ARCHETYPE: Capacitor Networks & Equivalent Capacitance

Every year, somewhere in a JEE-Main hall, a five-capacitor diamond splits the room in two. Half the candidates start writing equations for every plate — eight unknowns, five minutes gone, panic. The other half glance at the four arm values, multiply crosswise, smile, and delete one capacitor from the circuit. Same question, twenty seconds. The difference is not intelligence; the second group has simply met the network before. Capacitor networks are a game with exactly four levels, recycled by examiners across CBSE, ISC, JEE and NEET for decades: the mixed network you collapse step by step, the bridge you balance and simplify, the symmetric net you fold along its mirror line, and the infinite ladder that looks impossible until one idea — the ladder minus one rung is still the same ladder — turns it into a quadratic. Behind all four sit the two invariants you own from Capacitors in Series & Parallel: parallel plates share $V$ , series plates share $Q$ . This article treats the game itself as the concept: the reduction algorithm and its unwinding, the ratio rule that switches off a balanced bridge, the folding trick for symmetric nets, the self-similarity move for ladders — and the traps that cost marks in each. One archetype, mastered once, and an entire genus of exam problems becomes routine arithmetic.

Key takeaways

Reduction is a loop: collapse pure series/parallel clusters inside-out (rules from Capacitors in Series & Parallel), then unwind — shared $Q$ back through series steps, shared $V$ back through parallel steps.
Division rules do the unwinding: in series $V_i \propto 1/C_i$ (the smallest capacitor takes the biggest voltage); in parallel $Q_i \propto C_i$ (the biggest capacitor takes the most charge).
The bridge: if $C_1 C_4 = C_2 C_3$ (cross products of the arms), both middle nodes sit at the same potential and the bridge capacitor carries zero charge — erase it.
Symmetric networks fold: points that a mirror symmetry maps onto each other are at equal potential, so capacitors joining them are idle and equal-potential nodes may be merged.
The infinite ladder obeys self-similarity: removing one section leaves the same ladder, so $C_{\text{eq}} = x$ satisfies an algebra equation — always a quadratic, always one positive root.
Key idea: every network question is the same three moves — spot the structure (series/parallel, balance, symmetry, self-similarity), collapse to one $C_{\text{eq}}$ , then unwind the two invariants for the internal charges and voltages.

The core insight

A capacitor network reduces by repeated series/parallel collapse, and the irreducible cases surrender to structure: a balanced bridge ( $C_1C_4 = C_2C_3$ ) idles its middle capacitor, a symmetric net merges its equal-potential nodes, and an infinite ladder equals itself with one section removed.

Where this lives today

The touchscreen you may be reading this on is a live capacitor network: a grid of transparent electrodes forms hundreds of small capacitors, and your fingertip — a grounded conductor — perturbs the local values; the controller re-solves the division rules a hundred times per second to locate the touch. Motherboards gang decoupling capacitors in parallel (capacitances adding) to steady supply rails, while high-voltage stacks run capacitors in series and police the inverse voltage division with balancing networks — the smallest member really does take the largest voltage. And the infinite ladder is the standard circuit model of a long cable, and of the axon membranes carrying nerve impulses in your reading eye.

The setup

Geometry first, because in an archetype the geometry is the physics. Four shapes cover essentially every capacitor-combination question at this level, each with one structural feature that unlocks it.

Figure

The four exam networks and the one move that unlocks each

The battery across a network sees a single number, the equivalent capacitance $C_{\text{eq}} = Q/V$ : impose $V$ , count the charge $Q$ that flows in, take the ratio. Everything inside — however baroque — hides in that one number. Your job is to find $C_{\text{eq}}$ , then, if asked, reopen the box and distribute $Q$ and $V$ among the members. Two facts from Capacitors in Series & Parallel power the whole game. Parallel members share the same two nodes and one voltage; capacitances add: $C_{\text{eq}} = C_1 + C_2 + \cdots$ . Series members chain through private junctions; the stranded conductor between neighbours stays neutral, forcing one charge $Q$ through the chain, and reciprocals add: $1/C_{\text{eq}} = 1/C_1 + 1/C_2 + \cdots$ . We will not re-derive these — that was the previous module's work. We will drive them.

The master solution pattern

Every reducible network yields to the same loop, and board marking schemes are structured around its steps.

Step 1. Label the nodes, not the wires. Parallel means both ends on the same two nodes; series means the shared node touches nothing else. The drawing can lie — node labels tell the truth.

