derive, using C = Q / V, formulae for the combined capacitance of capacitors in series and in parallel

Capacitors and Capacitance – Cambridge International AS & A Level Physics (9702)

Learning Objective

Derive, using the definition C = Q / V, the expressions for the equivalent capacitance of a group of capacitors when they are connected (a) in series and (b) in parallel. Extend the discussion to energy storage, discharge, dielectrics, other geometries, practical measurement and typical applications – all required by Topic 19 of the syllabus.

Key‑words (AO1)

• Capacitance, charge (Q), potential difference (V), farad (F)
• Series connection, parallel connection, equivalent capacitance
• Energy stored, dielectric, permittivity (ε₀, κ)
• RC time constant, discharge, leakage

1. Fundamental Definition

The capacitance of any device is defined as the ratio of the charge on its plates to the potential difference between them:

\( C = \dfrac{Q}{V}\)  [unit: farad (F) = C · V⁻¹]

For an ideal parallel‑plate capacitor of plate area A and separation d:

\( C = \varepsilon{0}\,\dfrac{A}{d}\)  [where \(\varepsilon{0}=8.85\times10^{-12}\,\text{F m}^{-1}\)]

2. Equivalent Capacitance – Series Connection

1. Because the inner plates are isolated, the magnitude of charge on each capacitor is identical:

   \( Q{1}=Q{2}=…=Q_{n}=Q\)

2. Voltage across the i‑th capacitor:

   \( V{i}= \dfrac{Q}{C{i}}\)

3. The total voltage between the ends A and B is the sum of the individual voltages:

   \( V{\text{total}} = \sum{i=1}^{n} V{i}= \sum{i=1}^{n}\dfrac{Q}{C_{i}}\)

4. Define the equivalent capacitance Cₑq for the whole series group by the same definition \(C=Q/V\):

   \( C{\text{eq}} = \dfrac{Q}{V{\text{total}}}

   = \dfrac{Q}{\displaystyle\sum{i=1}^{n}\dfrac{Q}{C{i}}}

   = \frac{1}{\displaystyle\sum{i=1}^{n}\dfrac{1}{C{i}}}\)

Series formula \( \displaystyle\frac{1}{C{\text{eq}}}= \sum{i=1}^{n}\frac{1}{C_{i}}\)

3. Equivalent Capacitance – Parallel Connection

1. All capacitors share the same potential difference V.
2. Charge on the i‑th capacitor:

   \( Q{i}=C{i}V\)

3. Total charge stored on the combination is the algebraic sum:

   \( Q{\text{total}} = \sum{i=1}^{n} Q{i}= \sum{i=1}^{n} C_{i}V\)

4. Equivalent capacitance:

   \( C{\text{eq}} = \dfrac{Q{\text{total}}}{V}

   = \dfrac{\displaystyle\sum{i=1}^{n} C{i}V}{V}

   = \displaystyle\sum{i=1}^{n} C{i}\)

Parallel formula \( C{\text{eq}} = \displaystyle\sum{i=1}^{n} C_{i}\)

4. Worked Example – Series & Parallel

Problem

Three capacitors have the following values: \(C{1}=2\;\mu\text{F}\), \(C{2}=3\;\mu\text{F}\) and \(C_{3}=5\;\mu\text{F}\).

1. Find the equivalent capacitance when \(C{1}\) and \(C{2}\) are in parallel, and this combination is in series with \(C_{3}\).
2. If the whole network is connected across a 120 V supply, calculate the charge on each capacitor and the total stored energy.

Solution

1. First combine \(C{1}\) and \(C{2}\) in parallel:

   \[

   C{12}=C{1}+C_{2}=2\;\mu\text{F}+3\;\mu\text{F}=5\;\mu\text{F}.

   \]

   Now the parallel pair is in series with \(C_{3}=5\;\mu\text{F}\):

   \[

   \frac{1}{C{\text{eq}}}= \frac{1}{C{12}}+\frac{1}{C_{3}}

   =\frac{1}{5\;\mu\text{F}}+\frac{1}{5\;\mu\text{F}}

   =\frac{2}{5\;\mu\text{F}}

   \;\;\Longrightarrow\;\;

   C_{\text{eq}}= \frac{5\;\mu\text{F}}{2}=2.5\;\mu\text{F}.

   \]

2. Charge on the whole network:

   \[

   Q = C_{\text{eq}}V = (2.5\times10^{-6}\,\text{F})(120\,\text{V}) = 3.0\times10^{-4}\,\text{C}=300\;\mu\text{C}.

   \]

   In a series connection the same charge appears on each element, so

   \[

   Q{12}=Q{3}=300\;\mu\text{C}.

   \]

   The voltage across the parallel pair:

   \[

   V{12}= \frac{Q}{C{12}} = \frac{300\;\mu\text{C}}{5\;\mu\text{F}} = 60\;\text{V}.

   \]

   Hence the voltage across \(C_{3}\) is also 60 V (the total is 120 V).

   Energy stored in each capacitor:

   \[

   W{i}= \tfrac12 C{i}V_{i}^{2}.

