
The gate for which output is high, if at least one input is low is :
A. NAND
B. NOR
C. AND
D. OR
Answer
571.2k+ views
Hint: Logic gates are defined as the electronic circuits that have more than one input and only one output, whose relationship is dependent on an inherent logic of the logic gate. There are three basic logic gates: AND, OR and NOT. There are two more gates which are called Universal gates called NAND and NOR.
Complete step-by-step answer:
Let us analyse the operations of each of these logic gates briefly, to arrive at the right answer. a. NAND gate If A and B are two inputs to the NAND gate, the output follows the logic: $\overline {A.B} $
The truth table is as follows:
b. NOR gate If A and B are two inputs to the NOR gate, the output follows the logic: $\overline {A + B} $
The truth table is as follows:
c. AND gate If A and B are two inputs to the AND gate, the output follows the logic: $A.B$
The truth table is as follows:
d. OR gate If A and B are two inputs to the OR gate, the output follows the logic: $A + B$
The truth table is as follows:
As per the question, the logic gate where the output is High for at least one Low (0) input is –
NAND gate.
Because in the truth table, for the first three rows where 0 is one of the inputs, the output is always 1.
Hence, the correct option is Option A.
Note: NAND and NOR gates are called universal gates since all the basic and advanced gates can be constructed using a combination of NAND and NOR gates.
Sometimes, it becomes difficult to remember the conversion of Universal gates into the basic AND and OR gates. You can easily remember the whole thing by understanding this mnemonic.
NAND Gate:
To convert to AND gate, first negate together because AND means together and negate the result.
To convert to OR gate, first negate them individually because OR means individually and negate the result.
Complete step-by-step answer:
Let us analyse the operations of each of these logic gates briefly, to arrive at the right answer. a. NAND gate If A and B are two inputs to the NAND gate, the output follows the logic: $\overline {A.B} $
The truth table is as follows:
| A | B | $\overline {A.B} $ |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
b. NOR gate If A and B are two inputs to the NOR gate, the output follows the logic: $\overline {A + B} $
The truth table is as follows:
| A | B | $\overline {A + B} $ |
| 0 | 0 | 1 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
c. AND gate If A and B are two inputs to the AND gate, the output follows the logic: $A.B$
The truth table is as follows:
| A | B | $A.B$ |
| 0 | 0 | 0 |
| 0 | 1 | 0 |
| 1 | 0 | 0 |
| 1 | 1 | 1 |
d. OR gate If A and B are two inputs to the OR gate, the output follows the logic: $A + B$
The truth table is as follows:
| A | B | $A + B$ |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 1 |
As per the question, the logic gate where the output is High for at least one Low (0) input is –
NAND gate.
Because in the truth table, for the first three rows where 0 is one of the inputs, the output is always 1.
| A | B | $\overline {A.B} $ |
| 0 | 0 | 1 |
| 0 | 1 | 1 |
| 1 | 0 | 0 |
| 1 | 1 | 0 |
Hence, the correct option is Option A.
Note: NAND and NOR gates are called universal gates since all the basic and advanced gates can be constructed using a combination of NAND and NOR gates.
Sometimes, it becomes difficult to remember the conversion of Universal gates into the basic AND and OR gates. You can easily remember the whole thing by understanding this mnemonic.
NAND Gate:
To convert to AND gate, first negate together because AND means together and negate the result.
To convert to OR gate, first negate them individually because OR means individually and negate the result.
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