For the circuit show below, the Boolean polynomial is

Question

For the circuit show below, the Boolean polynomial is (A)  (∼ p ∨ q)∨(p ∨ ∼ q)  (B)  (∼ p ∧ q)∧(p ∧ q)  (C)  (∼ p ∧∼ q)∧(q ∧ p)  (D)  (∼ p ∧ q)∨(p ∧ ∼ q)

Tardigrade Admin · Accepted Answer

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)

Q. For the circuit show below, the Boolean polynomial is

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

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

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

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

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

Q. For the circuit show below, the Boolean polynomial is

(∼p∨q)∨(p∨∼q)

(∼p∧q)∧(p∧q)

(∼p∧∼q)∧(q∧p)

(∼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)

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

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

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

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

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