Combination of Capacitors

In practical circuits, a single capacitor is rarely sufficient. Engineers combine capacitors to achieve a desired capacitance value, withstand higher voltages, or store more charge. The process of replacing a group of capacitors with a single equivalent capacitor that produces the same electrical effect is called the combination of capacitors.

Real-Life Analogy: Think of capacitors like water tanks in a pipeline. Connecting tanks end-to-end (series) forces the same water flow (charge) through each, but the total pressure (voltage) is shared. Connecting tanks side-by-side (parallel) keeps the same pressure (voltage) across all, but accumulates more total water (charge).

The capacitance of a single capacitor that stores the same charge at the same voltage as the entire combination is called the equivalent capacitance of the combination.

The work done per unit charge in moving a charge from one plate of a capacitor to the other is called the potential difference between the plates.

Circuit Configuration

Capacitors are connected end-to-end, so there is only one path for charge to flow. The positive plate of one capacitor connects to the negative plate of the next.

 

Fig 1: Three capacitors C₁, C₂, C₃ in series. Same charge Q on each; voltages add up.

Key Properties

• Charge is the SAME on every capacitor: Q₁ = Q₂ = Q₃ = Q
• Voltage DIVIDES across capacitors: V = V₁ + V₂ + V₃
• Equivalent capacitance is always less than the smallest individual capacitor

Derivation

Since charge Q is the same on all capacitors:

Total voltage across the combination:

If C_S​ is the equivalent capacitance, then V = Q/C_S. Substituting:

\[{\frac{1}{C_S}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\cdots}\]

For n identical capacitors of capacitance C each: C_S = \[\frac {C}{n}\]

For two capacitors in series, the voltage across each is:

\[V_1=\frac{C_2}{C_1+C_2}\cdot V\]

\[V_2=\frac{C_1}{C_1+C_2}\cdot V\]

Physical Insight: The smaller the capacitor, the larger the voltage drop across it in a series combination. This is why identical series capacitors share voltage equally.

Circuit Configuration

Capacitors are connected so that all positive plates share one terminal and all negative plates share another terminal, providing multiple paths.

Fig 2: Capacitors C₁, C₂, C₃ in parallel. Same voltage V; charges add up.

Key Properties

• Voltage is the SAME across every capacitor: V₁ = V₂ = V₃ = V
• Charge DIVIDES among capacitors: Q = Q₁ + Q₂ + Q_3​
• The equivalent capacitance is always greater than the largest individual capacitor

Derivation

Since voltage V is the same across all:

Total charge: Q = Q₁ + Q₂ + Q₃ = (C₁ + C₂ + C₃)V

If C_P​ is the equivalent capacitance, then Q = C_PV. Substituting:

\[{C_P=C_1+C_2+C_3+\cdots}\]

For n identical capacitors of capacitance C each: C_P = nC

Physical Insight: Adding capacitors in parallel is like adding more storage tanks — the total storage capacity simply increases.

Question: When 10⁸ electrons are transferred from one conductor to another, a potential difference of 10 V appears between the conductors. Find the capacitance of the two conductors.

Step‑by‑step explanation

1. Identify what is given:

• Number of electrons moved: n = 10⁸
• Potential difference between conductors: V = 10 V
• Charge of one electron: e = 1.6 × 10⁻¹⁹ C

2. Find total charge transferred (Q):

• Each electron carries a charge e, so the total charge moved is: Q = ne = 10⁸ × 1.6 × 10⁻¹⁹ = 1.6 × 10⁻¹¹ C

3. Use the definition of capacitance:

• Capacitance is defined as: C = \[\frac {Q}{V}\]

4. Substitute values:

5. Interpretation:

• The pair of conductors behaves like a capacitor with a capacitance of 1.6 × 10⁻¹² F (1.6 pF).
• A small capacitance means a small charge produces a noticeable potential difference.

Question: In the circuit, the equivalent capacitance between A and B must be 1 μF. All other capacitors (C₁–C₅) are in μF. Find the unknown capacitance C.

Given:
C₁ = 8, C₂ = 4, C₃ = 1, C₄ = 4, C₅ = 4 (all in μF).

Step‑by‑step explanation:

1. First parallel combination: C₄ and C₅

• ​They are in parallel, so capacitances add: C₄₅ = C₄ + C₅ = 4 + 4 = 8 μF

2. Series combination: C₃​ and C₄₅ = 8

• Series formula: \[C_{\mathrm{series}}=\frac{C_3\cdot C_{45}}{C_3+C_{45}}=\frac{1\times8}{1+8}=\frac{8}{9}\mu\mathrm{F}\]

3. Parallel combination: C₁, C₂, and this series result:

• The capacitance 8 μF is in parallel with the series combination of C₁ and C₂. Their effective combination is
  \[\frac{C_1C_2}{C_1+C_2}+\frac{8}{9}\Rightarrow\frac{8\times4}{12}+\frac{8}{9}=\frac{32}{12}+\frac{8}{9}\]
• Simplify:
  \[\frac{32}{12}=\frac{8}{3},\quad\frac{8}{3}+\frac{8}{9}=\frac{24}{9}+\frac{8}{9}=\frac{32}{9}\mu\mathrm{F}\]
• So net capacitance of that part = \[\frac {32}{9}\] μF.

4. Series combination of this \[\frac {32}{9}\]​ μF with unknown C:

• This series combination is given to be 1 μF:
  \[\frac{\left(\frac{32}{9}\right)C}{\left(\frac{32}{9}\right)+C}\] ​= 1μF

5. Solve for C (conceptual explanation):

• The equation means: “Series combination of \[\frac {32}{9}\]​ μF and C gives 1 μF.”
• Rearranging, you’d solve algebraically for C. The important idea is to use the series formula backwards to find the unknown C once the total C_eq is specified.

Capacitors in Series:

Equivalent capacitance: \[\frac{1}{C_s}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}+\cdots\]

• Same voltage (V) across all capacitors
• Charge divides
• The equivalent capacitance is greater than the largest capacitor

Capacitors in Parallel:

\[C_p=C_1+C_2+C_3+\cdots\]

• Same voltage (V) across all capacitors
• Charge divides
• The equivalent capacitance is greater than the largest capacitor