Series and parallel capacitors — derivations: Worked examples

This page is a drill in coverage. The parent note built the two rules; here we hunt down every kind of question they can produce — including the weird edge cases (a zero-gap capacitor, an infinitely large one, an open branch) that trip people in exams.

Before we compute anything, we lay out a map of all the cases. Then every worked example is tagged with the map-cell it covers, so by the end you have hit each cell at least once.

The scenario matrix

Each row is a class of situation this topic can throw at you. The last column names the example that covers it.

#	Case class	What makes it tricky	Covered by
A	Pure parallel — find total & split charges	which capacitor grabs more charge	Ex 1
B	Pure series — find charge & split voltages	which capacitor takes more voltage	Ex 2
C	Mixed (series + parallel) reduction	reduce innermost group first	Ex 3
D	Energy across a combination	energy is not shared like charge	Ex 4
E	Degenerate: short-circuit plate ( $C \to \infty$ )	a huge cap: wire in series, short in parallel	Ex 5
F	Degenerate: open gap ( $C \to 0$ )	a tiny cap: break in series, dead branch in parallel	Ex 5
G	Limiting: equal capacitors, $n$ of them	series/parallel scaling with $n$	Ex 6
H	Real-world word problem	translate words → circuit	Ex 7
I	Exam twist — charge redistribution after reconnecting	charge conservation, new common $V$	Ex 8
J	Dielectric slab changes one $C$	recompute the network	Ex 9

Cells D, E, F, I are the ones students skip — we give them full attention.

Example 1 — Cell A: pure parallel, charge split

Worked example Two capacitors side by side

$C_{1} = 3 μ F$ , $C_{2} = 6 μ F$ in parallel across $V = 10 V$ . Find $C_{e q}$ , the total charge, and each capacitor's charge.

Forecast: Same voltage on both. Does the bigger or smaller capacitor grab more charge? Guess a number for $Q_{t o t a l}$ before reading on.

Step 1 — Combine. $C_{e q} = C_{1} + C_{2} = 3 + 6 = 9 μ F$ . Why this step? Parallel means same nodes → capacitances add directly.

Step 2 — Total charge. $Q_{t o t a l} = C_{e q} V = 9 \times 10 = 90 μ C$ . Why this step? The equivalent capacitor draws the same total charge from the battery, by its definition.

Step 3 — Split. Same $V = 10 V$ on each, so $Q_{1} = C_{1} V = 30 μ C$ , $Q_{2} = C_{2} V = 60 μ C$ . Why this step? Each branch independently obeys $Q = C V$ with the shared voltage.

Verify: $Q_{1} + Q_{2} = 30 + 60 = 90 = Q_{t o t a l}$ ✓. Bigger cap ( $C_{2}$ ) grabbed more charge: $Q \propto C$ when $V$ is common. Units: $μ F \times V = μ C$ ✓.

Figure below — read it like this: the two branches hang between the same pair of rails (that's what "parallel" means visually), so both feel the full $10 V$ . Follow the pink arrow: the wider capacitor $C_{2}$ collects the larger charge ( $60 μ C$ ), confirming $Q \propto C$ at fixed voltage.

Figure — Series and parallel capacitors — derivations

Example 2 — Cell B: pure series, voltage split

Worked example Two capacitors in a chain

$C_{1} = 2 μ F$ , $C_{2} = 4 μ F$ in series across $V = 12 V$ . Find $C_{e q}$ , the common charge, and each voltage.

Forecast: Same charge on both. Which one takes the bigger voltage — the small or the large capacitor? Commit to an answer now.

Step 1 — Combine. $\frac{1}{C _{e q}} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \Rightarrow C_{e q} = \frac{4}{3} \approx 1.33 μ F$ . Why this step? Series → reciprocals add (voltages add, charge is common).

Step 2 — Common charge. $Q = C_{e q} V = \frac{4}{3} \times 12 = 16 μ C$ — on both plates. Why this step? The trapped middle island forces identical $Q$ everywhere; the equivalent capacitor gives that $Q$ .

Step 3 — Split voltages. $V_{1} = \frac{Q}{C _{1}} = \frac{16}{2} = 8 V$ , $V_{2} = \frac{16}{4} = 4 V$ . Why this step? Apply $V = Q / C$ per capacitor with the common $Q$ .

Verify: $V_{1} + V_{2} = 8 + 4 = 12 = V$ ✓ (KVL around the loop). The smaller cap $C_{1}$ took the larger voltage: $V \propto 1/ C$ .

Figure below — read it like this: the yellow arrow points at the "isolated island" between the two caps — that trapped conductor is why the same $Q = 16 μ C$ marches through both. Then look at the blue callout: the narrower cap $C_{1}$ shows the bigger voltage label ( $8 V$ ), the geometric proof of $V \propto 1/ C$ .

