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Capacitors in series and parallel

Covers AS91526

Capacitors in parallel

$C_T = C_1 + C_2 + \dots$

Capacitances in parallel simply add.
Why: every capacitor has the same voltage across it, and the total charge stored is the sum of the individual charges. Since $Q_T = Q_1 + Q_2 = C_1V + C_2V = (C_1 + C_2)V$ , the effective capacitance is the sum.
Physically, connecting capacitors in parallel is equivalent to increasing the plate area, which increases capacitance.
The total is always larger than the largest individual capacitance.

In a parallel combination:

voltage is the same across each capacitor,
charge divides in proportion to capacitance: the largest capacitor holds the most charge.

Capacitors in series

$\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \dots$

Capacitances in series combine as reciprocals.
Why: the same charge appears on every capacitor (the charge pushed onto one plate of the first induces an equal charge along the chain), while the voltages add. Since $V_T = V_1 + V_2 = \dfrac{Q}{C_1} + \dfrac{Q}{C_2}$ , dividing through by $Q$ gives the reciprocal rule.
Physically, series connection is equivalent to increasing the plate separation, which decreases capacitance.
The total is always smaller than the smallest individual capacitance.

In a series combination:

charge is the same on each capacitor,
voltage divides inversely with capacitance: the smallest capacitor takes the largest share of the voltage.

The rules are the opposite way round from resistors

This is the trap, and it is worth stating explicitly:

	Resistors	Capacitors
Series	$R_T = R_1 + R_2$	$\dfrac{1}{C_T} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$
Parallel	$\dfrac{1}{R_T} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$	$C_T = C_1 + C_2$
Same in series	current	charge
Same in parallel	voltage	voltage

The reason for the reversal: resistance opposes flow so obstacles add in series, while capacitance measures capacity to store, and putting capacitors in series effectively lengthens the gap the charge must bridge.

Two useful shortcuts

Two capacitors in series: $C_T = \dfrac{C_1C_2}{C_1 + C_2}$ .
$n$ identical capacitors of value $C$ : $C_T = nC$ in parallel, $C_T = \dfrac{C}{n}$ in series.

Worked ExampleA parallel combination

Capacitors of $2.0$ µF, $3.0$ µF and $5.0$ µF are connected in parallel across a $12$ V supply. Find the total capacitance, the charge on each capacitor, and the total charge stored.

Step 1 — Total capacitance

$C_T = 2.0 + 3.0 + 5.0 = 10.0 \text{ µF} = 1.00 \times 10^{-5} \text{ F}$

Step 2 — Charge on each capacitor

Each has the full $12$ V across it:

$Q_1 = C_1V = (2.0 \times 10^{-6})(12) = 2.4 \times 10^{-5} \text{ C}$ $Q_2 = (3.0 \times 10^{-6})(12) = 3.6 \times 10^{-5} \text{ C}$ $Q_3 = (5.0 \times 10^{-6})(12) = 6.0 \times 10^{-5} \text{ C}$

Step 3 — Total charge

$Q_T = (2.4 + 3.6 + 6.0) \times 10^{-5} = 12.0 \times 10^{-5} = 1.20 \times 10^{-4} \text{ C}$

(Check: $Q_T = C_TV = (1.00 \times 10^{-5})(12) = 1.20 \times 10^{-4}$ C ✓)

$C_T = 10.0 \text{ µF}, \qquad Q_T = 1.20 \times 10^{-4} \text{ C}$

Worked ExampleA series combination

Capacitors of $6.0$ µF and $3.0$ µF are connected in series across a $9.0$ V supply. Find the total capacitance, the charge on each capacitor, and the voltage across each.

Step 1 — Total capacitance

$\frac{1}{C_T} = \frac{1}{6.0} + \frac{1}{3.0} = 0.1667 + 0.3333 = 0.5000 \text{ (µF)}^{-1}$ $C_T = \frac{1}{0.5000} = 2.0 \text{ µF}$

This is smaller than the smaller capacitor ( $3.0$ µF) — the check passes.

Step 2 — Charge (the same on both)

$Q = C_TV = (2.0 \times 10^{-6})(9.0) = 1.8 \times 10^{-5} \text{ C}$

Both capacitors carry this same charge.

Step 3 — Voltage across each

$V_1 = \frac{Q}{C_1} = \frac{1.8 \times 10^{-5}}{6.0 \times 10^{-6}} = 3.0 \text{ V}$ $V_2 = \frac{Q}{C_2} = \frac{1.8 \times 10^{-5}}{3.0 \times 10^{-6}} = 6.0 \text{ V}$

(Check: $3.0 + 6.0 = 9.0$ V ✓)

$C_T = 2.0 \text{ µF}, \quad Q = 1.8 \times 10^{-5} \text{ C}, \quad V_1 = 3.0 \text{ V}, \quad V_2 = 6.0 \text{ V}$

Capacitors in series and parallel

Covers AS91526

Capacitors in parallel

$C_T = C_1 + C_2 + \dots$

Capacitances in parallel simply add.
Why: every capacitor has the same voltage across it, and the total charge stored is the sum of the individual charges. Since $Q_T = Q_1 + Q_2 = C_1V + C_2V = (C_1 + C_2)V$ , the effective capacitance is the sum.
Physically, connecting capacitors in parallel is equivalent to increasing the plate area, which increases capacitance.
The total is always larger than the largest individual capacitance.

