOR gate, so the output is

$$Y = (A' + B')'$$

Using De Morgan's law,

$$(A' + B')' = A \cdot B$$

Therefore, the circuit performs the operation of an AND gate.

The correct option is C.

Given: The circuit contains inverter action on both inputs followed by a NOR combination.

Find: The equivalent logic gate.

Step 1: Trace the two inputs.

• Input $A$ goes through a NOT gate, giving $A'$.
• Input $B$ goes through a NOT gate, giving $B'$.

Step 2: Identify the final gate. The final combining gate is a NOR gate, so its output is

$$Y = (A' + B')'$$

Step 3: Simplify the expression. Applying De Morgan's theorem,

$$Y = (A')' \cdot (B')'$$

Double inversion gives

$$Y = A \cdot B$$

Step 4: Interpret the result. The Boolean expression $A \cdot B$ is the expression of an AND gate.

A truth-table check also confirms this:

• For $A=0, B=0$, output is $0$
• For $A=0, B=1$, output is $0$
• For $A=1, B=0$, output is $0$
• For $A=1, B=1$, output is $1$

So the output pattern is $0,0,0,1$, which matches an AND gate.

Therefore, the given circuit is equivalent to AND gate, i.e. option C.