Question:
Which of the following circuits correctly represents the following truth table ?
A B C 0 0 0 0 1 0 1 0 1 1 1 0
Which of the following circuits correctly represents the following truth table ?
| A | B | C |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
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From the table, \(C = A\cdot\bar{B}\). Reduce each circuit to a Boolean expression and compare.
Updated On: Nov 4, 2026
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The Correct Option is A
Solution and Explanation
The truth table has C = 1 only for A = 1, B = 0, so \(C = A\bar{B}\).
Option (1): the AND gate gives AB, the NOT gate gives \(\bar{A}\), and both go into a NOR gate: \(C = \overline{\bar{A} + AB} = A\cdot\overline{AB} = A(\bar{A}+\bar{B}) = A\bar{B}\). This matches.
Option (2): OR(A, B) ANDed with B gives \((A+B)B = B\). Not matching.
Option (3): OR of AB and \(\bar{B}\) gives \(AB + \bar{B}\), which is 1 for A = B = 0. Not matching.
Option (4): NAND gives \(\overline{AB}\), which is 1 for A = B = 0. Not matching.
Answer: (1) the circuit with AND(A, B) and NOT A feeding a NOR gate
Option (1): the AND gate gives AB, the NOT gate gives \(\bar{A}\), and both go into a NOR gate: \(C = \overline{\bar{A} + AB} = A\cdot\overline{AB} = A(\bar{A}+\bar{B}) = A\bar{B}\). This matches.
Option (2): OR(A, B) ANDed with B gives \((A+B)B = B\). Not matching.
Option (3): OR of AB and \(\bar{B}\) gives \(AB + \bar{B}\), which is 1 for A = B = 0. Not matching.
Option (4): NAND gives \(\overline{AB}\), which is 1 for A = B = 0. Not matching.
Answer: (1) the circuit with AND(A, B) and NOT A feeding a NOR gate
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