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MHT-CET Maths · Mathematical Logic

Switching Circuits

A circuit of switches is a logical statement in disguise: switches in series are AND, switches in parallel are OR, and the lamp glowing is the statement being true.

Why this matters

This is the smallest subtopic in the chapter at 12 PYQs and by a wide margin the hardest — 67% of them are HARD, against 14% for Negation — so it is where the chapter's difficulty actually lives, and it is worth knowing that before you budget your time. The good news is that nothing here is new. Once a circuit is translated into symbols, every question is answered by the algebra you already have, and the whole subtopic rests on two rules that take a minute to learn.

Concept 1 of 4

Series is And, Parallel is Or

Intuition

Current has to get from one end to the other. If the switches sit one after another, every one of them must be closed for current to pass — that is AND. If they sit on separate branches, any one of them closed is enough — that is OR.

Definition

Let pp be the statement 'switch S1S_1 is closed', and so on. Then:

  • Switches in series give pqp \wedge q — the lamp glows only when both are closed.
  • Switches in parallel give pqp \vee q — the lamp glows when at least one is closed.
  • A switch written S1S_1' is the negation p\sim p: it is closed exactly when S1S_1 is open.

'The lamp glows' is the statement being true; the circuit's truth table is the statement's truth table.

The translation rule

series    pqparallel    pqS1    p\text{series} \;\longrightarrow\; p \wedge q \qquad \text{parallel} \;\longrightarrow\; p \vee q \qquad S_1' \;\longrightarrow\; \sim p
  • pthe switch S1S_1 is closed
  • S1S_1'the complementary switch — closed exactly when S1S_1 is open
SeriesS₁S₂Lp ∧ qboth neededParallelS₁S₂Lp ∨ qeither will do

The lamp glowing is the statement being true. Series needs every switch closed, so it is AND; parallel needs only one, so it is OR. A switch marked S′ is the negation, closed exactly when S is open.

Worked example

A circuit has S1S_1 in series with a parallel pair S2S_2 and S3S_3. Write its symbolic form, and say when the lamp glows.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Mathematical LogicMODERATE
If p : switch S1S_{1} is closed, q: switch S2S_{2} is closed, r : switch S3S_{3} closed, then the symbolic form of the following switching circuit is equivalent to Switching Circuit:

[Q136 · 22 April Shift II · 2025]

Reading a branch before finishing it

A branch often contains its own series run before rejoining the parallel node. Resolve each branch completely into a single expression FIRST, then join the branches — reading left to right across the whole picture mixes the levels up.

Concept 2 of 4

Writing the Symbolic Form of a Printed Circuit

Intuition

Reading a circuit is a bracketing exercise. Find the points where the wire splits and rejoins: everything between a split and its rejoin is one parallel block, and blocks strung end to end are a series run.

Definition

A reliable procedure for any printed circuit:

  • Find every split-and-rejoin pair. Each becomes a bracket joined by \vee.
  • Inside a branch, switches met one after another are joined by \wedge.
  • Work innermost block first, replacing each finished block by a single symbol before moving out.
  • Mind the primes: S1S_1 and S1S_1' are the same switch, so they are pp and p\sim p, never two different letters.

Worked example

A circuit has two parallel branches. The upper is S1S_1 in series with S2S_2'; the lower is S1S_1' in series with S2S_2. Write its symbolic form and simplify in words what it says.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Mathematical LogicHARD
If pp\equiv The switch S1S_{1} is closed, qq\equiv The switch S2S_{2} is closed, rr \equiv switch S3S_{3} is closed, then symbolic form of following switching circuit is equivalent to

[Q121 · 25 April Shift II · 2025]

Giving a primed switch its own letter

S1S_1' is not a third switch. It is p\sim p, and treating it as a new letter destroys every simplification — the whole point of these circuits is that pp meeting p\sim p collapses a branch.

Concept 3 of 4

Simplifying a Circuit and Redrawing It

Intuition

Simplifying a circuit means using fewer switches for the same behaviour — which is an algebra problem, not a drawing problem. Translate to symbols, simplify with the laws, then draw the circuit the simplified expression describes.

Definition

The procedure is always the same three steps:

  • Translate the printed circuit into a statement pattern.
  • Simplify with the algebra of statements — complement, absorption and distribution do most of the work here.
  • Redraw the result, reading \wedge back as series and \vee back as parallel.

A branch simplifying to TT is a permanently closed path (the lamp always glows); a branch simplifying to FF is a dead branch that can be deleted.

As printedS₁S₂S₁S₃(p ∧ q) ∨ (p ∧ r)S₁ appears twicedistributivefactor out S₁SimplifiedS₁S₂S₃p ∧ (q ∨ r)one switch fewer

Both circuits are closed in exactly the same cases, so they are equivalent — and the right-hand one uses three switches instead of four. Simplify the expression first, then draw the circuit the answer describes.

Worked example

A circuit has two parallel branches: the upper is S1S_1 in series with S2S_2; the lower is S1S_1 in series with S3S_3. Simplify it and say how many switches the simpler circuit needs.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Mathematical LogicMODERATE
The new switching circuit for the following circuit by simplifying the given circuit is

[Q136 · 3rd May Shift 2 · 2023]

Trying to simplify by staring at the picture

Redrawing a circuit by eye, without writing the expression down, is where these questions are lost. The simplification is algebraic; the drawing is only the last step, and it follows mechanically from the simplified expression.

Deleting a repeated switch instead of factoring it

In (pq)(pr)(p \wedge q) \vee (p \wedge r), S1S_1 genuinely appears on both branches and both copies matter. Factoring gives p(qr)p \wedge (q \vee r), which is a different arrangement — not the result of rubbing one copy out.

Concept 4 of 4

Deciding Whether Two Circuits Are Equivalent

Intuition

Two circuits are equivalent when they are closed in exactly the same cases — which is just logical equivalence wearing a different costume. Translate both, simplify both, and compare; if the simplified expressions match, the circuits match.

Definition

To test a pair of circuits:

  • Translate each into a statement pattern.
  • Simplify both as far as they go.
  • They are equivalent exactly when the simplified patterns are logically equivalent.

To show two circuits are not equivalent, give one assignment of open and closed switches where one lamp glows and the other does not — a single case is enough. The recurring exam pair is (pq)(pr)(p \wedge q) \vee (p \wedge r) with p(qr)p \wedge (q \vee r).

Worked example

Five circuits are printed: (i) (pq)(pr)(p \wedge q) \vee (p \wedge r), (ii) p(qr)p \vee (q \wedge r), (iii) p(qr)p \wedge (q \vee r), (iv) pqrp \wedge q \wedge r, (v) (pq)r(p \wedge q) \vee r. Which pair is equivalent?
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Mathematical LogicMODERATE
Which one of the following is the pair of equivalent circuits? (i) through (v) as given

[Q150 · 2nd May Shift 2 · 2023]

Matching circuits by how they look

Two circuits drawn with the same number of switches in a similar arrangement can behave differently, and two that look nothing alike can be equivalent. The pictures are not the evidence — the simplified expressions are.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

  • Series is And, Parallel is Or

    The translation rule

    series    pqparallel    pqS1    p\text{series} \;\longrightarrow\; p \wedge q \qquad \text{parallel} \;\longrightarrow\; p \vee q \qquad S_1' \;\longrightarrow\; \sim p

Watch out for (5)

Drill every past-year question on this subtopic

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