The correct truth table for the following logic circuit is :
The Correct Option is B
Approach Solution - 1
The circuit consists of an AND gate, a NOT gate, and an OR gate. The output \(Y\) is determined as follows:
- The output of the AND gate is \(A \cdot B\).
- The output of the NOT gate is \(\overline{A \cdot B}\).
- The output \(Y\) is the OR of \(\overline{A \cdot B}\) and \(B\):
\[ Y = \overline{A \cdot B} + B \]
Step-by-Step Evaluation of Truth Table:
| A | B | A · B | A · B | Y = A · B + B |
|---|---|---|---|---|
| 0 | 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 1 |
| 1 | 1 | 1 | 0 | 1 |
Thus, the correct truth table is represented in Option (2).
Approach Solution -2
To determine the correct truth table for the given logic circuit, we need to analyze the circuit step by step. The circuit consists of an OR gate followed by a NOT gate and then an AND gate. Let's go through each component:
- The inputs to the circuit are A and B.
- The first gate is an OR gate. The output of the OR gate (\(Z\)) is given by: \(Z = A + B\) where \(+\) represents the OR operation.
- The output of the OR gate is then passed through a NOT gate, producing: \(\overline{Z} = \overline{A + B}\) where \(\overline{}\) denotes the NOT operation.
- The final gate is an AND gate that takes inputs from the output of the NOT gate and the original input \(B\). The final output (\(Y\)) is: \(Y = \overline{Z} \cdot B = \overline{A + B} \cdot B\)
Now, let's create a truth table for the circuit:
| A | B | Z = A + B | \(\overline{Z}\) | Y = \(\overline{Z} \cdot B\) |
| 0 | 0 | 0 | 1 | 0 |
| 0 | 1 | 1 | 0 | 0 |
| 1 | 0 | 1 | 0 | 0 |
| 1 | 1 | 1 | 0 | 0 |
Based on this analysis, the correct truth table is depicted in the following image:
This matches the correct truth table for the given logic circuit.
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