Example 3 — Cell C: mixed reduction

Worked example Series-with-a-parallel-block

$C_{1} = 2 μ F$ in series with the parallel pair $C_{2} = 2 μ F$ , $C_{3} = 4 μ F$ . Battery $= 18 V$ . Find every charge and voltage.

Forecast: Which do you reduce first — the series part or the parallel block? And will $C_{2}$ or $C_{3}$ hold more charge inside the block?

Step 1 — Inner parallel. $C_{23} = C_{2} + C_{3} = 2 + 4 = 6 μ F$ . Why this step? Always collapse the innermost group first so the remaining network is simple series/parallel.

Step 2 — Outer series. $\frac{1}{C _{e q}} = \frac{1}{C _{1}} + \frac{1}{C _{23}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6} \Rightarrow C_{e q} = 1.5 μ F$ . Why this step? $C_{1}$ and the block $C_{23}$ now sit in a single series chain.

Step 3 — Series charge. $Q = C_{e q} V = 1.5 \times 18 = 27 μ C$ , sitting on $C_{1}$ and on the whole $C_{23}$ block. Why this step? Series → same charge through $C_{1}$ and the block.

Step 4 — Split the series voltages. $V_{1} = \frac{27}{2} = 13.5 V$ , $V_{23} = \frac{27}{6} = 4.5 V$ . Why this step? $V = Q / C$ on each series member. Check: $13.5 + 4.5 = 18$ ✓.

Step 5 — Inside the block. Same $V_{23} = 4.5 V$ on $C_{2}$ and $C_{3}$ : $Q_{2} = 2 \times 4.5 = 9 μ C$ , $Q_{3} = 4 \times 4.5 = 18 μ C$ . Why this step? Parallel → shared voltage; each obeys $Q = C V$ .

Verify: $Q_{2} + Q_{3} = 9 + 18 = 27 = Q$ (the block's total charge equals the series charge) ✓.

Figure below — read it like this: the boxed region on the right is the collapsed block $C_{23} = 6 μ F$ ; treat it as a single capacitor in series with $C_{1}$ . Notice the $27 μ C$ label flows through $C_{1}$ and into the box unchanged (series = same charge), then splits inside the box between $C_{2}$ and $C_{3}$ which both carry the same $4.5 V$ .

Example 4 — Cell D: energy across a combination

Worked example Energy is NOT split like charge

Same as Ex 2: $C_{1} = 2 μ F$ , $C_{2} = 4 μ F$ in series across $12 V$ . Find the energy stored in each and in total.

Forecast: Which capacitor stores more energy — the one with the bigger voltage or the bigger capacitance?

Step 1 — Recall the energy formula. Energy $U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$ . Why this step? We need a stored-energy tool; any of these three forms works. Since $Q$ is common in series, $U = \frac{Q ^{2}}{2 C}$ is cleanest.

Step 2 — Per capacitor (using $Q = 16 μ C$ from Ex 2): $U_{1} = \frac{Q ^{2}}{2 C _{1}} = \frac{( 16 ) ^{2}}{2 \cdot 2} = 64 μ J, U_{2} = \frac{( 16 ) ^{2}}{2 \cdot 4} = 32 μ J .$ Why this step? Same $Q$ , so $U \propto 1/ C$ : the smaller cap stores more energy.

Step 3 — Total. $U_{t o t} = 64 + 32 = 96 μ J$ . Why this step? Energies of series parts add.

Verify: Compute from the equivalent capacitor: $U = \frac{1}{2} C_{e q} V^{2} = \frac{1}{2} \cdot \frac{4}{3} \cdot 1 2^{2} = 96 μ J$ ✓. (Units: $μ F \cdot V^{2} = μ J$ .) The smaller cap $C_{1}$ stores more energy — mirroring its larger voltage.

Example 5 — Cells E & F: degenerate capacitors (series AND parallel)

Before the numbers, a quick word on the notation of extremes, since we are about to use it heavily.

A capacitor with plates touching (zero gap) has $C \to \infty$ ; one with an infinite gap / broken plate has $C \to 0^{+}$ . We must show what each does in both wiring styles — series behaves opposite to parallel.

Worked example The two extreme members, tested in series

A chain: $C_{1} = 2 μ F$ in series with $C_{2}$ . (E) Let $C_{2} \to \infty$ (a shorted capacitor). (F) Let $C_{2} \to 0^{+}$ (an open gap). Find $C_{e q}$ each case.

Forecast: A huge capacitor barely resists charge — does it act like a wire or a break? And a zero capacitor?