   \]

   \[

   W_{12}= \tfrac12(5\;\mu\text{F})(60^{2})=9.0\times10^{-3}\,\text{J},

   \qquad

   W_{3}= \tfrac12(5\;\mu\text{F})(60^{2})=9.0\times10^{-3}\,\text{J}.

   \]

   Total energy:

   \[

   W{\text{total}} = W{12}+W_{3}=1.8\times10^{-2}\,\text{J}.

   \]

This example demonstrates the step‑by‑step use of the derived formulae (AO2) and the link between charge, voltage and stored energy.

5. Energy Stored in a Capacitor

1. Work required to move an infinitesimal charge dQ onto a plate at potential V is dW = V\,dQ.
2. Using the definition \(V = Q/C\) and integrating from 0 to Q:

   \[

   W = \int_{0}^{Q}\frac{Q'}{C}\,dQ' = \frac{1}{2}\frac{Q^{2}}{C}.

   \]

3. Expressed in terms of the applied voltage:

   \[

   W = \frac{1}{2} C V^{2}.

   \]

4. These expressions are a direct consequence of the first law of thermodynamics: the electrical work done on the capacitor is stored as electrostatic potential energy.

Numerical Example

A 10 µF capacitor is charged to 200 V. The stored energy is

\( W = \tfrac12(10\times10^{-6}\,\text{F})(200^{2}\,\text{V}^{2}) = 0.20\;\text{J}.\)

6. Discharging a Capacitor – RC Time Constant

• When a charged capacitor is connected across a resistor R, the charge and voltage decay exponentially:

  \[

  Q(t)=Q_{0}\,e^{-t/\tau},\qquad

  V(t)=V_{0}\,e^{-t/\tau},

  \]

  where the time constant \(\tau\) is

  \[

  \tau = RC\;\;[\text{s}].

  \]

• After a time \(t=\tau\) the charge (or voltage) has fallen to 37 % of its initial value.
• For a series RC circuit the instantaneous current is

  \[

  I(t)=\frac{V_{0}}{R}\,e^{-t/\tau}.

  \]

Example

A 4.7 µF capacitor discharges through a 10 kΩ resistor. The time constant is \(\tau = (10^{4}\,\Omega)(4.7\times10^{-6}\,\text{F}) = 0.047\;\text{s}\). After 0.047 s the voltage is 0.37 V₀.

7. Effect of a Dielectric

• Inserting a dielectric of relative permittivity \(\kappa\) between the plates multiplies the capacitance by \(\kappa\):

  \[

  C = \kappa\,\varepsilon_{0}\,\frac{A}{d}.

  \]

• Typical values: \(\kappa{\text{mica}}\approx5\), \(\kappa{\text{paper}}\approx3\), \(\kappa_{\text{plastic}}\approx2\).
• The dielectric reduces the electric field for a given charge, allowing a higher voltage before dielectric breakdown.

8. Capacitance of Other Geometries (Formulae Only)

These expressions illustrate that the definition \(C=Q/V\) is universal; geometry merely determines the proportionality constant.

9. Practical Measurement of Capacitance

• Bridge methods (e.g., Wheatstone‑type capacitance bridge) give high accuracy for laboratory investigations.
• Digital multimeters often include a capacitance‑measurement function; record the reading with its uncertainty (typically ±1 % + 1 digit).
• When measuring a network, first record the total capacitance, then compare with the theoretical value obtained from the derived series/parallel formulae – a common AO3 requirement.

10. Common Applications (linked to other syllabus topics)

• Energy storage – Flash cameras, defibrillators (Topic 22 – Medical Physics).
• RC timing circuits – Delay generators, waveform shaping (Topic 21 – Alternating Currents).
• Filtering & smoothing – Power‑supply rectifiers; a large parallel capacitor reduces ripple voltage (Topic 21).
• Coupling & decoupling – Pass AC signals while blocking DC in audio and radio circuits (Topic 20 – Fields and Waves).

11. Summary of Combined‑Capacitance Formulae

12. Exam‑Technique Checklist (AO1 & AO2)

1. Identify the connection type from the diagram – pure series, pure parallel, or a combination.
2. Write the fundamental definition \(C = Q/V\) before manipulating symbols.
3. Series networks: assume a common charge Q, express each voltage as \(Vi = Q/Ci\), then sum to obtain \(V_{\text{total}}\).
4. Parallel networks: assume a common voltage V, express each charge as \(Qi = CiV\), then sum to obtain \(Q_{\text{total}}\).
5. Calculate the equivalent capacitance, then use it to find energy (\(\tfrac12 C V^{2}\)) or the RC time constant (\(\tau = RC\)) if required.
6. Check limiting cases:

   • If any series capacitor → 0 F, the whole network → 0 F.
   • If any parallel capacitor → ∞ F, the whole network → ∞ F.

7. State units clearly and keep significant figures consistent with the data given.
8. When a dielectric is mentioned, multiply the calculated capacitance by its \(\kappa\) and comment on the effect on breakdown voltage.
9. For experimental questions, describe the measurement method, record uncertainties and compare measured and theoretical values.

These notes satisfy all outcomes for Topic 19 – Capacitors and Capacitance in the Cambridge International AS & A Level Physics syllabus, providing clear derivations, worked examples, extensions to energy and discharge, practical considerations, and exam‑focused guidance.