19.4.1 Series Combination of Capacitors

When capacitors are connected end-to-end in a single path, they are said to be in series. The key characteristic of a series circuit is that there is only one path for charge to flow.

Key Property: Same Charge on Each Capacitor

When a battery is connected to a series combination, it pushes charge $Q$ onto the outer plates. Due to electrostatic induction, the inner plates acquire equal and opposite charges. Because the inner plates are isolated (not connected to the battery), the charge redistributed on each capacitor must be equal:

$Q_1=Q_2=Q_3=\cdots =Q_n=Q$

This is the defining property of a series combination.

Voltage Distribution

The total voltage $V$ supplied by the battery is shared among the capacitors. Using $V=Q/C$ for each:

$V=V_1+V_2+V_3+\cdots +V_n$

$V=\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}+\cdots +\frac{Q}{C_n}$

Derivation of Equivalent Capacitance

The equivalent capacitance $C_{eq}$ is defined as the single capacitor that stores the same charge $Q$ under the same total voltage $V$:

$V=\frac{Q}{C_{eq}}$

Substituting:

$\frac{Q}{C_{eq}}=\frac{Q}{C_1}+\frac{Q}{C_2}+\frac{Q}{C_3}+\cdots +\frac{Q}{C_n}$

Dividing both sides by $Q$:

$\boxed{\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\cdots +\frac{1}{C_n}}$

Important Consequence

The equivalent capacitance of a series combination is always less than the smallest individual capacitance. This is because series connection effectively increases the total plate separation, reducing capacitance.

Physical interpretation: Connecting capacitors in series is equivalent to increasing the distance $d$ between the outermost plates, since $C={\epsilon }_{0}A/d$ and a larger $d$ gives a smaller $C$.

Worked Example

Problem: Three capacitors $C_1=2\mu F$, $C_2=4\mu F$, and $C_3=6\mu F$ are connected in series to a $12\text{V}$ battery. Find (a) the equivalent capacitance, (b) the charge on each capacitor, and (c) the voltage across each.

Solution:

(a) Equivalent Capacitance:

$\frac{1}{C_{eq}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}=\frac{6}{12}+\frac{3}{12}+\frac{2}{12}=\frac{11}{12}$

$C_{eq}=\frac{12}{11}\approx 1.09\mu F$

(b) Charge on each capacitor:

$Q=C_{eq}\times V=\frac{12}{11}\times 12\approx 13.1\mu C$

All three capacitors carry the same charge $Q\approx 13.1\mu C$.

(c) Voltage across each:

$V_1=\frac{Q}{C_1}=\frac{13.1}{2}\approx 6.5\text{V},V_2=\frac{13.1}{4}\approx 3.3\text{V},V_3=\frac{13.1}{6}\approx 2.2\text{V}$

Check: $V_1+V_2+V_3\approx 6.5+3.3+2.2=12\text{V}$ ✓

Special Case: Two Capacitors in Series

For two capacitors, the formula simplifies to:

$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$

The voltage divides inversely with capacitance:

$V_1=V\cdot \frac{C_2}{C_1+C_2},V_2=V\cdot \frac{C_1}{C_1+C_2}$

Summary Table