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A sequential circuit using D flip-flop and logic gates is shown in Figure, where $X$ and $Y$ are the inputs and $Z$ is the output. The circuit is

  1. $\text{S-R}$ Flip-flop with inputs $X = R$ and $Y=S$
  2. $\text{S-R}$ Flip-flop with inputs $X = S$ and $Y=R$
  3. $\text{J-K}$ Flip-flop with inputs $X = J$ and $Y=K$
  4. $\text{J-K}$ Flip-flop with inputs $X = K$ and $Y=J$
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Answer: (D)

explanation:-  let z=Qn 

now we will put X=k and Y=j

after solving first AND gate we got $\bar{k}Qn$

after solving second AND gate we got $J\bar{Qn}$

after solving OR gate we got $J\bar{Qn}\dotplus \bar{K}Qn$

$\because$ we knows that in D flip flop next output(Qn+1)=D

after solving D flip flop we got Qn+1=$J\bar{Qn}\dotplus \bar{K}Qn$

this is the characteristic equation of J-K flip flop…..hence option (D) is correct.

 

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ANS : D

After solving D flip flop we got Qn+1=JQn¯∔K¯QnJQn¯∔K¯Qn

this is the characteristic equation of J-K flip flop.

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ANS : D

J-KJ-K Flip-flop with inputs X=KX=K and Y=J
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