Capacitors in series and in parallel

General physics

Capacitor Combinations — Series and Parallel — Physics Lesson

🎯 Learning Objectives

Learning Objectives

Derive, using C = Q/V, formulae for the combined capacitance of capacitors in series and in parallel
Use the capacitance formulae for capacitors in series and in parallel
Apply conservation of charge and potential difference to capacitor networks
Solve complex problems involving mixed series-parallel capacitor circuits
Understand practical applications of capacitor combinations in electronic circuits
Calculate energy storage in capacitor combinations

🗣️ Language Objectives

Language Objectives

Use precise mathematical language to describe capacitor combinations and derivations
Explain the concept of equivalent capacitance using appropriate scientific terminology
Describe charge distribution and voltage across capacitor networks clearly
Communicate problem-solving strategies for complex capacitor circuits effectively
Write mathematical expressions and derive formulas systematically
Discuss practical applications of capacitor combinations using technical language

📝 Key Terms

Key Terms

English Term	Russian Translation	Kazakh Translation
Capacitance	Ёмкость	Сыйымдылық
Series Connection	Последовательное соединение	Тізбектей қосылу
Parallel Connection	Параллельное соединение	Параллель қосылу
Equivalent Capacitance	Эквивалентная ёмкость	Эквивалент сыйымдылық
Charge Distribution	Распределение заряда	Заряд бөлінуі
Potential Difference	Разность потенциалов	Потенциалдар айырмасы
Dielectric	Диэлектрик	Диэлектрик
Energy Storage	Накопление энергии	Энергия жинақтау

🃏 Topic Flashcards

Interactive Flashcards

Practice with these flashcards to memorize key concepts about capacitor combinations in series and parallel.

