PhysicsCBSE Class 12 · Electrostatic Potential and Capacitance

Combination of Capacitors and Energy Stored Class 12

Updated 15 October 2026 · Fact-checked

Capacitors in series carry the same charge, so 1/C = 1/C₁ + 1/C₂ + …; in parallel they share the same voltage, so C = C₁ + C₂ + …. A capacitor stores energy U = ½CV² = Q²/2C = ½QV. When two charged capacitors are joined, charge is conserved but some energy is always lost.

Understand Combination of Capacitors and Energy Stored

Electronic circuits rarely use one capacitor with exactly the value needed. Instead, capacitors are joined in series or in parallel to get the required capacitance or to share a high voltage.

Series combination. Capacitors joined end to end, one path for charge. When a voltage V is applied, the battery puts $+ Q$ on the first plate; induction then gives every capacitor the same charge Q (the inner plates are isolated, so each pair has equal and opposite charge). The voltages add: $V = V_{1} + V_{2} + V_{3} = Q (\frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}})$ . Writing $V = \frac{Q}{C}$ for the single equivalent capacitor gives $\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$ . The equivalent capacitance is smaller than the smallest one. Since $V_{i} = \frac{Q}{C _{i}}$ , the smallest capacitor takes the largest share of the voltage, which matters if each capacitor can only stand a certain voltage. For two capacitors, $C = \frac{C _{1} C _{2}}{C _{1} + C _{2}}$ .

Parallel combination. Capacitors joined across the same two points. Each has the same voltage V, and charges $Q_{i} = C_{i} V$ . The total charge drawn is $Q = Q_{1} + Q_{2} + Q_{3} = V (C_{1} + C_{2} + C_{3})$ , so $C = C_{1} + C_{2} + C_{3}$ , larger than the largest. The biggest capacitor holds the most charge.

Networks. Reduce step by step: combine capacitors that are clearly in series (one path, nothing branching off between them) or clearly in parallel (both ends joined to the same two points), redraw, and repeat. Then work backwards: total Q from $Q = C_{e q} V$ , then each voltage and charge.

Energy stored (formulae only). Charging a capacitor needs work against the field already built up; this work is stored as electrostatic energy in the field between the plates. For 2026-27 you quote the result without derivation: $U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$ . In terms of the field, the energy per unit volume (energy density) is $u = \frac{1}{2} ε_{0} E^{2}$ . In a series group the smallest C stores the most energy ( $U = \frac{Q ^{2}}{2 C}$ , same Q); in a parallel group the largest C stores the most ( $U = \frac{1}{2} C V^{2}$ , same V). The total energy of any group is $\frac{1}{2} C_{e q} V^{2}$ .

Charge sharing. If a capacitor $C_{1}$ at $V_{1}$ is joined (positive to positive) to $C_{2}$ at $V_{2}$ , charge flows until both are at a common potential. Charge is conserved: $V = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}}$ . Energy is not conserved in the capacitors: the loss is $Δ U = \frac{C _{1} C _{2} ( V _{1} - V _{2} ) ^{2}}{2 ( C _{1} + C _{2} )}$ , which appears as heat in the connecting wires and as electromagnetic radiation. If they are joined positive to negative, use $C_{1} V_{1} - C_{2} V_{2}$ for the net charge.

Combination of Capacitors and Energy Stored: formulas and key facts

Series combination: $\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \dots, C_{tw o} = \frac{C _{1} C _{2}}{C _{1} + C _{2}}$; Same charge on each; $V_{i} = \frac{Q}{C _{i}}$ .
Parallel combination: $C = C_{1} + C_{2} + \dots$; Same voltage on each; $Q_{i} = C_{i} V$ .
n identical capacitors: $C_{s} = \frac{C}{n}, C_{p} = n C, \frac{C _{p}}{C _{s}} = n^{2}$; Useful for quick checks.
Energy stored (formula only): $U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$; No derivation needed for 2026-27.
Energy density: $u = \frac{1}{2} ε_{0} E^{2}$; Energy per unit volume of the field, in J/m³.
Common potential on sharing: $V = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}}$; From conservation of charge.
Energy lost on sharing: $Δ U = \frac{C _{1} C _{2} ( V _{1} - V _{2} ) ^{2}}{2 ( C _{1} + C _{2} )}$; Always positive: some energy is always lost.

