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MCQEasyJEE 2025Logic Gates

JEE Physics 2025 Question with Solution

Choose the correct logic circuit for the given truth table having inputs AA and BB.

Question image
  • A
    Option 1 logic circuit diagram is missing from the scraped input and is required to identify the gate combination.
  • B
    Option 2 logic circuit diagram is missing from the scraped input and is required to identify the gate combination.
  • C
    Option 3 logic circuit diagram is missing from the scraped input and is required to identify the gate combination.
  • D
    Option 4 logic circuit diagram is missing from the scraped input and is required to identify the gate combination.

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The truth table for inputs AA and BB gives output YY as:

A=0,  B=0⇒Y=0A=0,  B=1⇒Y=0A=1,  B=0⇒Y=1A=1,  B=1⇒Y=1\begin{aligned} A=0,\; B=0 &\Rightarrow Y=0 \\ A=0,\; B=1 &\Rightarrow Y=0 \\ A=1,\; B=0 &\Rightarrow Y=1 \\ A=1,\; B=1 &\Rightarrow Y=1 \end{aligned}

Find: Which logic circuit matches this truth table.

From the table, the output depends only on AA. Whenever A=0A=0, the output is 00. Whenever A=1A=1, the output is 11. Therefore,

Y=AY = A

Now compare this with the Boolean expression of each circuit described in the solution.

For Circuit (1):

Y=(A+B)⋅BY = (A + B) \cdot B

This simplifies to

Y=BY = B

so it does not match Y=AY=A.

For Circuit (2):

Y=(A+B)⋅AY = (A + B) \cdot A

Using Boolean algebra,

Y=A⋅A+B⋅A=A+A⋅B=A(1+B)=A\begin{aligned} Y &= A \cdot A + B \cdot A \\ &= A + A \cdot B \\ &= A(1+B) \\ &= A \end{aligned}

So this matches the truth table.

For Circuit (3):

Y=(A+B)⋅BˉY = (A + B) \cdot \bar{B}

which becomes

Y=A⋅BˉY = A \cdot \bar{B}

This is not equal to AA for all input combinations.

For Circuit (4):

Y=(A+B)⋅AˉY = (A + B) \cdot \bar{A}

which becomes

Y=Aˉ⋅BY = \bar{A} \cdot B

This also does not match the truth table.

Therefore, the circuit in option (2) is correct. The correct option is B.

Expression from Minterms

Given: The output is 11 for (A,B)=(1,0)(A,B)=(1,0) and (1,1)(1,1).

Find: The Boolean form corresponding to the truth table.

Write the output as sum of minterms:

Y=ABˉ+ABY = A\bar{B} + AB

Factor AA:

Y=A(Bˉ+B)Y = A(\bar{B}+B)

Since

Bˉ+B=1\bar{B}+B = 1

we get

Y=AY = A

So the required circuit must realize Y=AY=A. According to the extracted solution, this corresponds to option (2), hence the correct option is B.

Common mistakes

  • A common mistake is to focus on both inputs equally and miss that YY is completely independent of BB. Check whether one variable alone determines the output before analyzing complicated gate combinations.

  • Another mistake is to stop at (A+B)⋅A(A+B)\cdot A without simplifying it. Use Boolean identities such as A(A+B)=AA(A+B)=A to match the circuit with the truth table correctly.

  • Students may confuse (A+B)⋅Bˉ(A+B)\cdot \bar{B} or (A+B)⋅Aˉ(A+B)\cdot \bar{A} with AA. Expand the expression first; these simplify to ABˉA\bar{B} and AˉB\bar{A}B, not AA.

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