Series and parallel capacitors — derivations

WHAT we are doing

We replace a network of capacitors by a single equivalent capacitor $C_{eq}$ that, when connected to the same battery, draws the same total charge at the same total voltage. That is the definition of equivalent: $C_{eq} \equiv \frac{Q_{total}}{V_{total}}$

Parallel — derivation from scratch

HOW (step by step):

Same voltage on each: $V_1=V_2=V_3=V$ . Why this step? They share the same nodes → equipotential wires.
Charge on each: $Q_1=C_1V,\ Q_2=C_2V,\ Q_3=C_3V$ . Why this step? Apply $Q=CV$ to each capacitor individually.
Battery supplies total charge = sum delivered to all branches: $Q = Q_1+Q_2+Q_3$ Why this step? Charge conservation at node A: the charge that left the battery splits among branches.
Substitute: $Q = C_1V + C_2V + C_3V = (C_1+C_2+C_3)V$
Compare to $Q=C_{eq}V$ :

Series — derivation from scratch

HOW (step by step):

Same charge on each: $Q_1=Q_2=Q_3=Q$ . Why this step? Charge induction on the isolated middle conductors (must stay net-neutral).
Voltage across each: $V_1=\dfrac{Q}{C_1},\ V_2=\dfrac{Q}{C_2},\ V_3=\dfrac{Q}{C_3}$ . Why this step? Rearrange $Q=CV$ for each.
Battery voltage = sum of drops (Kirchhoff's voltage law going around the loop): $V = V_1+V_2+V_3$ Why this step? Potential drops in series add to the total applied PD.
Substitute: $V = \frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}=Q\left(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right)$
But $V=\dfrac{Q}{C_{eq}}$ , so divide both sides by $Q$ :

Figure — Series and parallel capacitors — derivations

Forecast-then-Verify

Worked examples

Worked example Example 1 — Series, find charge and individual voltages

$C_1=2\ \mu F$ , $C_2=4\ \mu F$ in series across $V=12\ V$ .

Step 1: $\dfrac{1}{C_{eq}}=\dfrac12+\dfrac14=\dfrac34 \Rightarrow C_{eq}=\dfrac43\ \mu F\approx1.33\ \mu F$ . Why? Series → reciprocals add.

Step 2: $Q=C_{eq}V=\dfrac43\times12=16\ \mu C$ . This is the charge on both capacitors. Why? Series → same $Q$ everywhere; use the equivalent capacitor to get it.

Step 3: $V_1=\dfrac{Q}{C_1}=\dfrac{16}{2}=8\ V$ , $V_2=\dfrac{16}{4}=4\ V$ . Why? Apply $V=Q/C$ to each. Check: $8+4=12\ V$ ✓.

Note: The smaller capacitor takes the larger voltage. $V\propto 1/C$ .

Worked example Example 3 — Mixed network

$C_1=2\ \mu F$ in series with the parallel combo of $C_2=2\ \mu F$ and $C_3=4\ \mu F$ . Battery $=18\ V$ .

Step 1 (inner parallel): $C_{23}=2+4=6\ \mu F$ . Why? Reduce innermost group first.

Step 2 (series): $\dfrac{1}{C_{eq}}=\dfrac12+\dfrac16=\dfrac{4}{6}\Rightarrow C_{eq}=1.5\ \mu F$ . Why? $C_1$ and $C_{23}$ are now in series.

Step 3: Total charge $Q=1.5\times18=27\ \mu C$ — sits on $C_1$ and on the $C_{23}$ block (series). So $V_1=\dfrac{27}{2}=13.5\ V$ , $V_{23}=\dfrac{27}{6}=4.5\ V$ . Check: $13.5+4.5=18$ ✓. Both $C_2,C_3$ share $V_{23}=4.5\,V$ (they're parallel): $Q_2=2\times4.5=9\ \mu C$ , $Q_3=4\times4.5=18\ \mu C$ .

