Question

Logic gates are given the inputs A = 0 and B = 1 in case (a) and A = 1 and B = 0 in case (b). The gates giving output of y=1 for both the cases are

• A. OR and AND
• B. OR and NAND
• C. AND and NOR
• D. NOR and NAND
• E. AND and NAND

Answer 1

The Correct Option is B

We are given the following input cases:  
- Case (a): \(A = 0, B = 1\)  
- Case (b): \(A = 1, B = 0\)  

Now, let's analyze the behavior of different logic gates for these inputs:

1. OR gate:  
  - For \(A = 0, B = 1\), \(y = 1\)  
  - For \(A = 1, B = 0\), \(y = 1\)  
  Thus, the OR gate gives an output of \(y = 1\) for both cases.

2. AND gate:  
  - For \(A = 0, B = 1\), \(y = 0\)  
  - For \(A = 1, B = 0\), \(y = 0\)  
  The AND gate gives an output of \(y = 0\) for both cases.

3. NAND gate:  
  - For \(A = 0, B = 1\), \(y = 1\)  
  - For \(A = 1, B = 0\), \(y = 1\)  
  The NAND gate gives an output of \(y = 1\) for both cases.

Thus, the gates that give \(y = 1\) for both cases are the OR and NAND gates, which corresponds to option (B).

Answer 2

Let's evaluate the outputs for each logic gate with inputs:
Case (a): A = 0, B = 1
Case (b): A = 1, B = 0
 
OR Gate:
- Case (a): 0 OR 1 = 1
- Case (b): 1 OR 0 = 1
✅ Output is 1 in both cases

AND Gate:
- Case (a): 0 AND 1 = 0
- Case (b): 1 AND 0 = 0
❌ Output is 0 in both cases

NAND Gate:
- NAND is the inverse of AND
- Case (a): NOT(0 AND 1) = NOT(0) = 1
- Case (b): NOT(1 AND 0) = NOT(0) = 1
✅ Output is 1 in both cases

NOR Gate:
- NOR is the inverse of OR
- Case (a): NOT(0 OR 1) = NOT(1) = 0
- Case (b): NOT(1 OR 0) = NOT(1) = 0
❌ Output is 0 in both cases

✅ Correct Answer: OR and NAND