📚 Glossary

Glossary

Capacitance (C): The ability of a capacitor to store electric charge per unit potential difference, measured in Farads (F). Defined by the equation C = Q/V, where Q is charge and V is potential difference.
Translation
Russian: Ёмкость — способность конденсатора накапливать электрический заряд на единицу разности потенциалов, измеряется в Фарадах (Ф). Определяется уравнением C = Q/V, где Q — заряд, V — разность потенциалов.
Kazakh: Сыйымдылық — конденсатордың потенциалдар айырмасының бірлігіне электр зарядын жинақтау қабілеті, Фарадпен (Ф) өлшенеді. C = Q/V теңдеуімен анықталады, мұндағы Q — заряд, V — потенциалдар айырмасы.
Series Connection: A circuit arrangement where capacitors are connected end-to-end in a single path, so the same charge flows through each capacitor but the total voltage is divided among them.
Translation
Russian: Последовательное соединение — схема подключения, где конденсаторы соединены друг за другом в одном пути, так что одинаковый заряд протекает через каждый конденсатор, но общее напряжение распределяется между ними.
Kazakh: Тізбектей қосылу — конденсаторлар бір жолда ұштастыра қосылған схема, сондықтан әр конденсатор арқылы бірдей заряд өтеді, бірақ жалпы кернеу олардың арасында бөлінеді.
Parallel Connection: A circuit arrangement where capacitors are connected across the same two points, sharing the same potential difference but dividing the total charge among them.
Translation
Russian: Параллельное соединение — схема подключения, где конденсаторы подключены к одним и тем же двум точкам, имея одинаковую разность потенциалов, но разделяя общий заряд между собой.
Kazakh: Параллель қосылу — конденсаторлар бірдей екі нүктеге қосылған схема, олар бірдей потенциалдар айырмасына ие, бірақ жалпы зарядты өздері арасында бөледі.
Equivalent Capacitance: The single capacitance value that would store the same charge at the same voltage as a combination of capacitors. It simplifies complex capacitor networks for analysis.
Translation
Russian: Эквивалентная ёмкость — единственное значение ёмкости, которое накапливало бы тот же заряд при том же напряжении, что и комбинация конденсаторов. Упрощает анализ сложных сетей конденсаторов.
Kazakh: Эквивалент сыйымдылық — конденсаторлар комбинациясымен бірдей кернеуде бірдей зарядты жинақтайтын жалғыз сыйымдылық мәні. Күрделі конденсатор желілерін талдауды жеңілдетеді.
Charge Distribution: The way electric charge is allocated among capacitors in a circuit. In series, all capacitors have the same charge; in parallel, charge is distributed proportionally to their capacitances.
Translation
Russian: Распределение заряда — способ распределения электрического заряда между конденсаторами в цепи. При последовательном соединении все конденсаторы имеют одинаковый заряд; при параллельном — заряд распределяется пропорционально их ёмкостям.
Kazakh: Заряд бөлінуі — тізбектегі конденсаторлар арасында электр зарядының бөлінуі. Тізбектей қосылуда барлық конденсаторлар бірдей зарядқа ие; параллель қосылуда заряд олардың сыйымдылықтарына пропорционал бөлінеді.
Potential Difference (Voltage): The electrical potential energy difference per unit charge between two points in a circuit, measured in Volts (V). For capacitors, it’s the voltage across the plates.
Translation
Russian: Разность потенциалов (напряжение) — разность электрической потенциальной энергии на единицу заряда между двумя точками в цепи, измеряется в Вольтах (В). Для конденсаторов это напряжение между пластинами.
Kazakh: Потенциалдар айырмасы (кернеу) — тізбектегі екі нүкте арасындағы заряд бірлігіне келетін электрлік потенциалдық энергия айырмасы, Вольтпен (В) өлшенеді. Конденсаторлар үшін бұл пластиналар арасындағы кернеу.
Dielectric Material: An insulating material placed between capacitor plates that increases the capacitance by reducing the electric field strength for a given charge and voltage.
Translation
Russian: Диэлектрический материал — изолирующий материал, помещённый между пластинами конденсатора, который увеличивает ёмкость, уменьшая напряжённость электрического поля при данном заряде и напряжении.
Kazakh: Диэлектрлік материал — конденсатор пластиналары арасына орналастырылған оқшаулағыш материал, ол берілген заряд пен кернеуде электр өрісінің кернеулігін азайту арқылы сыйымдылықты арттырады.
Energy Storage: The ability of capacitors to store electrical energy in the electric field between their plates. The energy stored is given by U = ½CV² = ½QV = Q²/(2C).
Translation
Russian: Накопление энергии — способность конденсаторов накапливать электрическую энергию в электрическом поле между их пластинами. Накопленная энергия задаётся формулой U = ½CV² = ½QV = Q²/(2C).
Kazakh: Энергия жинақтау — конденсаторлардың пластиналары арасындағы электр өрісінде электр энергиясын жинақтау қабілеті. Жинақталған энергия U = ½CV² = ½QV = Q²/(2C) формуласымен беріледі.

📖 Theory: Capacitor Combinations - Series and Parallel

Theory: Deriving and Using Capacitance Formulas

Introduction to Capacitor Combinations

In electronic circuits, capacitors are often connected in various combinations to achieve desired capacitance values. Understanding how to calculate the equivalent capacitance is crucial for circuit analysis and design.

Kazakh Translation

Электрондық тізбектерде конденсаторлар қажетті сыйымдылық мәндеріне қол жеткізу үшін әртүрлі комбинацияларда жиі қосылады. Эквивалент сыйымдылықты есептеуді түсіну тізбекті талдау және жобалау үшін өте маңызды.

Capacitor series and parallel combinations

Basic series and parallel capacitor arrangements

Fundamental Capacitance Equation

The fundamental equation for capacitance is:

C = Q/V

Where:

C = Capacitance (Farads, F)
Q = Charge stored (Coulombs, C)
V = Potential difference (Volts, V)

Kazakh Translation

C = сыйымдылық (Фарад, Ф), Q = жинақталған заряд (Кулон, Кл), V = потенциалдар айырмасы (Вольт, В)

Capacitors in Series

When capacitors are connected in series, they form a single path for charge flow.

Kazakh Translation

Конденсаторлар тізбектей қосылғанда, олар заряд ағыны үшін жалғыз жол құрайды.

