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

Capacitors and Capacitance

Objective

Derive, using the definition \$C = \dfrac{Q}{V}\$, the formulae for the equivalent capacitance of a group of capacitors when they are connected (a) in series and (b) in parallel.

1. Review of a Single Capacitor

A capacitor stores electric charge \$Q\$ on its plates at a potential difference \$V\$. The capacitance \$C\$ is defined as

\$C = \frac{Q}{V}\$

where \$C\$ is measured in farads (F). For a parallel‑plate capacitor with plate area \$A\$ and separation \$d\$, the theoretical expression is \$C = \varepsilon_0 \dfrac{A}{d}\$, but the derivations below rely only on the definition above.

2. Capacitors Connected in Series

When capacitors are placed end‑to‑end, the same charge \$Q\$ flows onto each plate because the plates in the interior are isolated from the external circuit.

1. Let the voltage across the \$i^{\text{th}}\$ capacitor be \$Vi\$. By definition, \$Ci = \dfrac{Q}{Vi}\$, so \$Vi = \dfrac{Q}{C_i}\$.
2. The total voltage between the ends of the series combination is the sum of the individual voltages:

   \$V{\text{total}} = \sum{i=1}^{n} Vi = \sum{i=1}^{n} \frac{Q}{C_i}.\$

3. Define the equivalent capacitance \$C{\text{eq}}\$ for the whole series group by \$C{\text{eq}} = \dfrac{Q}{V{\text{total}}}\$. Substituting the expression for \$V{\text{total}}\$ gives

   \$C{\text{eq}} = \frac{Q}{\displaystyle\sum{i=1}^{n} \frac{Q}{Ci}} = \frac{1}{\displaystyle\sum{i=1}^{n} \frac{1}{C_i}}.\$

Thus, for capacitors in series,

\$\boxed{\displaystyle\frac{1}{C{\text{eq}}} = \sum{i=1}^{n} \frac{1}{C_i}}.\$

3. Capacitors Connected in Parallel

In a parallel configuration each capacitor experiences the same potential difference \$V\$ because their terminals are joined together.

1. For the \$i^{\text{th}}\$ capacitor, \$Qi = Ci V\$ (from \$C = Q/V\$).
2. The total charge stored on the combination is the sum of the individual charges:

   \$Q{\text{total}} = \sum{i=1}^{n} Qi = \sum{i=1}^{n} C_i V.\$

3. Define the equivalent capacitance \$C{\text{eq}}\$ by \$C{\text{eq}} = \dfrac{Q{\text{total}}}{V}\$. Substituting the expression for \$Q{\text{total}}\$ gives

   \$C{\text{eq}} = \frac{\displaystyle\sum{i=1}^{n} Ci V}{V} = \sum{i=1}^{n} C_i.\$

Hence, for capacitors in parallel,

\$\boxed{C{\text{eq}} = \sum{i=1}^{n} C_i}.\$

4. Summary of Combined Capacitance Formulae

5. Practical Implications for A‑Level Exams

• When a problem states that several capacitors are “connected in series”, always start by assuming a common charge \$Q\$ and write the voltage across each capacitor as \$Vi = Q/Ci\$.
• For a parallel network, write the charge on each capacitor as \$Qi = Ci V\$ and sum the charges.
• Remember to keep track of units: \$1\;\text{F} = 1\;\text{C/V}\$. Typical A‑Level values are in microfarads (\$\mu\text{F}\$) or nanofarads (nF).
• Check your final answer by confirming the dimensions are those of capacitance and that the limiting cases (e.g., one capacitor becoming very large or very small) behave as expected.