How to solve Combination of Capacitors and Energy Stored questions

Use this method for capacitor networks, energy and charge-sharing questions.

1Redraw the circuit neatly and mark the two terminals where the voltage is applied.
2Combine capacitors in clear series or parallel groups, redrawing after each step, until one equivalent capacitor remains.
3Find the total charge $Q = C_{e q} V$ .
4Work back through the steps: in series, each part has the same Q; in parallel, each branch has the same V.
5For energy, use $\frac{1}{2} C V^{2}$ for each capacitor or $\frac{1}{2} C_{e q} V^{2}$ for the group; check that the parts add to the total.
6For charge sharing, write total charge before = total charge after, find the common V, then compare energies before and after.
7Check: series voltages add to the supply; parallel charges add to the total.

Combination of Capacitors and Energy Stored 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.

Energy stored and released quickly

Suppose a camera flash unit charges a 200 μF capacitor to 300 V and then dumps its energy into the flash tube in about 1 ms.

(a) Find the energy stored. (b) Find the average power delivered during the flash. (c) Why is a capacitor used instead of drawing this power straight from the small battery?

Try it first, then show the solutionSolution

$U = \frac{1}{2} C V^{2} = \frac{1}{2} \times 200 \times 1 0^{- 6} \times (300)^{2} = 9$ J.
Average power = $\frac{9 J}{1 0 ^{- 3} s} = 9000$ W.
The battery can only supply energy slowly; the capacitor collects it over a few seconds and gives it out in a millisecond, at a much higher power.

Answer: (a) 9 J; (b) about 9 kW; (c) a capacitor can release stored energy far faster than a small battery can supply it.

Memory hook: A capacitor is a bucket: fill it slowly, tip it out in a flash.

Building a value from standard capacitors

Imagine you are fixing a circuit that needs 25 μF, but your box only has 10 μF capacitors.

(a) Show how four 10 μF capacitors can give 25 μF. (b) If 20 V is applied across the arrangement, find the charge on each capacitor.

Try it first, then show the solutionSolution

Two 10 μF in series give 5 μF. Put this pair in parallel with two more single 10 μF capacitors: $10 + 10 + 5 = 25$ μF.
Each single capacitor has 20 V: $Q = 10 \times 20 = 200$ μC.
The series pair (5 μF) has 20 V across it: $Q = 5 \times 20 = 100$ μC on each of its two capacitors (10 V each).

Answer: (a) two in parallel plus a series pair in parallel with them; (b) 200 μC on each single capacitor and 100 μC on each capacitor of the series pair.

Memory hook: Series halves, parallel adds: mix them to hit any value.

Tricks to remember Combination of Capacitors and Energy Stored

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

Capacitors are the opposite of resistors
Capacitors add in parallel and add as reciprocals in series; resistors do the reverse.
Use it when: Not mixing up the two sets of formulas.
Series shares voltage, parallel shares charge
In series, Q is the same and V divides; in parallel, V is the same and Q divides.
Use it when: Working back through a network.
Smallest suffers most in series
In series the smallest C has the biggest V and stores the most energy.
Use it when: Breakdown-voltage and energy-share questions.
Sharing always costs energy
Joining capacitors at different voltages: charge conserved, energy lost as heat in the wires.
Use it when: Charge-sharing questions.

Combination of Capacitors and Energy Stored: solved examples

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

Example 1: Series and parallel of three capacitors · 3 marks · Short answerBoard level

Capacitors of 2 μF, 3 μF and 6 μF are connected (a) in series and (b) in parallel across a 12 V supply. For each case find the equivalent capacitance and the charge and voltage of each capacitor.

Try it first, then show the solutionSolution

(a) $\frac{1}{C} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1$ , so C = 1 μF. Q = 12 μC on each. (1)
Voltages: $\frac{12}{2} = 6$ V, $\frac{12}{3} = 4$ V, $\frac{12}{6} = 2$ V (sum 12 V). (1)
(b) C = 2 + 3 + 6 = 11 μF; each has 12 V; charges 24 μC, 36 μC, 72 μC (total 132 μC). (1)

Answer: (a) 1 μF; 12 μC each; 6 V, 4 V, 2 V. (b) 11 μF; 12 V each; 24, 36 and 72 μC.

Example 2: Derivation: series and parallel formulas · 3 marks · Short answerBoard level

Derive the expressions for the equivalent capacitance of three capacitors connected (a) in series and (b) in parallel.