Key Characteristics of Series Connection:

Property	Series Connection	Mathematical Expression
Charge	Same on all capacitors	Q₁ = Q₂ = Q₃ = Q_total
Voltage	Adds up across capacitors	V_total = V₁ + V₂ + V₃
Individual Voltage	Inversely proportional to capacitance	V = Q/C

Series capacitor circuit showing charge and voltage distribution

Derivation of Series Capacitance Formula:

Step 1: Apply voltage law

V_total = V₁ + V₂ + V₃ + …

Step 2: Express each voltage using C = Q/V → V = Q/C

V_total = Q/C₁ + Q/C₂ + Q/C₃ + …

Step 3: Factor out common charge Q

V_total = Q(1/C₁ + 1/C₂ + 1/C₃ + …)

Step 4: For equivalent capacitance C_eq

V_total = Q/C_eq

Step 5: Equate the expressions

Q/C_eq = Q(1/C₁ + 1/C₂ + 1/C₃ + …)

Final Result:

1/C_eq = 1/C₁ + 1/C₂ + 1/C₃ + …

Kazakh Translation

Тізбектей қосылған конденсаторлар үшін эквивалент сыйымдылықты шығару: жалпы кернеу барлық конденсаторлардағы кернеулердің қосындысына тең, ал заряд барлығында бірдей. Нәтижесінде 1/Cэкв = 1/C₁ + 1/C₂ + 1/C₃ + … формуласы алынады.

Capacitors in Parallel

When capacitors are connected in parallel, they share the same potential difference.

Kazakh Translation

Конденсаторлар параллель қосылғанда, олар бірдей потенциалдар айырмасын бөліседі.

Key Characteristics of Parallel Connection:

Property	Parallel Connection	Mathematical Expression
Voltage	Same across all capacitors	V₁ = V₂ = V₃ = V_total
Charge	Adds up across capacitors	Q_total = Q₁ + Q₂ + Q₃
Individual Charge	Proportional to capacitance	Q = CV

Parallel capacitor circuit showing charge and voltage distribution

Derivation of Parallel Capacitance Formula:

Step 1: Apply charge conservation

Q_total = Q₁ + Q₂ + Q₃ + …

Step 2: Express each charge using C = Q/V → Q = CV

Q_total = C₁V + C₂V + C₃V + …

Step 3: Factor out common voltage V

Q_total = V(C₁ + C₂ + C₃ + …)

Step 4: For equivalent capacitance C_eq

Q_total = C_eqV

Step 5: Equate the expressions

C_eqV = V(C₁ + C₂ + C₃ + …)

Final Result:

C_eq = C₁ + C₂ + C₃ + …

Kazakh Translation

Параллель қосылған конденсаторлар үшін эквивалент сыйымдылықты шығару: жалпы заряд барлық конденсаторлардағы зарядтардың қосындысына тең, ал кернеу барлығында бірдей. Нәтижесінде Cэкв = C₁ + C₂ + C₃ + … формуласы алынады.

Summary of Formulas

Connection Type	Equivalent Capacitance	Key Principle	Memory Aid
Series	1/C_eq = 1/C₁ + 1/C₂ + …	Same charge, voltages add	Like resistors in parallel
Parallel	C_eq = C₁ + C₂ + …	Same voltage, charges add	Like resistors in series

Practice Questions

Question 1 (Easy):

Three capacitors with capacitances 2 μF, 4 μF, and 6 μF are connected in parallel. Calculate the equivalent capacitance.

Answer

For parallel connection: C_eq = C₁ + C₂ + C₃
C_eq = 2 μF + 4 μF + 6 μF = 12 μF
The equivalent capacitance is 12 μF.

Question 2 (Medium):

Two capacitors of 3 μF and 6 μF are connected in series. Calculate: (a) the equivalent capacitance, (b) the charge on each capacitor when 12 V is applied across the combination.