Step 1 — Case E, $C_{2} \to \infty$ . $\frac{1}{C _{e q}} = \frac{1}{2} + \frac{1}{\infty} \to \frac{1}{2} + 0 = \frac{1}{2}$ , so $C_{e q} \to 2 μ F = C_{1}$ . Why this step? $1/\infty \to 0$ : a giant series capacitor contributes no "difficulty", i.e. it behaves like a plain wire. The network is just $C_{1}$ .

Step 2 — Case F, $C_{2} \to 0^{+}$ . $\frac{1}{C _{e q}} = \frac{1}{2} + \frac{1}{0 ^{+}} \to \infty$ , so $C_{e q} \to 0$ . Why this step? A vanishing series capacitor is an open gap — no charge can get through the chain, so the whole series combo stores nothing.

Verify (sanity): In series, $C_{e q}$ is always below the smallest member. Case E: smallest finite member is $2$ , and adding a near-infinite one leaves us essentially at $2$ ✓. Case F: the smallest member is $\approx 0$ , and indeed $C_{e q} \to 0$ ✓.

Worked example The SAME extremes, now in parallel

Now put $C_{1} = 2 μ F$ in parallel with $C_{2}$ . (E) $C_{2} \to \infty$ . (F) $C_{2} \to 0^{+}$ . Find $C_{e q}$ each case.

Forecast: Parallel adds capacitances directly. Predict which extreme dominates and which vanishes — the roles should swap versus the series case.

Step 1 — Case E, $C_{2} \to \infty$ (parallel). $C_{e q} = C_{1} + C_{2} = 2 + \infty \to \infty$ . Why this step? Parallel adds. An infinite capacitor placed across the two nodes can absorb any charge with no voltage drop across itself: since $V = Q / C$ and $C \to \infty$ , its own $V \to 0$ for any finite $Q$ . So it behaves like a zero-voltage-drop link (a short) between those two nodes, and the pair's capacitance blows up. Note the precise statement: it enforces zero voltage across that branch, not "any voltage it likes". Contrast the series case, where the same $C_{2} \to \infty$ was a harmless wire.

Step 2 — Case F, $C_{2} \to 0^{+}$ (parallel). $C_{e q} = 2 + 0 = 2 μ F = C_{1}$ . Why this step? A vanishing parallel branch stores no charge; it's a dead / open branch you can erase. The network is just $C_{1}$ . Contrast the series case, where the same $C_{2} \to 0^{+}$ killed the whole chain.

Verify (sanity): In parallel, $C_{e q}$ is always above the largest member. Case E: adding a near-infinite one drives $C_{e q} \to \infty$ ✓ (above everything). Case F: adding a $\approx 0$ member leaves the largest, $2 μ F$ ✓.

Recall The four degenerate corners in one table

Series, $C \to \infty$ ::: acts like a plain wire; $C_{e q}$ unchanged by it. Series, $C \to 0^{+}$ ::: acts like an open break; kills the whole chain ( $C_{e q} \to 0$ ). Parallel, $C \to \infty$ ::: enforces zero voltage across that branch (a short); $C_{e q} \to \infty$ . Parallel, $C \to 0^{+}$ ::: acts like a dead/open branch; delete it, $C_{e q}$ unchanged.

Example 6 — Cell G: $n$ identical capacitors (limiting behaviour)

Example 7 — Cell H: real-world word problem

Worked example Camera flash bank

A camera flash needs a $12 μ F$ storage capacitor charged to $300 V$ , but you only have $4 μ F$ units rated to $300 V$ . How many, and wired how, reach $12 μ F$ ? How much energy is stored?

Forecast: To increase total capacitance you go parallel or series? How many $4 μ F$ units make $12 μ F$ ?

Step 1 — Choose topology. Need a bigger capacitance → parallel (parallel adds). Why this step? Series would shrink capacitance; parallel grows it. Each unit also then sees the full $300 V$ , matching its rating.

Step 2 — Count. $n \times 4 = 12 \Rightarrow n = 3$ units in parallel. Why this step? Parallel of $n$ equal caps is $n C$ (Ex 6).

Step 3 — Energy. $U = \frac{1}{2} C_{e q} V^{2} = \frac{1}{2} \times (12 \times 1 0^{- 6}) \times 30 0^{2} = 0.54 J$ . Why this step? Total stored energy uses the equivalent capacitance at the applied voltage.

Verify: Each unit stores $\frac{1}{2} \times (4 \times 1 0^{- 6}) \times 30 0^{2} = 0.18 J$ ; three of them $3 \times 0.18 = 0.54 J$ ✓. Voltage rating respected ( $300 V$ each) ✓.

Example 8 — Cell I: charge redistribution (exam twist)

The classic trap: charge a capacitor, disconnect the battery, then wire it to an uncharged capacitor. There is no battery to fix the voltage now — instead charge is conserved and the two find a new common voltage.