In a parallel combination:

voltage is the same across each capacitor,
charge divides in proportion to capacitance: the largest capacitor holds the most charge.

Capacitors in series

$\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2} + \dots$

Capacitances in series combine as reciprocals.
Why: the same charge appears on every capacitor (the charge pushed onto one plate of the first induces an equal charge along the chain), while the voltages add. Since $V_T = V_1 + V_2 = \dfrac{Q}{C_1} + \dfrac{Q}{C_2}$ , dividing through by $Q$ gives the reciprocal rule.
Physically, series connection is equivalent to increasing the plate separation, which decreases capacitance.
The total is always smaller than the smallest individual capacitance.

In a series combination:

charge is the same on each capacitor,
voltage divides inversely with capacitance: the smallest capacitor takes the largest share of the voltage.

The rules are the opposite way round from resistors

This is the trap, and it is worth stating explicitly:

	Resistors	Capacitors
Series	$R_T = R_1 + R_2$	$\dfrac{1}{C_T} = \dfrac{1}{C_1} + \dfrac{1}{C_2}$
Parallel	$\dfrac{1}{R_T} = \dfrac{1}{R_1} + \dfrac{1}{R_2}$	$C_T = C_1 + C_2$
Same in series	current	charge
Same in parallel	voltage	voltage

The reason for the reversal: resistance opposes flow so obstacles add in series, while capacitance measures capacity to store, and putting capacitors in series effectively lengthens the gap the charge must bridge.

Two useful shortcuts

Two capacitors in series: $C_T = \dfrac{C_1C_2}{C_1 + C_2}$ .
$n$ identical capacitors of value $C$ : $C_T = nC$ in parallel, $C_T = \dfrac{C}{n}$ in series.

Worked ExampleA parallel combination

Capacitors of $2.0$ µF, $3.0$ µF and $5.0$ µF are connected in parallel across a $12$ V supply. Find the total capacitance, the charge on each capacitor, and the total charge stored.

Step 1 — Total capacitance

$C_T = 2.0 + 3.0 + 5.0 = 10.0 \text{ µF} = 1.00 \times 10^{-5} \text{ F}$

Step 2 — Charge on each capacitor

Each has the full $12$ V across it:

$Q_1 = C_1V = (2.0 \times 10^{-6})(12) = 2.4 \times 10^{-5} \text{ C}$ $Q_2 = (3.0 \times 10^{-6})(12) = 3.6 \times 10^{-5} \text{ C}$ $Q_3 = (5.0 \times 10^{-6})(12) = 6.0 \times 10^{-5} \text{ C}$

Step 3 — Total charge

$Q_T = (2.4 + 3.6 + 6.0) \times 10^{-5} = 12.0 \times 10^{-5} = 1.20 \times 10^{-4} \text{ C}$

(Check: $Q_T = C_TV = (1.00 \times 10^{-5})(12) = 1.20 \times 10^{-4}$ C ✓)

$C_T = 10.0 \text{ µF}, \qquad Q_T = 1.20 \times 10^{-4} \text{ C}$

Worked ExampleA series combination

Capacitors of $6.0$ µF and $3.0$ µF are connected in series across a $9.0$ V supply. Find the total capacitance, the charge on each capacitor, and the voltage across each.

Step 1 — Total capacitance

$\frac{1}{C_T} = \frac{1}{6.0} + \frac{1}{3.0} = 0.1667 + 0.3333 = 0.5000 \text{ (µF)}^{-1}$ $C_T = \frac{1}{0.5000} = 2.0 \text{ µF}$

This is smaller than the smaller capacitor ( $3.0$ µF) — the check passes.

Step 2 — Charge (the same on both)

$Q = C_TV = (2.0 \times 10^{-6})(9.0) = 1.8 \times 10^{-5} \text{ C}$

Both capacitors carry this same charge.

Step 3 — Voltage across each

$V_1 = \frac{Q}{C_1} = \frac{1.8 \times 10^{-5}}{6.0 \times 10^{-6}} = 3.0 \text{ V}$ $V_2 = \frac{Q}{C_2} = \frac{1.8 \times 10^{-5}}{3.0 \times 10^{-6}} = 6.0 \text{ V}$

(Check: $3.0 + 6.0 = 9.0$ V ✓)

$C_T = 2.0 \text{ µF}, \quad Q = 1.8 \times 10^{-5} \text{ C}, \quad V_1 = 3.0 \text{ V}, \quad V_2 = 6.0 \text{ V}$