Answer

(a) Equivalent capacitance:
1/C_eq = 1/C₁ + 1/C₂ = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2
Therefore: C_eq = 2 μF

(b) Charge on each capacitor:
Q_total = C_eq × V_total = 2 μF × 12 V = 24 μC
In series, each capacitor has the same charge: Q₁ = Q₂ = 24 μC

Verification: V₁ = Q₁/C₁ = 24/3 = 8 V, V₂ = Q₂/C₂ = 24/6 = 4 V
Total voltage: 8 V + 4 V = 12 V ✓

Question 3 (Medium):

A 4 μF and 8 μF capacitor are connected in parallel, and this combination is then connected in series with a 6 μF capacitor. Find the equivalent capacitance of the entire circuit.

Answer

Step 1: Find parallel combination of 4 μF and 8 μF
C_parallel = 4 μF + 8 μF = 12 μF

Step 2: This 12 μF is now in series with 6 μF
1/C_eq = 1/12 + 1/6 = 1/12 + 2/12 = 3/12 = 1/4
Therefore: C_eq = 4 μF

The equivalent capacitance of the entire circuit is 4 μF.

Question 4 (Critical Thinking):

Design a capacitor network using only 6 μF capacitors to achieve an equivalent capacitance of 4 μF. Propose at least two different configurations and explain which would be more practical in terms of voltage rating and energy storage.

Answer

Configuration 1: Two 6 μF capacitors in series
1/C_eq = 1/6 + 1/6 = 2/6 = 1/3
C_eq = 3 μF (Close, but not exactly 4 μF)

Configuration 2: Complex series-parallel arrangement
Use three 6 μF capacitors: two in parallel (giving 12 μF), then in series with the third
Parallel: C_p = 6 + 6 = 12 μF
Series with third: 1/C_eq = 1/12 + 1/6 = 1/12 + 2/12 = 3/12
C_eq = 4 μF (Exactly what we need!)

Configuration 3: Alternative arrangement
Three capacitors in series (2 μF equivalent) in parallel with one capacitor in series with two parallel (3 μF equivalent)
This is more complex but achieves the target

Practical Analysis:
— Configuration 2 is most practical: uses only 3 capacitors, achieves exact value
— Voltage rating: Series combinations can handle higher voltages
— Energy storage: Parallel combinations store more energy
— Cost: Fewer components generally mean lower cost and higher reliability

🧠 Memorization Exercises

Exercises on Memorizing Terms

Exercise 1: Formula Recognition

Match each formula with the correct type of connection:

Formulas:

C_eq = C₁ + C₂ + C₃
1/C_eq = 1/C₁ + 1/C₂ + 1/C₃
Q_total = Q₁ + Q₂ + Q₃
V_total = V₁ + V₂ + V₃

Connection Types:

Parallel capacitance
Series voltage
Parallel charge
Series capacitance

Answer

1-A: C_eq = C₁ + C₂ + C₃ → Parallel capacitance
2-D: 1/C_eq = 1/C₁ + 1/C₂ + 1/C₃ → Series capacitance
3-C: Q_total = Q₁ + Q₂ + Q₃ → Parallel charge
4-B: V_total = V₁ + V₂ + V₃ → Series voltage

Exercise 2: Connection Properties

Fill in the table comparing series and parallel connections:

Property	Series	Parallel
Charge distribution	_______	_______
Voltage distribution	_______	_______
Equivalent capacitance	_______	_______

Answer

Charge distribution:
Series: Same on all capacitors
Parallel: Adds up (distributed according to capacitance)

Voltage distribution:
Series: Adds up (distributed inversely to capacitance)
Parallel: Same across all capacitors

Equivalent capacitance:
Series: 1/C_eq = 1/C₁ + 1/C₂ + … (reciprocal addition)
Parallel: C_eq = C₁ + C₂ + … (direct addition)

Exercise 3: Quick Calculations

Calculate the equivalent capacitance for these quick problems:

2 μF and 3 μF in parallel: C_eq = _______
4 μF and 12 μF in series: C_eq = _______
Three 6 μF capacitors in parallel: C_eq = _______
Two 8 μF capacitors in series: C_eq = _______
5 μF in series with (2 μF || 3 μF): C_eq = _______

