PhysicsISC Class 12 · Electrostatic Potential, Potential Energy and Capacitance

Capacitors in Series and Parallel: Derivations, Networks and Energy

Updated 15 October 2026 · Fact-checked

In series, capacitors carry the same charge and their p.d.s add, so $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$. In parallel, they share the same p.d. and their charges add, so $C = C_1 + C_2 + C_3$. The energy stored in any capacitor is $\frac{1}{2}CV^2 = \frac{1}{2}QV = \frac{Q^2}{2C}$.

In series the charge Q is the same and the p.d.s add; in parallel the p.d. V is the same and the charges add.

Illustration drawn for QuestionBuddy.

Understand Capacitors in Series and Parallel

Capacitance. A capacitor stores charge $\pm Q$ on two conductors with a p.d. $V$ between them; its capacitance is $C = \frac{Q}{V}$ (unit farad, F = C V⁻¹). For a parallel-plate capacitor, $C = \frac{ε _{0} A}{d}$ , and with a dielectric filling the gap, $K C$ .

Series combination. Capacitors joined end to end, with a single path for charge.
1. The battery moves charge $+ Q$ onto the first plate. It induces $- Q$ on the facing plate, which leaves $+ Q$ on the next capacitor's plate (the connecting wire and plates form an isolated conductor, so its net charge stays zero). So every capacitor in series carries the same charge Q.
2. The p.d.s add: $V = V_{1} + V_{2} + V_{3}$ , with $V_{i} = \frac{Q}{C _{i}}$ .
3. For the single equivalent capacitor, $V = \frac{Q}{C}$ . Hence $\frac{Q}{C} = \frac{Q}{C _{1}} + \frac{Q}{C _{2}} + \frac{Q}{C _{3}}$ , and dividing by Q,
$\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}} .$
The equivalent capacitance is smaller than the smallest one. For two: $C = \frac{C _{1} C _{2}}{C _{1} + C _{2}}$ . The p.d. divides in inverse ratio of capacitance: the smallest capacitor takes the largest share. (This was the 3-mark ISC 2024 derivation; the examiners penalised missing diagrams and the final step written as $C = \dots$ instead of $\frac{1}{C} = \dots$ .)

Parallel combination. Each capacitor connected directly across the same two points.
1. Every capacitor has the same p.d. V.
2. The charges add: $Q = Q_{1} + Q_{2} + Q_{3} = C_{1} V + C_{2} V + C_{3} V$ .
3. With $Q = C V$ for the equivalent capacitor, dividing by V,
$C = C_{1} + C_{2} + C_{3} .$
The equivalent is larger than the largest, and charge divides in direct ratio of capacitance.

Energy stored. The work done in charging a capacitor is stored as electric potential energy (the syllabus asks for the formula only):
$U = \frac{1}{2} C V^{2} = \frac{1}{2} Q V = \frac{Q ^{2}}{2 C},$
and the energy density in the field is $u = \frac{1}{2} ε_{0} E^{2}$ . In any network, the total energy equals the sum of the energies of the individual capacitors, and also $\frac{1}{2} C_{e q} V^{2}$ .

Networks. Reduce step by step: combine any group that is clearly in parallel (same two end points) or in series (same charge, nothing branching between), redraw, repeat. Then work back: find the total charge $Q = C_{e q} V$ , the p.d. across each block, and the charge on each capacitor.

Charge sharing. When a charged capacitor is joined to another (like plates together), charge flows until both have the same p.d., the common potential:
$V = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}}$ (charge is conserved).
Energy is lost as heat and radiation in the connecting wires:
$Δ U = \frac{C _{1} C _{2} ( V _{1} - V _{2} ) ^{2}}{2 ( C _{1} + C _{2} )} .$
If unlike plates are joined, use $C_{1} V_{1} - C_{2} V_{2}$ in the numerator.

Series or parallel? Series is used to share a high voltage among capacitors of low voltage rating; parallel is used to store more charge at the same voltage.