Try it first, then show the solutionSolution

(a) In series each capacitor carries the same charge Q (the inner plates get equal and opposite charges by induction). $V = V_{1} + V_{2} + V_{3} = \frac{Q}{C _{1}} + \frac{Q}{C _{2}} + \frac{Q}{C _{3}}$ . (1)
For the equivalent capacitor $V = \frac{Q}{C}$ , so $\frac{1}{C} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}}$ . (½)
(b) In parallel each capacitor has the same V, with $Q_{1} = C_{1} V$ , $Q_{2} = C_{2} V$ , $Q_{3} = C_{3} V$ . Total $Q = Q_{1} + Q_{2} + Q_{3} = (C_{1} + C_{2} + C_{3}) V$ . (1)
For the equivalent capacitor $Q = C V$ , so $C = C_{1} + C_{2} + C_{3}$ . (½)

Answer: Series: $\frac{1}{C} = \sum \frac{1}{C _{i}}$ ; parallel: $C = \sum C_{i}$ .

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

A 4 μF capacitor C₁ is connected in series with a parallel pair C₂ = 2 μF and C₃ = 6 μF, and the network is connected to a 12 V battery. Find (a) the equivalent capacitance, (b) the charge and voltage of each capacitor, and (c) the total energy stored.

Try it first, then show the solutionSolution

(a) $C_{2} ∥ C_{3} = 8$ μF; in series with 4 μF: $C = \frac{4 \times 8}{12} = \frac{8}{3} \approx 2.67$ μF. (1)
(b) $Q = \frac{8}{3} \times 12 = 32$ μC on C₁; $V_{1} = \frac{32}{4} = 8$ V. The pair has 12 − 8 = 4 V: $Q_{2} = 8$ μC, $Q_{3} = 24$ μC. (1)
(c) $U = \frac{1}{2} C V^{2} = \frac{1}{2} \times \frac{8}{3} \times 1 0^{- 6} \times 144 = 1.92 \times 1 0^{- 4}$ J (check: $1.28 \times 1 0^{- 4} + 1.6 \times 1 0^{- 5} + 4.8 \times 1 0^{- 5}$ J). (1)

Answer: (a) 8/3 μF; (b) C₁: 32 μC, 8 V; C₂: 8 μC, 4 V; C₃: 24 μC, 4 V; (c) $1.92 \times 1 0^{- 4}$ J.

Example 4: Charge sharing and energy loss · 5 marks · Long answerHOTS / competency

A 4 μF capacitor charged to 100 V is disconnected from the supply and connected across an uncharged 6 μF capacitor.
(a) Find the common potential. (2)
(b) Find the energy before and after, and the loss. (2)
(c) Where does the lost energy go? (1)

Try it first, then show the solutionSolution

(a) Charge is conserved: $Q = 4 \times 100 = 400$ μC. Common $V = \frac{400}{4 + 6} = 40$ V. (2)
(b) Before: $U_{i} = \frac{1}{2} \times 4 \times 1 0^{- 6} \times 10 0^{2} = 0.02$ J. After: $U_{f} = \frac{1}{2} \times 10 \times 1 0^{- 6} \times 4 0^{2} = 0.008$ J. (1)
Loss = 0.012 J; check: $\frac{C _{1} C _{2} ( V _{1} - V _{2} ) ^{2}}{2 ( C _{1} + C _{2} )} = \frac{24 \times 1 0 ^{- 12} \times 1 0 ^{4}}{2 \times 1 0 ^{- 5}} = 0.012$ J. (1)
(c) As heat in the connecting wires (and some electromagnetic radiation) while the charge flows. (1)

Answer: (a) 40 V; (b) 0.02 J before, 0.008 J after, 0.012 J lost; (c) heat in the wires and radiation.

Example 5: Energy density in a capacitor · 3 marks · Short answerHarder board level

A parallel plate air capacitor has plates of area 0.02 m², 1 mm apart, and is connected to 500 V. Find (a) the field between the plates, (b) the energy density, and (c) the total energy, checking it with $\frac{1}{2} C V^{2}$ .