Step 2. Collapse the innermost pure cluster (series or parallel), redraw the smaller circuit, and repeat until one capacitor remains. Compute $Q = C_{\text{eq}} V$ .

Step 3. Unwind. Walk backwards through the collapses: a series collapse hands its single $Q$ to every member; a parallel collapse hands its single $V$ to every branch; $V_i = Q_i/C_i$ fills in the rest.

Step 4. Audit. Path voltages must sum to $V$ ; charges on any interior island must sum to zero. Both checks are free; failing either means an upstream error. The unwinding runs on two division rules, worth carrying as slogans. In series, all members hold the same $Q$ , so each voltage is $V_i = Q/C_i$ — inverse to capacitance:

V_i \;=\; V \cdot \frac{C_{\text{other}}}{C_1 + C_2} \qquad \text{(two in series)} \qquad (1)

where: $C_{\text{other}}$ is the other capacitor's value — the cross-over form that avoids reciprocal arithmetic. Read it physically: the smallest capacitor eats the biggest voltage, because at fixed $Q$ a small $C$ means a large $Q/C$ . That sentence decides which capacitor in a stack breaks down first, and it is the reverse of resistor intuition. In parallel, all members feel the same $V$ , so charge divides in direct proportion:

Q_i \;=\; C_i V \;=\; Q_{\text{tot}}\cdot\frac{C_i}{C_1 + C_2 + \cdots} \qquad (2)

where: $Q_{\text{tot}}$ is the total delivered to the group. The big capacitor drinks the most charge — the wider vessel fills more. Note the duality: series is equal- $Q$ with $V \propto 1/C$ ; parallel is equal- $V$ with $Q \propto C$ .

structure	recognize it by	the move	then
mixed network	pure series/parallel clusters visible	collapse inside-out	unwind $Q$ , $V$ by (1), (2)
bridge	five capacitors, diamond + crossbar	test $C_1C_4 = C_2C_3$	balanced → erase $C_5$ , two series branches in parallel
symmetric net	equal values mirrored about an axis	fold: merge equal-potential nodes, delete idle links	reduce the half-circuit
infinite ladder	repeating section, "infinite" in the stem	set $C_{\text{eq}} = x$ , remove one section, equate	solve the quadratic, keep the positive root

Figure

Balance test: when both branches divide the voltage alike, the crossbar goes dead

Derivation AWhy a Balanced Bridge Ignores Its Middle Capacitor

We derive the balance condition for the capacitor Wheatstone bridge — five capacitors: $C_1$ and $C_2$ from the top terminal $A$ to nodes $a$ and $b$ , $C_3$ and $C_4$ from $a$ and $b$ down to terminal $B$ , and $C_5$ across $a$ – $b$ . The strategy is to suppose $C_5$ carries no charge, work out what that forces on the arms using the series voltage-division rule, and read off the condition on $C_1 \ldots C_4$ that makes the supposition self-consistent. Watch for the inverse division: it is $C_3/(C_1+C_3)$ , not $C_1/(C_1+C_3)$ , that gives the upper arm's voltage share.

Step 1. Suppose $Q_5 = 0$ . Then no charge has been pushed through the crossbar, so branch $A$ – $a$ – $B$ is an honest two-capacitor series chain, and so is $A$ – $b$ – $B$ .

Step 2. Voltage across $C_1$ in the left chain, by the series division rule (1): $V_1 = V\cdot\dfrac{C_3}{C_1 + C_3}$ . The potential at node $a$ is therefore $V_a = V_A - V\cdot\dfrac{C_3}{C_1+C_3}$ . Why the "wrong-way" ratio: series members share $Q$ , so the voltage lands in inverse proportion to capacitance — the $C_3$ in the numerator is the other arm's value.

Step 3. Same for the right chain: $V_b = V_A - V\cdot\dfrac{C_4}{C_2+C_4}$ .

Step 4. The supposition $Q_5 = 0$ is consistent exactly when $C_5$ feels no voltage: $V_a = V_b$ , i.e.

\frac{C_3}{C_1+C_3} = \frac{C_4}{C_2+C_4} \quad\Longleftrightarrow\quad C_1 C_4 = C_2 C_3

Cross-multiply and the mixed terms $C_3C_4$ cancel, leaving the cross-product form — equivalently the ratio form $\dfrac{C_1}{C_3} = \dfrac{C_2}{C_4}$ : each branch divides voltage in the same proportion.

