Question:

For given logic gate circuit choose correct truth table.

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Simplifying the Boolean expression before building the truth table can save a lot of time and reduce the chances of calculation errors. Remember key Boolean identities like the absorption law (\(A + AB = A\)) and De Morgan's theorems.

Updated On: Jan 25, 2026

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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
We need to determine the output C for all possible binary inputs A and B for the given logic circuit and find the corresponding truth table.
Step 2: Key Formula or Approach:
We will first identify the gates and write the Boolean expression for the final output C in terms of the inputs A and B. Then, we will construct a truth table by evaluating this expression for all four possible input combinations.
Step 3: Detailed Explanation:
Let's analyze the circuit diagram.
- Gate 1: The first gate has inputs A and B. It is a D-shaped gate, which represents an AND gate. Let's call its output D. So, \(D = A \cdot B\).
- Gate 2: The second gate has inputs A and D. It is a gate with a curved input side followed by a 'NOT' bubble at the output, which represents a NOR gate. Its output is C. So, \(C = \overline{A + D}\).
Now, substitute the expression for D into the expression for C:
\[ C = \overline{A + (A \cdot B)} \] We can simplify this Boolean expression using absorption law: \(X + X \cdot Y = X\).
Here, \(A + A \cdot B = A(1 + B)\). Since \(1+B=1\) in Boolean algebra, \(A(1+B) = A\).
So, the expression simplifies to:
\[ C = \overline{A} \] This means the output C depends only on the input A; it is the logical NOT of A.
Let's construct the truth table for \(C = \overline{A}\):
- When A = 0, B = 0: \(C = \overline{0} = 1\).
- When A = 0, B = 1: \(C = \overline{0} = 1\).
- When A = 1, B = 0: \(C = \overline{1} = 0\).
- When A = 1, B = 1: \(C = \overline{1} = 0\).
The resulting truth table is: