A 2-input NOR gate has one input permanently connected to logic LOW (0). How does the output behave with respect to the other input?
Output is the complement of the other input
A NOR gate produces the logical OR of its inputs followed by inversion. For a 2-input NOR gate with inputs A and B the output is:
Output = NOT(A + B) = (A + B)'
Here one input is permanently tied to logic LOW (0). Substituting B = 0 and using the OR identity A + 0 = A:
Output = NOT(A + 0) = NOT(A) = Ā
So the gate simply inverts the remaining input. The verification table makes this clear:
| A | B = 0 | Output = (A + B)' |
|---|---|---|
| 0 | 0 | 1 |
| 1 | 0 | 0 |
When A = 0 the output is 1, and when A = 1 the output is 0 — the output is always the complement of the other input, which is the correct answer. This is a standard technique: a NOR gate with one input grounded behaves as a NOT (inverter) gate, just as a NAND gate with one input tied HIGH also acts as an inverter.
The output being always LOW would only occur if an input were held HIGH (since any HIGH input forces a NOR output to 0), not LOW. The output being always HIGH is impossible here because when A = 1 the sum is 1 and the inverted result is 0. And the output does not follow the other input directly — that describes a buffer, whereas the NOR here inverts the signal.
Similar Questions
Signal A passes through a NOT gate and then through an AND gate where the second input is permanently set to logic 1. What is the final output expression?
Important Questions from Logic Gates
Which logical gate is included in a binary adder circuit to enable binary addition and subtraction?
A Circuit that operates in such a way that its output is high only when all its inputs are high
______ is also known as Inverter.
The minimum number of NAND gates required to reduce the expression ((A + B)C) D is:
In which of the following gates, the output is high if and only if all inputs are high?