Capacitors in Series and Parallel: formulas and key facts

Capacitance: $C = \frac{Q}{V}, C_{pp} = \frac{ε _{0} A}{d}$; Unit farad.
Series: $\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$; Same Q; V divides as $\frac{1}{C}$ .
Two in series: $C = \frac{C _{1} C _{2}}{C _{1} + C _{2}}$; Product over sum.
Parallel: $C = C_{1} + C_{2} + C_{3}$; Same V; Q divides as C.
Energy: $U = \frac{1}{2} C V^{2} = \frac{1}{2} Q V = \frac{Q ^{2}}{2 C}$; Formula only in ISC.
Common potential: $V = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}}$; Like plates joined.
Energy lost on sharing: $Δ U = \frac{C _{1} C _{2} ( V _{1} - V _{2} ) ^{2}}{2 ( C _{1} + C _{2} )}$; Always positive.

How to solve Capacitors in Series and Parallel questions

Simplify first, then work backwards for individual charges and voltages.

1Redraw the circuit and label junction points; spot groups that are in parallel (same two ends) or in series (no branch between).
2Replace each group by its equivalent and redraw until one capacitor remains.
3Find the total charge $Q = C_{e q} V$ .
4Work back: in a series chain each element has Q; find each block's p.d. as $\frac{Q}{C _{b l oc k}}$ ; in a parallel block share that p.d.
5Find individual charges $Q_{i} = C_{i} V_{i}$ and energies $\frac{1}{2} C_{i} V_{i}^{2}$ .
6Check: p.d.s around the circuit add to the supply, and charges at a junction balance.

Capacitors in Series and Parallel in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Building a value from identical capacitors

Suppose a technician needs a 3 µF capacitor for a repair but has only 2 µF capacitors.

How can three of them give 3 µF?

Try it first, then show the solutionSolution

Two in series give $\frac{2 \times 2}{2 + 2} = 1$ µF.
Put that pair in parallel with the third: $1 + 2 = 3$ µF.

Answer: Connect two in series and the third in parallel with the pair.

Memory hook: Series makes smaller, parallel makes bigger.

Series to share voltage

Imagine an electronics project using two 10 F supercapacitors, each rated at 2.7 V, connected in series and charged from a 5.4 V supply.

Find the combined capacitance, the voltage on each and the total energy stored.

Try it first, then show the solutionSolution

$C = \frac{10 \times 10}{20} = 5$ F; identical capacitors share the voltage equally: 2.7 V each (within their rating).
$U = \frac{1}{2} \times 5 \times 5. 4^{2} = 72.9$ J.

Answer: 5 F; 2.7 V each; about 73 J.

Memory hook: Series: same charge, voltages add.

Tricks to remember Capacitors in Series and Parallel

Short memory aids. Say each one aloud once, then use it on a question the same day.

Opposite of resistors
Capacitors in parallel add like resistors in series, and vice versa.
Use it when: Remembering both formulas.
Series: same Q; parallel: same V
"Series shares the queue (Q), parallel shares the voltage."
Use it when: Starting any network.
Smallest C, biggest V
In series, the smallest capacitor takes the largest p.d.
Use it when: Checking voltage division and ratings.
Joining always wastes energy
Charge is conserved on sharing; energy is not.
Use it when: Common potential problems.

Capacitors in Series and Parallel: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Series against parallel · 1 mark · MCQBoard level

Two identical capacitors of capacitance C are connected first in series and then in parallel. The ratio $C_{ser i es} : C_{p a r a l l e l}$ is:
(a) 1 : 1
(b) 1 : 2
(c) 1 : 4
(d) 2 : 1

Try it first, then show the solutionSolution

$\frac{C}{2} : 2 C = 1 : 4$ .

Answer: (c).

Example 2: Two in series · 2 marks · Short answerBoard level

Capacitors of 4 µF and 12 µF are connected in series across 24 V. Find the equivalent capacitance, the charge on each and the p.d. across each.

Try it first, then show the solutionSolution

$C = \frac{4 \times 12}{16} = 3$ µF; $Q = 3 \times 24 = 72$ µC on each. (1)
$V_{4} = \frac{72}{4} = 18$ V, $V_{12} = \frac{72}{12} = 6$ V (total 24 V). (1)

Answer: 3 µF; 72 µC each; 18 V and 6 V.

Example 3: Derivation: parallel combination · 3 marks · Long answerBoard level

Three capacitors $C_{1}$ , $C_{2}$ , $C_{3}$ are connected in parallel across a p.d. V. Draw the circuit and derive the equivalent capacitance.