Step 5. When the condition holds, erase $C_5$ (a capacitor at zero voltage stores nothing and influences nothing): $C_{\text{eq}}$ is just the two series chains in parallel,

C_{\text{eq}} = \frac{C_1C_3}{C_1+C_3} + \frac{C_2C_4}{C_2+C_4}

Why erasing is legal: with $V_a = V_b$ you may equally well replace $C_5$ by a solid wire (merging $a$ and $b$ ) — both surgeries give the same $C_{\text{eq}}$ , a classic self-check. Unequal cross products permit neither surgery: that network is irreducible, Kirchhoff's Rules territory. *Verified: $C_1 = 2, C_2 = 4, C_3 = 3, C_4 = 6\ \mu\mathrm{F}$ : cross products $2\times6 = 4\times3 = 12$ — balanced. Both node drops compute to $6\ \mathrm{V}$ on a $10\ \mathrm{V}$ supply; $C_{\text{eq}} = 1.2 + 2.4 = 3.6\ \mu\mathrm{F}$ , matching the merged-node route.*

Go deeper · Derivation B · 1 minThe Infinite Ladder Solved by Self-Similarity

Derivation BThe Infinite Ladder Solved by Self-Similarity

We derive the equivalent capacitance of the infinite ladder whose repeating section is a series arm $C_1$ followed by a parallel rung $C_2$ — the standard JEE-Advanced construction. The strategy is the self-similarity argument: an infinite ladder with one section chopped off the front is still the same infinite ladder, so the unknown $C_{\text{eq}} = x$ must satisfy an equation in which $x$ appears on both sides. The algebra is one quadratic; the physics is the single sentence "infinity minus one is infinity." Watch out for the sign choice at the end — a capacitance must be positive.

Step 1. Call the capacitance of the whole ladder, seen from its input terminals, $x$ .

Step 2. Peel off the first section. What remains beyond it is an identical infinite ladder: capacitance $x$ again. The original ladder is therefore $C_1$ in series with the combination ( $C_2$ in parallel with $x$ ). Why this is rigorous and not a trick: the ladder is defined by its repeating structure, and removing one period leaves a network indistinguishable from the original — the same argument that sums geometric series.

Step 3. Write the self-consistency equation using the two rules:

x = \frac{C_1\,(C_2 + x)}{C_1 + C_2 + x}

Step 4. Cross-multiply and tidy: $x(C_1 + C_2 + x) = C_1C_2 + C_1x$ , so

x^2 + C_2\,x - C_1 C_2 = 0 \qquad\Longrightarrow\qquad x = \frac{-C_2 + \sqrt{C_2^2 + 4C_1C_2}}{2}

where: the quadratic always has one positive and one negative root; keep the positive one — the negative root solves the algebra but not the physics.

Step 5. The all-equal case, $C_1 = C_2 = C$ , gives $x^2 + Cx - C^2 = 0$ , hence $x = C\,\dfrac{\sqrt5 - 1}{2} \approx 0.62\,C$ — the golden ratio, hiding in a capacitor bank. A finite but long ladder converges to this value within a few sections, which is why "infinite" is a fair model. Why $x$ came out smaller than $C$ : every extra section adds series links, and series stacking can only pull capacitance down; the parallel rungs fight back, and the golden ratio is the truce. *Verified: $C_1 = 2\ \mu\mathrm{F}$ , $C_2 = 1\ \mu\mathrm{F}$ : $x^2 + x - 2 = 0$ gives $x = 1\ \mu\mathrm{F}$ exactly. Self-check: $2$ in series with $(1 + 1) = 2$ is $1$ ✓ — the ladder reproduces itself.*

The folding move for symmetric networks

Symmetry is the third unlock, and it needs no new formula — only a principle: nodes that a symmetry of the circuit exchanges must be at the same potential (if swapping them changes nothing, no measurement could tell them apart). Two consequences follow instantly: a capacitor joining twin nodes is idle — erase it; and twin nodes may be soldered into one — merge them, and hidden parallel combinations pop into view. The all-equal bridge is the simplest case: the left–right mirror swaps $a$ and $b$ , idling $C_5$ with no cross products needed, so $C_{\text{eq}} = C$ . The same fold cracks the square-with-diagonals net and the cube-of-capacitors JEE favourite: fold, merge, count.