Try it first, then show the solutionSolution

(a) $E = \frac{V}{d} = \frac{500}{1 0 ^{- 3}} = 5 \times 1 0^{5}$ V/m. (1)
(b) $u = \frac{1}{2} ε_{0} E^{2} = \frac{1}{2} \times 8.854 \times 1 0^{- 12} \times 2.5 \times 1 0^{11} \approx 1.11$ J/m³. (1)
(c) Volume = $0.02 \times 1 0^{- 3} = 2 \times 1 0^{- 5}$ m³, so $U \approx 2.21 \times 1 0^{- 5}$ J. Check: $C = \frac{ε _{0} A}{d} \approx 1.77 \times 1 0^{- 10}$ F and $\frac{1}{2} C V^{2} \approx 2.21 \times 1 0^{- 5}$ J. (1)

Answer: (a) $5 \times 1 0^{5}$ V/m; (b) about 1.11 J/m³; (c) about $2.21 \times 1 0^{- 5}$ J by both methods.

Example 6: Designing for a high voltage · 4 marks · Case-basedHarder board level

Read the passage and answer.

A technician has many 2 μF capacitors, each of which breaks down above 100 V. She needs an arrangement of 2 μF that can safely work across 300 V.
(a) How many capacitors must be placed in series in each row, and why? (1)
(b) What is the capacitance of one such row? (1)
(c) How many rows in parallel give 2 μF? (1)
(d) What is the total number of capacitors used? (1)

Try it first, then show the solutionSolution

(a) Identical capacitors in series share the voltage equally, so 3 in series take 100 V each across 300 V. (1)
(b) One row: $\frac{2}{3}$ μF. (1)
(c) Rows in parallel add: $n \times \frac{2}{3} = 2$ gives n = 3 rows. (1)
(d) 3 × 3 = 9 capacitors. (1)

Answer: (a) 3 per row; (b) 2/3 μF; (c) 3 rows; (d) 9 capacitors.

Where students lose marks in Combination of Capacitors and Energy Stored

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

Mistake: Using resistor rules for capacitors.
The formulas look alike but are swapped.
Wrong working
Series: C = C₁ + C₂
Right working
Series: 1/C = 1/C₁ + 1/C₂
Do this: Remember: capacitors add directly in parallel.
Usually costs: The whole numerical.
Mistake: Giving every capacitor in series the supply voltage.
Confusing series with parallel.
Wrong working
Q on 2 μF = 2 × 12 = 24 μC (in series across 12 V)
Right working
Q = C_eq V = 1 × 12 = 12 μC on each in series
Do this: In series, find Q from the equivalent capacitance first.
Usually costs: 1 to 2 marks.
Mistake: Assuming energy is conserved when capacitors share charge.
Applying energy conservation instead of charge conservation.
Do this: Use charge conservation for the common V; then calculate the energy loss.
Usually costs: The main marks of a charge-sharing question.
Mistake: Mixing μF and F in energy calculations.
Leaving C in μF gives an energy 10⁶ times too big.
Wrong working
U = ½ × 4 × 100² = 20000 J
Right working
U = ½ × 4 × 10⁻⁶ × 100² = 0.02 J
Do this: Write C in farad before using an energy formula.
Usually costs: The final mark.
Mistake: Deriving the energy formula when only the formula is required.
Old notes include the derivation, which takes time.
Do this: For 2026-27, quote the energy formulas and use them; spend the time on the numerical.
Usually costs: Time, not marks, but it can cost a later question.

Exam tips

Show the reduction of a network step by step with a redrawn diagram at each stage.
Always check that series voltages add up to the supply voltage.
In charge-sharing problems, state 'charge is conserved' explicitly; it earns a step mark.
Combination and energy questions have appeared as 1- to 5-mark items in recent board papers (for example 2023 55/5/1 Q35, 2024 55/1/1 Q2, 2025 55/7/1 Q31(a) and 2026 55/5/1 Q22).
Quote the energy formula that uses the quantity held constant; it shortens the working.

