The simplified switching circuit for the foll | vedclass

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(A) Let $S_1, S_2, S_3$ represent the switches. The circuit consists of two parallel branches connected in series with the lamp $L$.Branch $1$: $S_1$

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(A) Let $S_1, S_2, S_3$ represent the switches. The circuit consists of two parallel branches connected in series with the lamp $L$.Branch $1$: $S_1$ is in series with the parallel combination of $S_2$ and $S_3$. The logical expression is $S_1 \land (S_2 \lor S_3)$.Branch $2$: $S_3'$, $S_2'$, and $S_1$ are in series. The logical expression is $S_3' \land S_2' \land S_1$.Since the branches are in parallel, the total expression is $Y = (S_1 \land (S_2 \lor S_3)) \lor (S_3' \land S_2' \land S_1)$.Using the distributive law: $Y = (S_1 \land S_2) \lor (S_1 \land S_3) \lor (S_3' \land S_2' \land S_1)$.Factor out $S_1$: $Y = S_1 \land (S_2 \lor S_3 \lor (S_3' \land S_2'))$.Using the law $(A \lor (B \land C)) = (A \lor B) \land (A \lor C)$, we have $(S_2 \lor (S_3' \land S_2')) = (S_2 \lor S_3') \land (S_2 \lor S_2') = (S_2 \lor S_3') \land 1 = (S_2 \lor S_3')$.So, $Y = S_1 \land (S_3 \lor S_2 \lor S_3') = S_1 \land ((S_3 \lor S_3') \lor S_2) = S_1 \land (1 \lor S_2) = S_1 \land 1 = S_1$.Thus, the circuit simplifies to a single switch $S_1$ in series with the lamp $L$, which corresponds to image $(a)$.

The simplified switching circuit for the following circuit is

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