Limiting cases

case	result	meaning
$N$ equal $C$ in parallel	$NC$	charge storage scales with the bank
$N$ equal $C$ in series	$C/N$	gaps stack; each member takes $V/N$
bridge with $C_1C_4 = C_2C_3$	$C_5$ idle	the diamond is secretly two independent branches
all-equal bridge	$C_{\text{eq}} = C$	symmetry idles the crossbar; each branch is $C/2$
ladder, $C_1 = C_2 = C$	$C(\sqrt5-1)/2 \approx 0.62\,C$	the golden-ratio truce between series and parallel
one series member huge	it drops out ( $V_i \to 0$ )	a big series capacitor is nearly a plain wire
one series member tiny	it dominates: $C_{\text{eq}} \to C_{\text{small}}$ , $V_{\text{small}} \to V$	the runt takes the whole voltage — the breakdown risk

Worked examples

Board level — collapse and unwind (CBSE/ISC pattern, 3 marks)

A $10\ \mu\mathrm{F}$ capacitor is connected in series with a parallel pair of $5\ \mu\mathrm{F}$ and $15\ \mu\mathrm{F}$ . The combination is placed across a $30\ \mathrm{V}$ battery. Find the equivalent capacitance, the charge drawn from the battery, and the charge on each capacitor.

Step 1. Collapse the parallel pair: $C_{p} = 5 + 15 = 20\ \mu\mathrm{F}$ .

Step 2. Collapse the series pair: $C_{\text{eq}} = \dfrac{10 \times 20}{10 + 20} = \boxed{6.67\ \mu\mathrm{F}}$ (i.e. $20/3\ \mu\mathrm{F}$ ).

Step 3. Charge from the battery: $Q = C_{\text{eq}}V = \dfrac{20}{3} \times 30 = \boxed{200\ \mu\mathrm{C}}$ .

Step 4. Unwind the series step — both series members carry the full $200\ \mu\mathrm{C}$ : the $10\ \mu\mathrm{F}$ has $V_1 = 200/10 = 20\ \mathrm{V}$ ; the pair shares $V_p = 200/20 = 10\ \mathrm{V}$ . Audit: $20 + 10 = 30$ ✓.

Step 5. Unwind the parallel step — both members feel $10\ \mathrm{V}$ : $Q_5 = 5 \times 10 = \boxed{50\ \mu\mathrm{C}}$ , $Q_{15} = 15 \times 10 = \boxed{150\ \mu\mathrm{C}}$ . Audit: $50 + 150 = 200\ \mu\mathrm{C}$ ✓.

Board level — series division and the breakdown question (ISC/A-Level pattern)

Capacitors of $2\ \mu\mathrm{F}$ and $6\ \mu\mathrm{F}$ are joined in series across a $24\ \mathrm{V}$ supply. Find the charge on each, the voltage across each, and state which capacitor is closer to breakdown if both are rated $20\ \mathrm{V}$ .

Step 1. Equivalent capacitance: $C_{\text{eq}} = \dfrac{2\times6}{2+6} = 1.5\ \mu\mathrm{F}$ ; charge $Q = 1.5 \times 24 = \boxed{36\ \mu\mathrm{C}}$ — the same $36\ \mu\mathrm{C}$ on each capacitor (series invariant).

Step 2. Voltages by $V = Q/C$ : across the $2\ \mu\mathrm{F}$ : $36/2 = \boxed{18\ \mathrm{V}}$ ; across the $6\ \mu\mathrm{F}$ : $36/6 = \boxed{6\ \mathrm{V}}$ . Audit: $18 + 6 = 24$ ✓. (Cross-over check via equation (1): $V_2 = 24 \times \frac{6}{8} = 18\ \mathrm{V}$ ✓.)

Step 3. The smaller capacitor sits at $18\ \mathrm{V}$ , only $2\ \mathrm{V}$ below its rating — it fails first. Writing this sentence, with the inverse-division reason, is what the final mark is for.

Competitive level — spot the balanced bridge (JEE-Main pattern, ~2 minutes)

In the bridge network, $C_1 = 2\ \mu\mathrm{F}$ and $C_2 = 4\ \mu\mathrm{F}$ run from terminal $A$ to nodes $a$ and $b$ ; $C_3 = 3\ \mu\mathrm{F}$ and $C_4 = 6\ \mu\mathrm{F}$ run from $a$ and $b$ to terminal $B$ ; $C_5 = 5\ \mu\mathrm{F}$ joins $a$ to $b$ . Find the equivalent capacitance between $A$ and $B$ and the charge drawn from a $10\ \mathrm{V}$ source.

Step 1. Test the cross products before touching any algebra: $C_1C_4 = 2\times6 = 12$ and $C_2C_3 = 4\times3 = 12$ . Balanced — the $5\ \mu\mathrm{F}$ is idle. Erase it.

