Question:
The truth table for the given circuit is
The truth table for the given circuit is
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Work out the output of each gate in turn; the final output turns out to follow input B.
Updated On: Nov 12, 2026
A B E 0 0 1 0 1 0 1 0 1 1 1 0 A B E 0 0 1 0 1 0 1 0 0 1 1 0 A B E 0 0 0 0 1 1 1 0 0 1 1 1 A B E 0 0 0 0 1 1 1 0 1 1 1 0
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Write the Boolean expression gate by gate and then fill the truth table.
Step 2: Output of each gate
The upper AND gate gives \(C = A\cdot B\).
The lower AND gate receives the NOT of A and B, giving \(D = \bar{A}\cdot B\).
The OR gate gives \(E = A\cdot B + \bar{A}\cdot B = B\).
Step 3: Truth table
For \((A, B) = (0,0), (0,1), (1,0), (1,1)\), the output is \(E = 0, 1, 0, 1\), which is option (C).
Final Answer:
The output E follows B, giving 0, 1, 0, 1 for the four input rows. \[ \boxed{\text{(C) }\text{truth table with output E = 0, 1, 0, 1}} \]
Write the Boolean expression gate by gate and then fill the truth table.
Step 2: Output of each gate
The upper AND gate gives \(C = A\cdot B\).
The lower AND gate receives the NOT of A and B, giving \(D = \bar{A}\cdot B\).
The OR gate gives \(E = A\cdot B + \bar{A}\cdot B = B\).
Step 3: Truth table
For \((A, B) = (0,0), (0,1), (1,0), (1,1)\), the output is \(E = 0, 1, 0, 1\), which is option (C).
Final Answer:
The output E follows B, giving 0, 1, 0, 1 for the four input rows. \[ \boxed{\text{(C) }\text{truth table with output E = 0, 1, 0, 1}} \]
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