Answer

1. 2 μF + 3 μF = 5 μF
2. 1/C = 1/4 + 1/12 = 3/12 + 1/12 = 4/12, so C = 3 μF
3. 6 μF + 6 μF + 6 μF = 18 μF
4. 1/C = 1/8 + 1/8 = 2/8, so C = 4 μF
5. First: 2 μF || 3 μF = 5 μF, Then: 1/C = 1/5 + 1/5 = 2/5, so C = 2.5 μF

📺 Educational Video

Video Tutorial: Capacitor Combinations — Series and Parallel

Additional Resources:

🔬 Problem Solving Examples

Worked Examples

Example 1: Series-Parallel Combination

Problem: Find the equivalent capacitance between points A and B in the circuit where C₁ = 2 μF, C₂ = 4 μF, C₃ = 6 μF, and C₄ = 3 μF. C₁ and C₂ are in parallel, and this combination is in series with C₃, and the whole thing is in parallel with C₄.

🎤 Audio Solution

Detailed Solution with Pronunciation

Step 1: Find parallel combination of C₁ and C₂ (pronounced: PAR-uh-lel kom-bi-NAY-shun)

C₁₂ = C₁ + C₂ = 2 μF + 4 μF = 6 μF

Step 2: Find series combination of C₁₂ and C₃

1/C₁₂₃ = 1/C₁₂ + 1/C₃ = 1/6 + 1/6 = 2/6

Therefore: C₁₂₃ = 3 μF

Step 3: Find parallel combination of C₁₂₃ and C₄

C_eq = C₁₂₃ + C₄ = 3 μF + 3 μF = 6 μF

Verification: Check the calculation steps

Parallel combinations add directly: 2 + 4 = 6 μF

Series combinations: 1/6 + 1/6 = 1/3, so C = 3 μF

Final parallel: 3 + 3 = 6 μF

📝 Quick Solution

Brief Solution

Given: C₁ = 2 μF, C₂ = 4 μF, C₃ = 6 μF, C₄ = 3 μF

Step 1: C₁ || C₂

C₁₂ = 2 + 4 = 6 μF

Step 2: C₁₂ in series with C₃

1/C₁₂₃ = 1/6 + 1/6 = 1/3

C₁₂₃ = 3 μF

Step 3: C₁₂₃ || C₄

C_eq = 3 + 3 = 6 μF

Example 2: Energy and Charge Distribution

Problem: Two capacitors, 3 μF and 6 μF, are connected in series and attached to a 12 V battery. Calculate: (a) equivalent capacitance, (b) charge on each capacitor, (c) voltage across each capacitor, (d) energy stored in each capacitor.

🎤 Audio Solution

Detailed Solution with Pronunciation

Part (a): Equivalent capacitance

For series: 1/C_eq = 1/C₁ + 1/C₂

1/C_eq = 1/3 + 1/6 = 2/6 + 1/6 = 3/6 = 1/2

Therefore: C_eq = 2 μF

Part (b): Charge on each capacitor

Total charge: Q = C_eq × V = 2 μF × 12 V = 24 μC

In series, same charge on both: Q₁ = Q₂ = 24 μC

Part (c): Voltage across each capacitor

V₁ = Q₁/C₁ = 24 μC / 3 μF = 8 V

V₂ = Q₂/C₂ = 24 μC / 6 μF = 4 V

Check: V₁ + V₂ = 8 V + 4 V = 12 V ✓

Part (d): Energy stored in each capacitor

U₁ = ½C₁V₁² = ½ × 3 × 8² = ½ × 3 × 64 = 96 μJ

U₂ = ½C₂V₂² = ½ × 6 × 4² = ½ × 6 × 16 = 48 μJ

Total energy: U = 96 + 48 = 144 μJ

📝 Quick Solution

Brief Solution

Given: C₁ = 3 μF, C₂ = 6 μF, V = 12 V, series connection

(a) 1/C_eq = 1/3 + 1/6 = 1/2, so C_eq = 2 μF

(b) Q = C_eqV = 2 × 12 = 24 μC (same for both)

(c) V₁ = 24/3 = 8 V, V₂ = 24/6 = 4 V

(d) U₁ = ½ × 3 × 8² = 96 μJ
U₂ = ½ × 6 × 4² = 48 μJ

🔬 Investigation Task

Interactive Simulation

Use this PhET simulation to investigate capacitor combinations and their behavior:

Investigation Questions:

How does adding capacitors in parallel affect the total capacitance and charge storage?
What happens to the voltage across individual capacitors in a series combination?
How does the plate area and separation affect capacitance in combinations?
Compare energy storage efficiency between series and parallel arrangements.