Step 2. Left branch: $2$ and $3$ in series $= \dfrac{2\times3}{5} = 1.2\ \mu\mathrm{F}$ . Right branch: $4$ and $6$ in series $= \dfrac{4\times6}{10} = 2.4\ \mu\mathrm{F}$ .

Step 3. The branches share nodes $A$ and $B$ : parallel. $C_{\text{eq}} = 1.2 + 2.4 = \boxed{3.6\ \mu\mathrm{F}}$ .

Step 4. Charge: $Q = 3.6 \times 10 = \boxed{36\ \mu\mathrm{C}}$ , splitting $12\ \mu\mathrm{C}$ through the left branch and $24\ \mu\mathrm{C}$ through the right. Timing note: the cross-product test is a ten-second gate — run it the instant you see a diamond with a crossbar. If it balances, the problem is two lines; if it does not, expect the question to supply extra symmetry or to be worth leaving for last. Budget ~2 min balanced, and be suspicious of any bridge question in JEE-Main that is not balanced.

Competitive level — the infinite ladder (JEE-Advanced stretch, ~3 minutes)

An infinite ladder is built from repeating sections: each section has a $2\ \mu\mathrm{F}$ capacitor in the series arm followed by a $1\ \mu\mathrm{F}$ capacitor as a rung to the return rail. Find the equivalent capacitance of the ladder.

Step 1. Let the whole ladder equal $x$ . Chop off the first section: the remainder is the same infinite ladder, $x$ again.

Step 2. Structure of the first section: $2\ \mu\mathrm{F}$ in series with ( $1\ \mu\mathrm{F}$ parallel $x$ ):

x = \frac{2\,(1 + x)}{2 + 1 + x}

Step 3. Cross-multiply: $x(3 + x) = 2 + 2x$ , so $x^2 + x - 2 = 0$ , factoring as $(x+2)(x-1) = 0$ .

Step 4. Keep the positive root: $x = \boxed{1\ \mu\mathrm{F}}$ . Self-check: $2$ in series with $(1 + 1) = 2\ \mu\mathrm{F}$ gives $1\ \mu\mathrm{F}$ — feeding the answer back reproduces it ✓. Timing note: the self-similarity equation is always the same shape — series arm in series with (rung parallel $x$ ). Write it directly; deriving it fresh under exam pressure wastes ninety seconds. If the quadratic refuses to factor, use the formula and keep the positive root; a negative or complex "capacitance" means an equation-setup slip, not exotic physics.

Exam traps

The resistor-rule swap. Series capacitances do not add. Sanity gate every combine: series result below the smallest member, parallel above the largest.

Trusting the drawing. Only node connectivity defines series and parallel. Redraw messy circuits with labelled nodes — examiners deliberately sketch parallel pairs to look like chains.

Voltage division the wrong way round. In series the small capacitor takes the large voltage — equation (1) puts the other capacitor's value in the numerator. An answer where the biggest capacitor of a chain carries the biggest voltage is wrong before you check the arithmetic.

Erasing an unbalanced bridge. $C_5$ may be removed only when $C_1C_4 = C_2C_3$ or when symmetry idles it; otherwise you have invented a circuit — unbalanced bridges need Kirchhoff's rules (E3).

Forgetting to unwind. Most marks often sit in the internal charges, not $C_{\text{eq}}$ . Walk backwards — series steps hand back $Q$ , parallel steps hand back $V$ — and run both audits; they are worth their thirty seconds on every board paper.

Keeping the negative root. Ladder quadratics always offer one negative root; capacitance is positive — discard it explicitly (A-Level and JEE-Advanced mark schemes look for the justification).

Unit slips. Work in $\mu\mathrm{F}$ –V– $\mu\mathrm{C}$ throughout; mixing $\mu\mathrm{F}$ with farads mid-solution scatters factors of $10^6$ .

Origins · 1 minCharles Wheatstone and how this was discovered

Charles Wheatstone

1802–1875 · King's College London

Wheatstone was a musical-instrument maker's son who became Victorian Britain's great engineer of measurement — co-inventor of the electric telegraph, inventor of the concertina and the stereoscope. The diamond circuit carrying his name was actually invented by Samuel Christie in 1833; Wheatstone said so plainly, but his 1843 Bakerian Lecture showed what the diamond was for: comparing unknown components by hunting the balance point where the crossbar goes dead. That null idea — measure by making a detector read zero — anchored precision electrical measurement for a century, and it is exactly the condition you exploit when the cross products match. Every student who erases an idle $C_5$ is running Wheatstone's null argument in reverse.

