During a laboratory test, a trainee builds an RS flip-flop using NOR gates (active-HIGH inputs). Initially, the output Q = 0. The trainee applies S = 1 (Set input) and R = 0 (Reset input). After this, both inputs are returned to S = 0 and R = 0. What will happen to the output Q?
Q will remain at 1 after inputs become 0, holding the last state.
In this problem, we are dealing with an RS flip-flop created using NOR gates and analyzing how the output Q changes based on specific inputs. Let's break down the problem step-by-step:
- The RS flip-flop is made using NOR gates, which means it is a simple bistable multivibrator.
- Initially, the output Q is given as 0 (Q = 0).
- The inputs provided are S = 1 (Set input) and R = 0 (Reset input). When S = 1 and R = 0 are applied to an RS flip-flop:
- The flip-flop will set the output to Q = 1, because the set input (S) is high and the reset input (R) is low.
- After setting S = 1 and R = 0, both inputs are returned to S = 0 and R = 0. For a NOR gate-based RS flip-flop:
- When S = 0 and R = 0, the flip-flop holds its last state. Therefore, the output Q will remain 1.
Therefore, after applying the sequence of inputs, the output Q remains at 1, holding the last state when the inputs become S = 0 and R = 0.
Thus, the correct answer is: Q will remain at 1 after inputs become 0, holding the last state.
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