Practice questions on Combination of Capacitors and Energy Stored

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
Two identical capacitors have a combined capacitance $C_{s}$ in series and $C_{p}$ in parallel. $\frac{C _{s}}{C _{p}}$ is
(a) 1 : 2 (b) 2 : 1 (c) 1 : 4 (d) 4 : 1
Hint
C/2 and 2C.
Answer
(c) 1 : 4.
Q2 · 2 marks · Very short answerBoard level
Write three expressions for the energy stored in a charged capacitor, and say which one to use when a capacitor is disconnected from the battery and its plates are moved apart.
Hint
Which quantity stays fixed when the battery is removed?
Answer
$U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$ . Once disconnected, Q is fixed, so use $\frac{Q ^{2}}{2 C}$ : as the plates separate, C falls and U rises (the work done pulling the plates apart is stored).
Q3 · 2 marks · Very short answerHarder board level
List all the different capacitances you can get using three 3 μF capacitors, using all three each time.
Hint
There are four arrangements.
Answer
All parallel: 9 μF. All series: 1 μF. Two parallel (6 μF) in series with one: 2 μF. Two series (1.5 μF) in parallel with one: 4.5 μF.
Q4 · 3 marks · Short answerHarder board level
Capacitors of 2 μF and 4 μF are connected in series across 90 V. Find the charge, the voltage across each and the energy stored in each. Which stores more energy?
Hint
Same Q in series.
Answer
$C = \frac{8}{6} = \frac{4}{3}$ μF; Q = 120 μC. $V_{1} = 60$ V, $V_{2} = 30$ V. $U_{1} = \frac{1}{2} \times 2 \times 1 0^{- 6} \times 3600 = 3.6 \times 1 0^{- 3}$ J; $U_{2} = 1.8 \times 1 0^{- 3}$ J. The smaller (2 μF) capacitor stores more.
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In a series combination of capacitors, the capacitor with the smallest capacitance has the largest potential difference across it.
Reason (R): All capacitors in series carry the same charge, and $V = \frac{Q}{C}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Fix Q and vary C.
Answer
(a).
Q6 · 3 marks · Short answerHOTS / competency
A 3 μF capacitor charged to 200 V and a 6 μF capacitor charged to 100 V are joined with the positive plate of each connected to the negative plate of the other. Find the final voltage and the energy lost.
Hint
Net charge = difference of the charges.
Answer
Charges 600 μC and 600 μC; joined oppositely, the net charge is 600 − 600 = 0, so the final voltage is 0. All the energy is lost: $\frac{1}{2} \times 3 \times 1 0^{- 6} \times 20 0^{2} + \frac{1}{2} \times 6 \times 1 0^{- 6} \times 10 0^{2} = 0.06 + 0.03 = 0.09$ J.
Q7 · 2 marks · Very short answerHOTS / competency
A capacitor C is charged to V and then connected across an identical uncharged capacitor. Show that half of the original energy is lost.
Hint
Common V = V/2.
Answer
Common potential $= \frac{C V}{2 C} = \frac{V}{2}$ . Final energy $= \frac{1}{2} (2 C) (\frac{V}{2})^{2} = \frac{1}{4} C V^{2}$ , which is half of the initial $\frac{1}{2} C V^{2}$ . So half is lost.

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 capacitance of C₁ and C₂ in series?Show answer
$\frac{C _{1} C _{2}}{C _{1} + C _{2}}$ .
2.What is the same for capacitors in parallel?Show answer
The potential difference.
3.Energy stored in a capacitor in terms of Q and C?Show answer
$\frac{Q ^{2}}{2 C}$ .
4.Energy density of an electric field?Show answer
$\frac{1}{2} ε_{0} E^{2}$ .
5.Common potential when C₁ at V₁ is joined to C₂ at V₂?Show answer
$\frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}}$ .
6.n identical capacitors: ratio of parallel to series capacitance?Show answer
n².

Combination of Capacitors and Energy Stored: frequently asked questions

Why is the charge the same on capacitors in series?

Only the outer plates touch the battery. The inner plates between two capacitors form an isolated conductor with zero net charge, so whatever charge one side gains, the other side loses. By induction every capacitor ends up with the same ±Q.

Is the derivation of energy stored in a capacitor in the 2026-27 syllabus?

No. The syllabus says 'energy stored in a capacitor (no derivation, formulae only)'. Learn the three forms of the formula and the energy density ½ε₀E².

Where does energy go when two charged capacitors are joined?

Charge flows through the connecting wires until both capacitors reach the same potential. The lost energy is turned into heat in the wires and some electromagnetic radiation.

Why do we connect capacitors in series?

To share a large voltage among several capacitors, each with a lower voltage rating, or to get a smaller capacitance than any one available.

Which capacitor stores more energy in parallel?

The one with the larger capacitance, because all have the same V and U = ½CV².