History

1745

— Leyden jars are wired into parallel "batteries" within months of invention — the first capacitor networks, built for bigger sparks

1833

— Samuel Christie presents the diamond bridge circuit to the Royal Society, buried in a paper on the resistance of wires

1843

— Wheatstone's Bakerian Lecture turns Christie's diamond into the standard null-measurement instrument of the telegraph age

1873

— Maxwell's Treatise recasts multi-conductor systems as a matrix of capacitance coefficients — the general theory behind every reduction trick

1960

s — Ladder networks of capacitors and resistors become the standard model of transmission lines and nerve fibres, making the self-similarity argument daily engineering

Going further

Texts. NCERT Physics Class 12, Ch. 2 · HC Verma, Concepts of Physics Vol. 2, Ch. 31 · SS Krotov, Problems in Physics (ladder and symmetry classics) · Cambridge 9702 coursebook, capacitance chapter.

Provenance. All results are standard. A verifier should recompute: Example 1's $C_{\text{eq}} = 20/3\ \mu\mathrm{F}$ , $Q = 200\ \mu\mathrm{C}$ , split $50/150\ \mu\mathrm{C}$ ; Example 2's $36\ \mu\mathrm{C}$ , $18/6\ \mathrm{V}$ ; Example 3's balance $12 = 12$ , $C_{\text{eq}} = 3.6\ \mu\mathrm{F}$ , $Q = 36\ \mu\mathrm{C}$ ; Example 4's quadratic $x^2 + x - 2 = 0 \Rightarrow x = 1\ \mu\mathrm{F}$ ; the golden-ratio ladder $x = C(\sqrt5-1)/2$ .

Prerequisites. Capacitance & the Parallel-Plate Capacitor · Capacitors in Series & Parallel. Leads to. Dielectric Slabs — Battery Connected or Disconnected? · Charge Sharing & the Lost Energy · Kirchhoff's Rules (coming in E3) · Capacitors in Series and Parallel (UG deep dive).

How do I quickly tell series from parallel in a messy diagram?

Interrogate the nodes, not the picture. Label every junction: parallel means both terminals on the same two nodes; series means a shared node touching nothing else. If a pair passes neither test, do not combine them — look for a bridge balance or a symmetry instead. Redrawing the circuit with labelled nodes and straightened wires resolves ninety percent of "impossible" networks before any formula is used.

Why does the smallest capacitor get the biggest voltage in series?

Because series members carry the same charge $Q$ — the stranded conductor between them stays neutral, as you saw in Capacitors in Series & Parallel. Each voltage is then $V_i = Q/C_i$ : same numerator, smallest denominator wins. Picture the same charge crammed into a smaller vessel producing more "pressure." Mismatched series stacks therefore fail at the runt first — and the equal-voltage assumption is the marked trap in rating questions.

Is the cross-product bridge condition something I memorize or something I can rebuild?

Both, ideally. Memorize $C_1C_4 = C_2C_3$ for speed, but rebuild it whenever doubt strikes: assume the crossbar idle, write each branch's node potential with the series division rule, set them equal — one line. If memory and rebuild conflict, trust the rebuild: the common slip is pairing the wrong arms, and the pairing rule is diagonally opposite arms multiply together.

What if the bridge is not balanced?

Then no series/parallel reduction exists — the middle capacitor genuinely carries charge and the branches talk to each other; you need the systematic node-and-loop method (Kirchhoff's rules, arriving in the Current Electricity chapter). At this level an unbalanced bridge in an MCQ usually means re-check the values: JEE-Main examiners almost always set balanced ones, precisely to reward the ten-second test.

Why is removing one section of an infinite ladder legal? Surely the ladder gets shorter.

Infinity is the point: the ladder has no last rung, so "all of it" and "all of it minus one section" are structurally identical — just as the geometric series $1 + r + r^2 + \cdots$ equals $1 + r(\text{itself})$ . For a finite ladder the argument is approximate, but convergence is fast: five or six sections land within a percent of the infinite answer, which is why the model is trusted for long cables.

In the unwinding step, how do I know whether to hand back Q or V?

Replay your collapses in reverse, letting each collapse's invariant travel backwards: a collapsed series chain hands its $Q$ to every member ( $V_i = Q/C_i$ ); a collapsed parallel group hands back its $V$ ( $Q_i = C_iV$ ). A small margin sketch of each reduction stage makes the replay mechanical, and the two audits — path voltages sum to the supply, island charges sum to zero — certify the result.