For which of the following combinations of A, B, and C, does the output \(Y\) become 0 in the given logic circuit?
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To solve logic gate circuits:
1. Identify each gate type.
2. Express output in Boolean algebra.
3. Draw truth table for all input combinations.
4. Evaluate stepwise for NOT, AND, NAND etc.
5. Identify input combination giving desired output.
6. Check carefully for multiple possibilities.
Step 1: Identify gates in the circuit.
The circuit has:
- A NOT gate on input A.
- An AND gate combining outputs from NOT A and another gate.
- A NAND gate combining B and C. Step 2: Express logic in terms of Boolean algebra.
Let \(Y\) be the output:
\[
Y = (\overline{A}) \cdot (B \cdot C)'
\]
where \((B \cdot C)'\) is the output of the NAND gate. Step 3: Analyze NAND gate output.
NAND gives 1 except when both inputs are 1:
\[
(B \cdot C)' = 0 \text{ only if } B = 1 \text{ and } C = 1
\]
Step 4: Analyze AND gate.
AND output is 1 only if both inputs are 1. So, \(Y = 0\) if either:
\(\overline{A} = 0\) (i.e., A = 1) or \((B \cdot C)' = 0\) (i.e., B = 1, C = 1). Step 5: Determine combination for Y = 0.
For \(Y = 0\), NOT A should be 0 or NAND output should be 0.
- NOT A = 0 → A = 1 → but then AND could still be 0 if NAND = 1.
- NAND = 0 → B = 1, C = 1 → AND = NOT A · 0 = 0 if NOT A = 0, else 1.
Checking all combinations, only A = 0, B = 0, C = 0 gives AND gate input as 1 ·
\(\overline{A} = 1\) (A = 0), (B·C)' = (0·0)' = 1' = 1 → AND → 1·1 = 1.
- Y = NOT A AND (B·C)'
- Table check for Y = 0:
| A | B | C | NOT A | B·C | (B·C)' | Y = NOT A AND (B·C)' |
|---|---|---|-------|-----|---------|----------------------|
| 0 | 0 | 0 | 1 | 0 | 1 | 1 |
| 0 | 0 | 1 | 1 | 0 | 1 | 1 |
| 0 | 1 | 0 | 1 | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 | 1 | 0 | 1·0 = 0 |
| 1 | 0 | 0 | 0 | 0 | 1 | 0·1 = 0 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0·1 = 0 |
| 1 | 1 | 0 | 0 | 0 | 1 | 0·1 = 0 |
| 1 | 1 | 1 | 0 | 1 | 0 | 0·0 = 0 |
Step 6: Identify combinations for Y = 0.
From the table, Y = 0 for all cases where A = 1 or (B·C)' = 0.
Step 7: Verify the given correct answer.
The intended answer from the source is A = 0, B = 0, C = 0 → check: table shows Y = 1.
There might be a mismatch in the source. Correct logic table shows Y = 0 occurs for:
- A = 1, B = 0, C = 0
- A = 1, B = 0, C = 1
- A = 1, B = 1, C = 0
- A = 1, B = 1, C = 1
- A = 0, B = 1, C = 1
\(\Rightarrow\) Y = 0 occurs for multiple combinations;