Try it first, then show the solutionSolution

Diagram: each capacitor connected between the same two points A and B, across the source; label $Q_{1}, Q_{2}, Q_{3}$ and V. Each has p.d. V. (1)
Charges: $Q_{1} = C_{1} V$ , $Q_{2} = C_{2} V$ , $Q_{3} = C_{3} V$ ; total charge drawn $Q = Q_{1} + Q_{2} + Q_{3}$ . (1)
For the equivalent capacitor $Q = C V$ : $C V = (C_{1} + C_{2} + C_{3}) V$ , so $C = C_{1} + C_{2} + C_{3}$ . (1)

Answer: $C = C_{1} + C_{2} + C_{3}$ .

Example 4: A mixed network · 3 marks · Short answerHarder board level

A 6 µF capacitor is connected in series with a parallel pair of 2 µF and 4 µF, across a 12 V supply. Find the charge on each capacitor and the total energy stored.

Try it first, then show the solutionSolution

Parallel pair: 6 µF; with 6 µF in series: $C_{e q} = 3$ µF; $Q = 36$ µC. (1)
The 6 µF carries 36 µC (p.d. 6 V); the pair has p.d. 6 V, so $Q_{2} = 12$ µC, $Q_{4} = 24$ µC. (1)
$U = \frac{1}{2} \times 3 \times 1 0^{- 6} \times 144 = 2.16 \times 1 0^{- 4}$ J. (1)

Answer: 36 µC, 12 µC, 24 µC; 216 µJ.

Example 5: Common potential and lost energy · 4 marks · Long answerHOTS / competency

A 5 µF capacitor charged to 60 V is connected (like plates together) to a 10 µF capacitor charged to 30 V. Find the common potential and the energy lost.

Try it first, then show the solutionSolution

Charges: 300 µC and 300 µC; total 600 µC is conserved. (1)
Common potential $V = \frac{600}{15} = 40$ V. (1)
Energy before: $\frac{1}{2} (5) (6 0^{2}) + \frac{1}{2} (10) (3 0^{2})$ µJ $= 9000 + 4500 = 13500$ µJ; after: $\frac{1}{2} (15) (4 0^{2}) = 12000$ µJ. (1)
Loss = 1500 µJ = 1.5 mJ; check: $\frac{5 \times 10}{2 \times 15} (30)^{2} = 1500$ µJ. (1)

Answer: 40 V; 1.5 mJ lost as heat (and radiation) in the wires.

Example 6: Meeting a voltage rating · 4 marks · Case-basedHOTS / competency

Suppose a circuit needs a capacitance of 6 µF that can withstand 1000 V, but only 2 µF capacitors rated at 250 V are available.
(i) How many must be put in series in each row? (1 mark)
(ii) What is the capacitance of one row? (1 mark)
(iii) How many rows in parallel, and how many capacitors in all? (1 mark)
(iv) Find the energy stored at 1000 V. (1 mark)

Try it first, then show the solutionSolution

(i) Identical capacitors in series share the voltage equally: $\frac{1000}{250} = 4$ in each row.
(ii) $\frac{2}{4} = 0.5$ µF.
(iii) $\frac{6}{0.5} = 12$ rows; $12 \times 4 = 48$ capacitors.
(iv) $U = \frac{1}{2} \times 6 \times 1 0^{- 6} \times 100 0^{2} = 3$ J.

Answer: (i) 4; (ii) 0.5 µF; (iii) 12 rows, 48 capacitors; (iv) 3 J.

Where students lose marks in Capacitors in Series and Parallel

The slips examiners see most often, and what to write instead.

Mistake: No circuit diagram
The ISC 2024 examiners noted that many did not draw the series circuit in the derivation.
Do this: Draw and label $C_{1}, C_{2}, C_{3}$ , $V_{1}, V_{2}, V_{3}$ and ±Q on each plate.
Usually costs: 1 mark.
Mistake: Final step written wrongly
Also in 2024: the result was written as $C = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$ .
Wrong working
$C = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$
Right working
$\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$
Do this: Divide through by Q carefully and check the units of both sides.
Usually costs: 1 mark.
Mistake: Not realising p.d.s add in series
2024 candidates did not see that charge stays the same while potentials add; some confused series and parallel.
Do this: Series: same Q, $V = \sum V_{i}$ . Parallel: same V, $Q = \sum Q_{i}$ .
Usually costs: 1 to 2 marks.
Mistake: Assuming energy is conserved on sharing
Charge is conserved, but some energy is always dissipated.
Do this: Compute energies before and after separately.
Usually costs: 1 mark.

