Consider statements p: S_1 is closed; q: S_2 | vedclass

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(D) The given circuit consists of two main parts connected in series with switch $S_3$. Let the first part be $C_1$ and the second part be $C_2$. The

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(D) The given circuit consists of two main parts connected in series with switch $S_3$. Let the first part be $C_1$ and the second part be $C_2$. The circuit is $C_1 \wedge r \wedge C_2$.For $C_1$: It has three parallel branches: $(p' \wedge q')$,$p$,and $q$. So,$C_1 = (p' \wedge q') \vee p \vee q$.Using the law of absorption and distributive laws:$C_1 = (p' \wedge q') \vee (p \vee q) = (p' \vee (p \vee q)) \wedge (q' \vee (p \vee q)) = (T \vee q) \wedge (p \vee (q' \vee q)) = T \wedge (p \vee T) = T \wedge T = T$.For $C_2$: It has two parallel branches: $(p \wedge q)$ and $(p' \wedge q)$. So,$C_2 = (p \wedge q) \vee (p' \wedge q)$.Using the distributive law:$C_2 = (p \vee p') \wedge q = T \wedge q = q$.Thus,the total circuit expression is $T \wedge r \wedge q = q \wedge r$.The simplified circuit is a series connection of switches $S_2$ and $S_3$,which corresponds to the logical statement $(q \wedge r)$.Therefore,the correct option is $D$.

Consider statements $p: S_1$ is closed; $q: S_2$ is closed; $r: S_3$ is closed. The simplified equivalent circuit diagram and its logical statement for the switching circuit is respectively.

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