Common mistakes (Steel-manned)

Active recall

Recall Try before peeking

State the parallel rule and derive it in two lines.
State the series rule and the physical reason the charges are equal.
In series, which capacitor has the larger voltage drop and why?
Why is series $C_{eq}$ always less than the smallest member?

Recall Feynman: explain to a 12-year-old

Think of capacitors as buckets for electric charge.

Parallel = put buckets side by side under one tap, all filled to the same water level (voltage). More buckets = more total water you can hold → capacity adds up.
Series = stack buckets in a single pipe so the same trickle of water passes through each one. Each narrow bucket makes the water work harder (more pressure drop), so stacking them makes the whole thing harder to fill → total capacity goes down. That's why we add the "difficulties" ( $1/C$ ), not the buckets.

Parallel capacitors share the same...

voltage (potential difference) across each.

Series capacitors share the same...

charge

Q

on each.

Formula for capacitors in parallel

C_{eq}=\sum_i C_i

(they add).

Formula for capacitors in series

\frac{1}{C_{eq}}=\sum_i \frac{1}{C_i}

Two-capacitor series shortcut

C_{eq}=\dfrac{C_1C_2}{C_1+C_2}

(product over sum).

Why is series equivalent smaller than the smallest?

Adding reciprocals only increases

1/C_{eq}

, so

C_{eq}

drops below every member.

In series, which capacitor takes the larger voltage?

The smaller one, since

V=Q/C \propto 1/C

with

Q

common.

In parallel, which capacitor takes the larger charge?

The larger one, since

Q=CV \propto C

with

V

common.

Physical reason series charges are equal

The isolated middle conductors stay net-neutral, inducing the same

Q

on each plate.

Capacitor rules vs resistor rules

They are swapped: capacitors-in-series behave like resistors-in-parallel and vice versa.

Connections

Capacitance and the relation Q = CV — the single equation every step rests on.
Energy stored in a capacitor — $U=\tfrac12 CV^2$ ; redistributes when capacitors are combined.
Resistors in series and parallel — the mirror-image algebra; compare and contrast.
Kirchhoff's voltage law — justifies "voltages add" in the series loop.
Charge conservation — justifies "charges add" at a node in parallel.
Dielectrics in capacitors — how inserting a slab changes individual $C_i$ before combining.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, capacitor ek charge ka "bucket" hai jisme $Q=CV$ chalta hai. Do hi rules yaad rakho, baaki sab khud derive ho jaata hai. Parallel mein dono plates same do wires (nodes) se judi hoti hain, isliye har capacitor par same voltage padta hai. Same V hone se charges add hote hain: $Q=Q_1+Q_2+Q_3=(C_1+C_2+C_3)V$ , matlab $C_{eq}=C_1+C_2+C_3$ . Parallel hamesha sabse bade capacitor se bhi bada banta hai, kyunki tumne plate area badha diya.

Series mein beech wale plates ek isolated island hain jo neutral rehte hain, isliye induction se same charge $Q$ har capacitor se guzarta hai. Ab voltages add hote hain (KVL): $V=\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}$ , jisse $\frac{1}{C_{eq}}=\frac1{C_1}+\frac1{C_2}+\frac1{C_3}$ . Series ka $C_{eq}$ hamesha sabse chhote se bhi chhota hota hai.

Sabse common galti: log resistor ki aadat se capacitor ko series mein add kar dete hain — yeh galat hai! Capacitor ke rules resistor ke ulte hote hain. Aur ek baat: series mein chhota capacitor zyada voltage leta hai ( $V\propto 1/C$ ), jabki parallel mein bada capacitor zyada charge leta hai ( $Q\propto C$ ). Exam mein pehle forecast karo (parallel bada, series chhota), phir calculate karke verify karo — answer galat hua toh turant pakad lega.

Series and parallel capacitors — derivations

WHAT we are doing

Parallel — derivation from scratch

Series — derivation from scratch

Forecast-then-Verify

Worked examples

Common mistakes (Steel-manned)

Active recall

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Electromagnetism

Connections