In a positive edge triggered JK flip-flop, J = 1, K = 0 and clock pulse is rising, Q will be

Question

Question

This question was previously asked in

UKPSC JE Electrical 2013 Official Paper I (Held on 7 Nov 2015)

Answer 1

Concept:

Characteristic table of J-K flip-flop is,

Here, Q(n) is the present state of flip flop and Q(n + 1) is the next state of flip flow.

+ve edge-triggered means the state of flip-flop will only change at the rising edge of the clock.

Therefore, from the characteristic table;

when J = 1, K = 0, Q(n) = 1

Therefore, the correct option will be (2).