Brief Answers

1. Parallel connection increases total capacitance (C_eq = C₁ + C₂ + …) and total charge storage proportionally
2. In series, voltage divides inversely proportional to capacitance values (larger C gets smaller V)
3. Larger plate area increases capacitance; smaller separation increases capacitance — effects are multiplicative in combinations
4. Parallel arrangements store more total energy at same voltage; series arrangements allow higher working voltages

👥 Group/Pair Activity

Collaborative Learning Activity

Work with your partner or group to complete this capacitor network analysis challenge:

Discussion Points:

How do engineers choose between series and parallel capacitor arrangements in real circuits?
What are the advantages and disadvantages of each connection type?
How do capacitor tolerances affect the performance of series and parallel combinations?
What safety considerations are important when working with capacitor networks?

Group Challenge Activities:

Design capacitor networks to achieve specific capacitance values using standard capacitor values
Calculate power factor correction requirements for AC circuits using capacitor combinations
Investigate timing circuits using RC combinations in series and parallel
Analyze energy storage systems using capacitor banks in various configurations

✏️ Individual Assessment

Structured Questions - Individual Work

Question 1 (Analysis):

A complex capacitor network consists of five capacitors arranged as follows: C₁ = 4 μF and C₂ = 6 μF in parallel, this combination in series with C₃ = 3 μF, and this entire group in parallel with C₄ = 8 μF and C₅ = 12 μF which are themselves in series.

Draw the circuit diagram and identify all series and parallel sections.
Calculate the equivalent capacitance of the entire network.
If 24 V is applied across the network, find the total charge stored.
Calculate the voltage across C₃ and explain your reasoning.
Determine the energy stored in C₄ and compare it with the energy in C₁.

Answer

a) Circuit has two main parallel branches: [(C₁||C₂) in series with C₃] || [C₄ in series with C₅]

b) Left branch: C₁||C₂ = 4+6 = 10 μF, then (10 μF in series with 3 μF) = 30/13 μF ≈ 2.31 μF
Right branch: C₄ in series with C₅: 1/C = 1/8 + 1/12 = 5/24, so C = 4.8 μF
Total: C_eq = 2.31 + 4.8 = 7.11 μF

c) Q_total = C_eq × V = 7.11 × 24 = 170.6 μC

d) Voltage across left branch = 24 V. Charge in left branch = 2.31 × 24 = 55.4 μC
This same charge flows through C₃, so V₃ = Q/C₃ = 55.4/3 = 18.5 V

e) Charge through right branch = 4.8 × 24 = 115.2 μC (same through C₄)
V₄ = 115.2/8 = 14.4 V, Energy in C₄ = ½ × 8 × (14.4)² = 829 μJ
V₁ = (24 — 18.5) = 5.5 V, Energy in C₁ = ½ × 4 × (5.5)² = 60.5 μJ

Question 2 (Synthesis):

Design a capacitor bank for a flash photography system that needs to store 50 J of energy at 400 V. You have access to capacitors with values 100 μF, 200 μF, and 500 μF, each rated at 100 V maximum.

Calculate the total capacitance required for the energy storage specification.
Design a network using the available capacitors that can safely operate at 400 V.
Calculate how many of each capacitor type you need and their arrangement.
Verify that your design meets both energy and voltage requirements.
Analyze the cost-effectiveness if 100 μF costs $1, 200 μF costs $1.8, and 500 μF costs $4.

Answer

a) Required capacitance: E = ½CV², so C = 2E/V² = 2×50/400² = 625 μF

b) For 400 V operation with 100 V rated capacitors, need 4 capacitors in series per string
Series reduces capacitance by factor of 4, so need 4 parallel strings
Total: 16 capacitors