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The given following circuit is equivalent to
Options:
- A
- B
- C
- D
Solution:
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Verified Answer
The correct answer is:
Let $\mathrm{p}$ : The switch $\mathrm{S}_1$
$\mathrm{q}$ : The switch $\mathrm{S}_2$
The symbolic form is
$$
(p \wedge \sim q) \vee(\sim p \wedge q) \vee(\sim p \wedge \sim q)
$$
$\begin{aligned} & \equiv(p \wedge \sim q) \vee[\sim p \wedge(q \vee \sim q)] \\ & \text {...[Distributive Law] } \\ & \equiv(p \wedge \sim q) \vee[\sim p \wedge t] \\ & \text {...[Complement Law] } \\ & \equiv(p \wedge \sim q) \vee \sim p \\ & \text {...[Identity Law] } \\ & \equiv \sim p \vee(p \wedge \sim q) \\ & \text { [Commutative Law] } \\ & \equiv(\sim p \vee p) \wedge(\sim p \vee \sim q) \\ & \text {.[Distributive Law] } \\ & \equiv \mathrm{t} \wedge(\sim \mathrm{p} \vee \sim \mathrm{q}) \\ & \text {.[Complement Law] } \\ & \equiv \sim p \vee \sim q \\ & \text {.[Identity Law] } \\ & \end{aligned}$
$\mathrm{q}$ : The switch $\mathrm{S}_2$
The symbolic form is
$$
(p \wedge \sim q) \vee(\sim p \wedge q) \vee(\sim p \wedge \sim q)
$$
$\begin{aligned} & \equiv(p \wedge \sim q) \vee[\sim p \wedge(q \vee \sim q)] \\ & \text {...[Distributive Law] } \\ & \equiv(p \wedge \sim q) \vee[\sim p \wedge t] \\ & \text {...[Complement Law] } \\ & \equiv(p \wedge \sim q) \vee \sim p \\ & \text {...[Identity Law] } \\ & \equiv \sim p \vee(p \wedge \sim q) \\ & \text { [Commutative Law] } \\ & \equiv(\sim p \vee p) \wedge(\sim p \vee \sim q) \\ & \text {.[Distributive Law] } \\ & \equiv \mathrm{t} \wedge(\sim \mathrm{p} \vee \sim \mathrm{q}) \\ & \text {.[Complement Law] } \\ & \equiv \sim p \vee \sim q \\ & \text {.[Identity Law] } \\ & \end{aligned}$
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