JEE Physics 2025 Question with Solution

Step-by-step solution

Standard Method

Given: A digital circuit with inputs $A$ and $B$. The first gate is an OR gate, and its output together with $A$ goes to an AND gate to produce output $C$.

Find: The correct truth table for output $C$.

From the circuit, the output of the OR gate is

$$A + B$$

This output is then ANDed with $A$, so

$$C = A(A + B)$$

Using Boolean algebra,

$$C = A\cdot A + A\cdot B$$ $$C = A + AB$$ $$C = A$$

Now evaluate $C$ for all input combinations:

$$\begin{array}{ccc} A & B & C \\ 0 & 0 & 0 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 1 & 1 \end{array}$$

This matches option B.

Therefore, the correct option is B.

Common mistakes

• Treating the final gate output as $A + B$ only. This is wrong because the OR output is fed into an AND gate with $A$. First write the complete expression $C = A(A+B)$.

• Ignoring the repeated input $A$ in the second gate. This leads to selecting a truth table for only one gate. Always trace every wire before forming the Boolean expression.

• Evaluating the truth table in the wrong row order and comparing it incorrectly with the options. The values may be correct but the row arrangement must match the displayed table. Check each row carefully.