
Which logic gate is represented by the following combination of logic gates?
$\left( a \right){\text{ OR}}$
$\left( b \right){\text{ NAND}}$
$\left( c \right){\text{ AND}}$
$\left( d \right){\text{ NOR}}$
Answer
560.4k+ views
Hint: In this particular type of question use the concept that the output of the not gate is the reverse of the input i.e. if input is 1 than the output is zero and if the input is zero then the output is one after that the two output of the not gate goes to nor gate so use these concepts to reach the solution of the question.
Complete answer:
In this particular type of question use the concept that the output of the not gate is the reverse of the input i.e. if input is 1 than the output is zero and if the input is zero then the output is one after that the two output of the not gate goes to nor gate so use these concepts to reach the solution of the question.
As we all know, when there are two inputs the possible cases can be constructed as ${2^2} = 4$.
I.e. the inputs of A and B are 00, 01, 10, and 11.
Now construct the truth table of the given circuit which is shown below in the table,
Now the truth tables of the given options are also given below
OR logic gate
AND logic gate
NOR logic gate
NAND logic gate
Now as we see that the output of the truth table of the given logic circuit matches with the output of the truth table of the AND logic gate.
So this is the required answer.
So, the correct answer is “Option C”.
Note:
Whenever we face such types of questions the key concept we have to remember is that always recall the truth tables of the standard logic gates such as, OR, AND, NAND, NOR which is all written above, so first find out the truth table of the given logic circuit and then check the output of the logic circuit follows which truth table of the given options logic gate as above.
Complete answer:
In this particular type of question use the concept that the output of the not gate is the reverse of the input i.e. if input is 1 than the output is zero and if the input is zero then the output is one after that the two output of the not gate goes to nor gate so use these concepts to reach the solution of the question.
As we all know, when there are two inputs the possible cases can be constructed as ${2^2} = 4$.
I.e. the inputs of A and B are 00, 01, 10, and 11.
Now construct the truth table of the given circuit which is shown below in the table,
| A | B | $\bar A$ | $\bar B$ | $\bar A + \bar B$ | $Y = \overline {\bar A + \bar B} $ |
| 0 | 0 | 1 | 1 | 1 + 1 = 1 | $\bar 1 = 0$ |
| 0 | 1 | 1 | 0 | 1 + 0 = 1 | $\bar 1 = 0$ |
| 1 | 0 | 0 | 1 | 0 + 1 = 1 | $\bar 1 = 0$ |
| 1 | 1 | 0 | 0 | 0 + 0 = 0 | $\bar 0 = 1$ |
Now the truth tables of the given options are also given below
OR logic gate
| A | B | Y’ = A + B |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
AND logic gate
| A | B | Y’ = AB |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
NOR logic gate
| A | B | Y’ = $\overline {A + B} $ |
| 0 | 0 | $\bar 0 = 1$ |
| 0 | 1 | $\bar 1 = 0$ |
| 1 | 0 | $\bar 1 = 0$ |
| 1 | 1 | $\bar 1 = 0$ |
NAND logic gate
| A | B | Y’ = $\overline {AB} $ |
| 0 | 0 | $\bar 0 = 1$ |
| 0 | 1 | $\bar 0 = 1$ |
| 1 | 0 | $\bar 0 = 1$ |
| 1 | 1 | $\bar 1 = 0$ |
Now as we see that the output of the truth table of the given logic circuit matches with the output of the truth table of the AND logic gate.
So this is the required answer.
So, the correct answer is “Option C”.
Note:
Whenever we face such types of questions the key concept we have to remember is that always recall the truth tables of the standard logic gates such as, OR, AND, NAND, NOR which is all written above, so first find out the truth table of the given logic circuit and then check the output of the logic circuit follows which truth table of the given options logic gate as above.
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