Question
Question: The truth table for the following combination of gates is 
(A)
| A | B | Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
(B)
| A | B | Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
(C)
| A | B | Y |
|---|---|---|
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
(D)
| A | B | Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Solution
Hint
We need to find the output of the first AND gate and then it goes as the input to the OR gate. So we get the output in terms of the input A and B. So for various values of A and B we can find the output Y and form the truth table.
Formula Used: In this solution we will be using the following formulas,
For AND gate Y=A⋅B where Y is output and A, B are inputs.
For OR gate Y=A+B where Y is output and A, B are inputs.
Complete step by step answer
To solve this problem, we need to first find the output to this combination given by Y. Now for the AND gate the inputs given are A and B. So the output of the AND gate will be given by the formula,
Y′=A⋅B
This output of the AND gate goes on to become one of the inputs of the OR gate. So the two inputs of the OR gate are A and A⋅B.
Hence the output Y will be, Y=A+(A⋅B)
Now when the value of A is 0 and the value of B is zero, then Y’ will be 0. So the output of Y will also be 0.
When A is 0 and B is 1, then also the output of the Y’ will be 0 and the final output will again be 0.
When the input A is 1 and B is 0, then like the previous case, the output of Y’ will be zero. But since A is 1, the final output will be 1 as it will depend on A.
When both the inputs are 1, then the output of Y’ will be 1 and consecutively, Y will also be 1. So the truth table can be written as,
| A | B | Y |
|---|---|---|
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
Hence the correct answer is option (A).
Note
In the case of AND gates the output comes to be 1 only when both the inputs are 1 but in case of OR gate, the output is 1 even when only one of the inputs is 1. So the final output here is dependent on the value of A. If A is 1 the output will always be 1 irrespective of Y’. But when A is zero, Y’ will also be zero irrespective of value of B. So the output is solely dependent on A.