Two switches in parallel

Two Switches in Parallel

Now wire $S_1$ and $S_2$ side by side, each offering its own path to the lamp $L$ (a parallel connection), with $p$ for $S_1$ and $q$ for $S_2$. Current reaches the lamp as long as at least one path is closed — the behaviour of disjunction $p \lor q$.

The lamp is dark only when both switches are off; a parallel connection is the circuit picture of OR.

With series = AND, parallel = OR, and a complementary switch = NOT now established, any circuit built from switches can be translated into a logical expression, and any logical expression can equally be built as a circuit — which is exactly what the worked examples below do.

Worked Example 1 — reading circuits into symbolic form. Three circuits (using switches $S_1, S_2$, and in the third case $S_3$, all controlling a lamp $L$) are to be written symbolically and given their input-output tables.

(i) With $p$ for $S_1$ and $q$ for $S_2$, this circuit's expression is $(p \lor q) \lor (\sim p \land \sim q)$. Building the table column by column:

Every row gives 1 — the last column is all ones, so this circuit is a tautology: however the two switches are set, the lamp always glows.

(ii) With $p$ for $S_1$, $q$ for $S_2$, $r$ for $S_3$, the expression is $[(p \land q) \lor (\sim p \land \sim q)] \land r$. Here $(p\land q)\lor(\sim p\land \sim q)$ is 1 exactly when $p$ and $q$ agree (both on or both off), and the whole expression additionally needs $r$ on:

The lamp glows only when $p$ and $q$ agree AND $r$ is on (rows 1 and 7).

(iii) With $p, q, r$ for $S_1, S_2, S_3$, the expression is $(p \lor q) \land q \land (r \lor \sim p)$:

Notice $(p\lor q)\land q$ always equals $q$ itself (an instance of the Absorption Law), so the lamp really only needs $q$ on together with $r$ on or $p$ off — the table confirms it lights only where $q=1$ and ($r=1$ or $p=0$).

Worked Example 2 — building circuits from expressions. Given three logical expressions (with $p,q,r$ for $S_1,S_2,S_3$), each is translated into a switch network by reading $\land$ as switches wired in series and $\lor$ as branches wired in parallel; no input-output table is needed here, only the circuit layout.

(i) $[p \lor (\sim p \land q)] \lor [(\sim q \land r) \lor \sim p]$ — since disjunction only ever adds parallel branches, this circuit is four paths side by side between the two terminals: a lone switch $p$; a series pair $\sim p, q$; a series pair $\sim q, r$; and a lone switch $\sim p$.

(ii) $(p \land q \land r) \lor [\sim p \lor (q \land \sim r)]$ — three parallel branches: $p, q, r$ all in series; a lone switch $\sim p$; and $q, \sim r$ in series.

(iii) $[(p \land r) \lor (\sim q \land \sim r)] \lor (\sim p \land \sim r)$ — three parallel branches: $p, r$ in series; $\sim q, \sim r$ in series; and $\sim p, \sim r$ in series.

Worked Example 3 — simplifying to fewer switches. A circuit with switches $S_1 (p)$ and $S_2 (q)$ has expression $(p \land \sim q) \lor (\sim p \land q) \lor (\sim p \land \sim q)$ — four switch-occurrences in all. The law chain below finds an equivalent circuit using as few switches as possible:

$(p \land \sim q) \lor (\sim p \land q) \lor (\sim p \land \sim q)$

$\equiv (p \land \sim q) \lor [\sim p \land (q \lor \sim q)]$ — Distributive Law, factoring $\sim p$ out of the last two terms

$\equiv (p \land \sim q) \lor (\sim p \land t)$ — Complement Law, $q \lor \sim q \equiv t$

$\equiv (p \land \sim q) \lor \sim p$ — Identity Law, $\sim p \land t \equiv \sim p$

$\equiv \sim p \lor (p \land \sim q)$ — Commutative Law

$\equiv (\sim p \lor p) \land (\sim p \lor \sim q)$ — Distributive Law

$\equiv t \land (\sim p \lor \sim q)$ — Complement Law, $\sim p \lor p \equiv t$

$\equiv \sim p \lor \sim q$ — Identity Law, $t \land X \equiv X$

So this four-switch circuit is logically identical to just two switches, $\sim p$ and $\sim q$ (the complements of $S_1$ and $S_2$), wired in parallel — the minimum arrangement for this expression.

Worked Example 4 — expressing, tabulating, and interpreting. With $p$ for $S_1$ and $q$ for $S_2$, a given circuit's expression is $(p \lor q) \land (\sim p) \land (\sim q)$:

Every row is 0. Algebraically this checks out too: $(p\lor q)\land\sim p\land\sim q \equiv (p\land\sim p\land\sim q)\lor(q\land\sim p\land\sim q) \equiv c \lor c \equiv c$, a contradiction. Interpretation: the lamp never glows, no matter how the two switches are set.

Worked Example 5 — simplifying to a constant. With $p$ for $S_1$ and $q$ for $S_2$, a circuit's expression is $p \land (\sim p \lor \sim q) \land q$:

$p \land (\sim p \lor \sim q) \land q$

$\equiv [p \land (\sim p \lor \sim q)] \land q$ — Associative Law

$\equiv [(p \land \sim p) \lor (p \land \sim q)] \land q$ — Distributive Law

$\equiv [c \lor (p \land \sim q)] \land q$ — Complement Law, $p \land \sim p \equiv c$

$\equiv (p \land \sim q) \land q$ — Identity Law, $c \lor X \equiv X$

$\equiv p \land (\sim q \land q)$ — Associative Law

$\equiv p \land c$ — Complement Law, $\sim q \land q \equiv c$

$\equiv c$ — Identity Law, $X \land c \equiv c$

Conclusion: the expression collapses to the contradiction $c$, so the lamp will not glow regardless of how the switches are set.

Worked Example 6 — symbolic form, table, and simplification together. With $p, q, r$ for $S_1, S_2, S_3$: …