The simplified switching circuit for the following circuit is

Let $p, q, r$ represent the switches $S_1, S_2, S_3$ respectively.

The Boolean expression for the top branch is $p \wedge (q \vee r)$.

The Boolean expression for the bottom branch is $\sim r \wedge \sim q \wedge p$, which can be rewritten as $p \wedge (\sim q \wedge \sim r)$.

The total circuit is the parallel combination of these two branches, so the expression is:

$[p \wedge (q \vee r)] \vee [p \wedge (\sim q \wedge \sim r)]$

Using De Morgan's Law, $\sim q \wedge \sim r \equiv \sim(q \vee r)$.

Substituting this, we get:

$[p \wedge (q \vee r)] \vee [p \wedge \sim(q \vee r)]$

Using the distributive law, we can factor out $p$:

$p \wedge [(q \vee r) \vee \sim(q \vee r)]$

Since $A \vee \sim A \equiv T$ (where $T$ represents a closed circuit or tautology), we have:

$p \wedge T \equiv p$

This means the simplified circuit consists only of the switch $S_1$.

Answer: