If p : switch $\mathrm{s}_1$ is closed, q : switch $\mathrm{s}_2$ is closed then correct interpretation from…
- The lamp is always on
- The lamp is always off
- Symbolic form is $\mathrm{p} \vee(\sim \mathrm{p} \wedge \sim \mathrm{q}) \vee \mathrm{q}$
- is equivalent to $\mathrm{p} \vee \mathrm{q}$
Solution
Interpretation: Represent switch S1 closed as $p$ and S2 closed as $q$, with S1' as $\sim p$ and S2' as $\sim q$. The lamp glows when there is a complete conductive path.
The circuit consists of S1 in series with a parallel combination of S1' and S2', followed by S2 in series. The condition for the lamp to be on is $p \land (\sim p \lor \sim q) \land q$.
Simplifying the expression:
$p \land (\sim p \lor \sim q) = (p \land \sim p) \lor (p \land \sim q) = F \lor (p \land \sim q) = p \land \sim q$
Substituting gives $(p \land \sim q) \land q = p \land (\sim q \land q) = p \land F = F$.
The expression simplifies to false, indicating the lamp never glows regardless of switch positions.
Option B ($\boxed{\text{B}}$) is correct.
Asked in: MHT CET 2025 (05 May Shift 2)