Question

The truth table for the given circuit is

• A.

  A	B	Y
  1	1	1
  1	0	0
  0	1	1
  0	0	1

• B.

  A	B	Y
  1	1	1
  1	0	1
  0	1	1
  0	0	1

• C.

  A	B	Y
  1	1	1
  1	0	1
  0	1	1
  0	0	0

• D.

  A	B	Y
  1	1	1
  1	0	1
  0	1	0
  0	0	1

Answer 1

The Correct Option is B

Correct Answer: B

Explanation: 

The given truth table represents the output behavior of a logic gate based on the inputs A and B. We need to identify the type of gate based on the given outputs.

The first table shows the following output for inputs A and B:

A	B	Y
1	1	1
1	0	0
0	1	1
0	0	1

Upon analyzing the truth table, we can observe that the output is 1 when either A is 1 and B is 0, or when A is 0 and B is 1, which matches the behavior of an Exclusive OR (XOR) gate.

The second table represents the truth table for an OR gate, the third table represents the truth table for an AND gate, and the fourth table represents the truth table for a NAND gate. Therefore, the correct answer is B, which corresponds to the XOR gate.

Answer 2

The circuit consists of two logic gates: a NAND gate and an OR gate. For the inputs A and B, the NAND gate first operates on the values of A and B. The output of the NAND gate is then passed as one of the inputs to the OR gate, and the other input to the OR gate is B. The OR gate produces the final output Y.

Now, let's analyze the truth table:

- For A = 1 and B = 1, the output of the NAND gate is 0, and the OR gate gives the output 1 (0 OR 1 = 1). Thus, Y = 1.  
- For A = 1 and B = 0, the output of the NAND gate is 1, and the OR gate gives the output 1 (1 OR 0 = 1). Thus, Y = 1.  
- For A = 0 and B = 1, the output of the NAND gate is 1, and the OR gate gives the output 1 (1 OR 1 = 1). Thus, Y = 1.  
- For A = 0 and B = 0, the output of the NAND gate is 1, and the OR gate gives the output 0 (1 OR 0 = 0). Thus, Y = 0.  

Thus, the correct truth table is \(\texttt{(B)}\).