The output of gate is low when at least one of its input is high. This is true for :-

Options

A. NOR
B. OR
C. AND
D. NAND

Answer

To solve the problem, we need to determine which logic gate has an output that is low when at least one of its inputs is high. We will analyze each gate one by one.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We need to find a logic gate where the output is low (0) when at least one of its inputs is high (1).

2. **Identify the Gates**:
- The gates we need to consider are: NOR gate, OR gate, AND gate, and NAND gate.

3. **Boolean Expressions**:
- Write down the Boolean expressions for each gate:
- **NOR Gate**: \( y = \overline{A + B} \)
- **OR Gate**: \( y = A + B \)
- **AND Gate**: \( y = A \cdot B \)
- **NAND Gate**: \( y = \overline{A \cdot B} \)

4. **Evaluate Each Gate**:
- **NOR Gate**:
- If \( A = 0, B = 1 \):
- \( y = \overline{0 + 1} = \overline{1} = 0 \) (Output is low)
- If \( A = 1, B = 0 \):
- \( y = \overline{1 + 0} = \overline{1} = 0 \) (Output is low)
- If \( A = 1, B = 1 \):
- \( y = \overline{1 + 1} = \overline{1} = 0 \) (Output is low)
- Conclusion: The NOR gate satisfies the condition.

- **OR Gate**:
- If \( A = 0, B = 1 \):
- \( y = 0 + 1 = 1 \) (Output is high)
- Conclusion: The OR gate does not satisfy the condition.

- **AND Gate**:
- If \( A = 0, B = 1 \):
- \( y = 0 \cdot 1 = 0 \) (Output is low)
- If \( A = 1, B = 1 \):
- \( y = 1 \cdot 1 = 1 \) (Output is high)
- Conclusion: The AND gate does not satisfy the condition.

- **NAND Gate**:
- If \( A = 0, B = 1 \):
- \( y = \overline{0 \cdot 1} = \overline{0} = 1 \) (Output is high)
- Conclusion: The NAND gate does not satisfy the condition.

5. **Final Conclusion**:
- The only gate that satisfies the condition where the output is low when at least one input is high is the **NOR gate**.

### Answer:
The output of the gate is low when at least one of its input is high for the **NOR gate**.

---