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

A capacitor stores electric charge. Think of it like a water tank: the charge (Q) is the amount of water, the voltage (V) is the water pressure, and the capacitance (C) is the size of the tank.

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

For a parallel‑plate capacitor:

• Area of plates: \$A\$
• Distance between plates: \$d\$
• Permittivity of the dielectric: \$\varepsilon\$

Then \$C = \varepsilon \frac{A}{d}\$

When capacitors are connected side‑by‑side, the voltage across each is the same, but the charges add.

1. Let \$C1, C2, \dots, C_n\$ be the individual capacitances.
2. Same voltage \$V\$ on each.
3. Charges: \$Qi = Ci V\$.
4. Total charge: \$Q{\text{tot}} = \sum{i=1}^{n} Qi = V \sum{i=1}^{n} C_i\$.
5. Using \$C{\text{eq}} = Q{\text{tot}}/V\$ gives \$C{\text{eq}} = \sum{i=1}^{n} C_i\$.

🔌 Example: Two 4 µF capacitors in parallel give \$C_{\text{eq}} = 4\,\mu\text{F} + 4\,\mu\text{F} = 8\,\mu\text{F}\$.

When capacitors are connected end‑to‑end, the charge on each is the same, but the voltages add.

1. Let \$C1, C2, \dots, C_n\$ be the individual capacitances.
2. Same charge \$Q\$ on each.
3. Voltages: \$Vi = Q/Ci\$.
4. Total voltage: \$V{\text{tot}} = \sum{i=1}^{n} Vi = Q \sum{i=1}^{n} \frac{1}{C_i}\$.
5. Using \$C{\text{eq}} = Q/V{\text{tot}}\$ gives \$\frac{1}{C{\text{eq}}} = \sum{i=1}^{n} \frac{1}{C_i}\$.

⚡️ Example: Two 4 µF capacitors in series give \$\frac{1}{C{\text{eq}}} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\$ so \$C{\text{eq}} = 2\,\mu\text{F}\$.

• 🔍 Remember the key difference: In parallel, voltages are equal; in series, charges are equal.
• 🧮 Quick formula check: For two capacitors, \$C{\text{eq,parallel}} = C1 + C2\$ and \$C{\text{eq,series}} = \frac{C1 C2}{C1 + C2}\$.
• 📚 Use the water‑tank analogy: It helps visualise why charges add in parallel and voltages add in series.
• ✏️ Practice: Work through at least 3 problems of each type before the exam.