For the circuit show below, the Boolen polynomial is

(\sim p \lor q) \lor (p \lor \sim q)

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(\sim p \land q) \land (p \land \sim q)

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(\sim p \land \sim q) \land (q \land p)

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(\sim p \land q) \lor (p \land \sim q)

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Solution

The correct option is D $(\sim p \land q) \lor (p \land \sim q)$
Since, $\sim p$ and $q$ both switch also are in series
$\Rightarrow (\sim p \land q)$ ...(i)
and $p$ and $\sim q$ both switch also are in series
$\Rightarrow (p \land \sim q)$ ...(ii)
Both Eqs. (i) and (ii) are parallel switch
$\Rightarrow (\sim p \land q) \lor (p \land \sim q)$

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Similar questions

The negation of $p \land q \to p \lor q i s$

If $\frac{{(p + \frac{1}{q})}^{p - q} {(p - \frac{1}{q})}^{p + q}}{{(q + \frac{1}{p})}^{p - q} {(q - \frac{1}{p})}^{p + q}} = {(\frac{p}{q})}^{x}, then x =_________.$

The dual of statement $p \lor (q \land r) \equiv (p \lor q) \land (p \lor r)$ is

Q. Let (i)

(p \lor q) \lor (p \lor \sim q)

,
(ii)

(p \land q) \land (p \lor \sim q)

,
(iii)

(p \lor q) \land (p \lor \sim q)

,
(iv)

(p \lor q) \lor (p \land \sim q)

which one is tautology

Q. The proposition

\sim (p \Leftrightarrow q)

is equivalent to

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For the circuit show below, the Boolen polynomial is

The correct option is D (∼p∧q)∨(p∧∼q)Since, ∼p and q both switch also are in series⇒(∼p∧q) ...(i)and p and ∼q both switch also are in series⇒(p∧∼q) ...(ii)Both Eqs. (i) and (ii) are parallel switch⇒(∼p∧q)∨(p∧∼q)