Exam tips

Write µF values throughout and convert to F only when calculating energy in joules.
For networks, redraw after each simplification.
State "charge is the same in series" or "p.d. is the same in parallel" as a reason in your working.
For energy, choose the form that uses the quantity that stays constant.

Practice questions on Capacitors in Series and Parallel

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
When capacitors are connected in parallel, each has the same:
(a) charge
(b) potential difference
(c) energy
(d) capacitance
Hint
They share the same two end points.
Answer
(b).
Q2 · 1 mark · Very short answerBoard level
Three 3 µF capacitors are connected so that two are in parallel and the third is in series with that pair. Find the equivalent capacitance.
Hint
6 µF in series with 3 µF.
Answer
2 µF.
Q3 · 2 marks · Short answerHarder board level
A 2 µF capacitor in series with an unknown capacitor gives 1.5 µF. Find the unknown capacitance.
Hint
$\frac{1}{C} = \frac{1}{1.5} - \frac{1}{2}$ .
Answer
6 µF.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In a series combination, the smallest capacitor has the largest potential difference.
Reason (R): All capacitors in series carry the same charge, and $V = \frac{Q}{C}$ .
(a) Both true, R explains A. (b) Both true, R does not explain A. (c) A true, R false. (d) A false, R true.
Hint
Same Q, so V ∝ 1/C.
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
Two identical capacitors are connected to the same battery, first in series and then in parallel. Find the ratio of the total energies stored.
Hint
$U = \frac{1}{2} C_{e q} V^{2}$ .
Answer
$\frac{C}{2} : 2 C = 1 : 4$ .
Q6 · 3 marks · Short answerHOTS / competency
Four 4 µF capacitors: two in series form one branch, which is in parallel with a third capacitor; this group is in series with the fourth, across 10 V. Find the equivalent capacitance and the charge on each capacitor.
Hint
Group: 2 + 4 = 6 µF; then 6 and 4 in series.
Answer
$C_{e q} = 2.4$ µF; Q = 24 µC (on the fourth, p.d. 6 V); group p.d. 4 V: single capacitor 16 µC, each in the series pair 8 µC.
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a camera flash unit has two 100 µF capacitors and a 300 V charging circuit.
(i) Find the energy stored when they are in parallel. (1 mark)
(ii) Find the energy stored if they were in series across the same 300 V. (1 mark)
(iii) Find the charge delivered by the parallel bank on full discharge. (1 mark)
(iv) Which arrangement would you choose for a brighter flash, and why? (1 mark)
Hint
$U = \frac{1}{2} C_{e q} V^{2}$ .
Answer
(i) $\frac{1}{2} (200 \times 1 0^{- 6}) (30 0^{2}) = 9$ J. (ii) $\frac{1}{2} (50 \times 1 0^{- 6}) (30 0^{2}) = 2.25$ J. (iii) 0.06 C. (iv) Parallel: at the same voltage it stores four times the energy.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Equivalent of $C_{1}$ and $C_{2}$ in series?Show answer
$\frac{C _{1} C _{2}}{C _{1} + C _{2}}$ .
2.What is the same for capacitors in series?Show answer
The charge.
3.10 µF at 100 V: energy stored?Show answer
0.05 J.
4.Energy density of an electric field?Show answer
$\frac{1}{2} ε_{0} E^{2}$ .
5.Common potential of $C_{1}$ at $V_{1}$ and $C_{2}$ at $V_{2}$ ?Show answer
$\frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}}$ .

Capacitors in Series and Parallel: frequently asked questions

Why is the charge the same on capacitors in series?

The inner plates and the wire joining them form an isolated conductor with zero net charge. When one plate gets −Q, the other must get +Q, so the same charge appears on every capacitor.

Do I need to derive the energy formula?

No. The ISC 2027 syllabus asks for the formula for energy stored (no derivation), along with energy density.

Where does the energy go when capacitors share charge?

It is dissipated as heat in the resistance of the connecting wires and as electromagnetic radiation during the brief current surge.

Why use capacitors in series at all if the capacitance falls?

To divide a high voltage among capacitors with lower voltage ratings, so that none breaks down.

How was this tested recently?

In 2024 a 3-mark question asked for the derivation of three capacitors in series, with a labelled circuit diagram.