MHTCET 19 Apr 2026 Shift 1 Paper

Concept:Convert the switching circuit into a Boolean expression, then simplify it using laws of Boolean algebra.Explanation:Let $p, q, r$ represent switches $S_1, S_2, S_3$ respectively.In the upper branch, $S_1$ is in series with the parallel combination of $S_2$ and $S_3$, giving $p \wedge (q \vee r)$.In the lower branch, $S_1$ is in series with not $S_2$ and not $S_3$, giving $p \wedge (\sim q \wedge \sim r)$.Since the branches are in parallel, combine them using OR:$$[p \wedge (q \vee r)] \vee [p \wedge (\sim q \wedge \sim r)]$$Using De Morgan's law, $\sim q \wedge \sim r = \sim(q \vee r)$.So the expression becomes:$$[p \wedge (q \vee r)] \vee [p \wedge \sim(q \vee r)]$$Since $p$ is common, factor it out:$$p \wedge [(q \vee r) \vee \sim(q \vee r)]$$Now $(q \vee r)$ and its negation $\sim(q \vee r)$ are complementary, so their OR is always true:$$(q \vee r) \vee \sim(q \vee r) = T$$Therefore:$$p \wedge T = p$$So the entire circuit is equivalent to only switch $S_1$.Answer:The simplified switching circuit is equivalent to a single switch $S_1$.Select the option that shows only switch $S_1$ in the circuit.