The characteristic equation of a JK flip flop is :

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Accepted Answer

Correct answer: B — Concept: The characteristic equation of a flip-flop expresses the next state $Q_{n+1}$ as a Boolean function of the present state $Q_n$ and the flip-flop's control inputs. It is obtained by writing out the flip-flop's truth table for every combination of inputs and present state, then reducing that table to a sum-of-products expression -- for a JK flip-flop the cleanest route is to split the table on the present state $Q_n$ itself. Application: Write the JK truth table over J, K, $Q_n$ -> $Q_{n+1}$: J=0,K=0 holds ($Q_{n+1}=Q_n$); J=0,K=1 resets ($Q_{n+1}=0$); J=1,K=0 sets ($Q_{n+1}=1$); J=1,K=1 toggles ($Q_{n+1}=\overline{Q_n}$). Isolate the rows where the present state $Q_n=0$: $Q_{n+1}$ equals J in every one of them, so the $Q_n=0$ branch reduces to $Q_{n+1}=J$. Isolate the rows where the present state $Q_n=1$: $Q_{n+1}$ equals the complement of K in every one of them (K=0 keeps $Q_{n+1}=1$, K=1 flips it to $Q_{n+1}=0$), so the $Q_n=1$ branch reduces to $Q_{n+1}=\overline{K}$. Recombine the two branches with the present state itself acting as the selector -- $\overline{Q_n}$ selects the J branch and $Q_n$ selects the $\overline{K}$ branch -- giving $Q_{n+1}=J\cdot\overline{Q_n}+\overline{K}\cdot Q_n$. Cross-check: Substituting the hold case J=K=0 gives $Q_{n+1}=0\cdot\overline{Q_n}+1\cdot Q_n=Q_n$ (state unchanged), and substituting the toggle case J=K=1 gives $Q_{n+1}=1\cdot\overline{Q_n}+0\cdot Q_n=\overline{Q_n}$ (state inverted) -- both match the defining hold/toggle behaviour of a JK flip-flop, confirming the derivation.

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A. $Q_{n+1}=J\cdot Q_n+K\cdot Q_n$

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B. $Q_{n+1}=J\cdot\overline{Q}_n+\overline{K}\cdot Q_n$

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C. $Q_{n+1}=Q_n\,J\cdot K$

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D. $Q_{n+1}=(J+K)Q_n$