Worked example Sharing charge between two capacitors

$C_{1} = 2 μ F$ charged to $V_{0} = 10 V$ . Battery removed. Then $C_{1}$ is connected in parallel to an uncharged $C_{2} = 3 μ F$ . Find the final voltage and final charges.

Forecast: Total charge can't change. The voltage will settle to some value between $0$ and $10 V$ — guess above or below $5 V$ ?

Step 1 — Initial charge. $Q_{0} = C_{1} V_{0} = 2 \times 10 = 20 μ C$ ; $C_{2}$ starts with $0$ . Why this step? Fix the conserved quantity: total charge $= 20 μ C$ before and after.

Step 2 — After connecting, common voltage. In parallel they share one voltage $V_{f}$ . Total charge splits as $Q_{1}^{'} = C_{1} V_{f}$ , $Q_{2}^{'} = C_{2} V_{f}$ , and $Q_{1}^{'} + Q_{2}^{'} = Q_{0}$ : $(C_{1} + C_{2}) V_{f} = Q_{0} \Rightarrow V_{f} = \frac{20}{2 + 3} = 4 V .$ Why this step? No battery → charge conservation replaces "voltage fixed by battery". The parallel connection forces a shared $V_{f}$ .

Step 3 — Final charges. $Q_{1}^{'} = 2 \times 4 = 8 μ C$ , $Q_{2}^{'} = 3 \times 4 = 12 μ C$ . Why this step? Each obeys $Q = C V_{f}$ at the new common voltage.

Verify: $Q_{1}^{'} + Q_{2}^{'} = 8 + 12 = 20 = Q_{0}$ ✓ (charge conserved). $V_{f} = 4 V < 5 V$ because the bigger uncharged cap pulls the voltage down more. Energy check (bonus): initial $U_{i} = \frac{1}{2} \cdot 2 \cdot 1 0^{2} = 100 μ J$ ; final $U_{f} = \frac{1}{2} (2 + 3) \cdot 4^{2} = 40 μ J$ . Energy dropped — some is lost as heat/radiation while charge flows. Charge is conserved, energy is not.

Example 9 — Cell J: dielectric changes one capacitor

Inserting a dielectric of constant $κ$ (the dielectric constant, a pure number $\geq 1$ telling how much better an insulator stores charge than vacuum) multiplies that capacitor's value: $C \to κ C$ . Then re-solve the network.

Worked example Slab in a series pair

$C_{1} = 2 μ F$ series with $C_{2} = 2 μ F$ across $12 V$ . A dielectric $κ = 3$ fills $C_{2}$ , making it $C_{2}^{'} = 6 μ F$ . Find the new $C_{e q}$ and the new voltage split.

Forecast: Filling $C_{2}$ makes it bigger. In series, a bigger cap takes less voltage — so does $V_{2}$ rise or fall?

Step 1 — New value. $C_{2}^{'} = κ C_{2} = 3 \times 2 = 6 μ F$ . Why this step? A dielectric of constant $κ$ multiplies the capacitance.

Step 2 — Recombine. $\frac{1}{C _{e q}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6} \Rightarrow C_{e q} = 1.5 μ F$ . Why this step? Still series; just re-add reciprocals with the new $C_{2}^{'}$ .

Step 3 — Charge & voltages. $Q = 1.5 \times 12 = 18 μ C$ ; $V_{1} = \frac{18}{2} = 9 V$ , $V_{2} = \frac{18}{6} = 3 V$ . Why this step? Series common charge, then $V = Q / C$ per cap.

Verify: $V_{1} + V_{2} = 9 + 3 = 12$ ✓. Before the slab ( $C_{2} = 2 μ F$ ) the split was $6 V$ each; after inserting it, $C_{2}$ (now bigger) drops to $3 V$ and $C_{1}$ rises to $9 V$ — consistent with $V \propto 1/ C$ ✓. Units: $μ C / μ F = V$ ✓.

Active recall

Recall One-liners for each cell

Parallel: bigger cap gets more...? ::: charge (same $V$ , $Q \propto C$ ). Series: bigger cap gets more...? ::: less voltage ( $V \propto 1/ C$ ); charge is equal. A series capacitor with $C \to \infty$ behaves like a...? ::: plain wire ( $1/ C \to 0$ ). A series capacitor with $C \to 0^{+}$ behaves like a...? ::: open break (no charge passes). A parallel capacitor with $C \to \infty$ enforces...? ::: zero voltage across that branch (a short); $C_{e q} \to \infty$ . A parallel capacitor with $C \to 0^{+}$ behaves like a...? ::: dead/open branch you can delete. $n$ equal caps $C$ in series give $C_{e q} =$ ? ::: $C / n$ . After disconnecting the battery and reconnecting caps, what is conserved? ::: charge (not voltage, not energy). A dielectric $κ$ changes a capacitor's value how? ::: multiplies it: $C \to κ C$ .