# Capacitors in Series and Parallel Circuits ## Capacitors in Series - Consider two parallel plate capacitors $C_1$ and $C_2$ connected in series with a total potential difference (p.d.) $V$ across them

Capacitors in series have different p.d. across them but store the same charge

- In a series circuit, p.d. is shared between all components - If the capacitors store the same charge on their plates but have different p.d.s, then: - The p.d. across $C_1$ is $V_1$ - The p.d. across $C_2$ is $V_2$ - The total potential difference $V$ is the sum of the individual p.d.s: $$V = V_1 + V_2$$ - Rearranging the capacitance equation for p.d., $V$, means $V_1$ and $V_2$ can be written as: $$V_1 = \frac{Q}{C_1} \quad \text{and} \quad V_2 = \frac{Q}{C_2}$$ - The total p.d. $V$ is defined by the total capacitance: $$V = \frac{Q}{C_{total}}$$ - Substituting these into the equation $V = V_1 + V_2$ gives: $$\frac{Q}{C_{total}} = \frac{Q}{C_1} + \frac{Q}{C_2}$$ - Since the current is the same through all components in a series circuit, the charge $Q$ is the same through each capacitor and cancels out

For capacitors in series: 1/C_total = 1/C₁ + 1/C₂ + 1/C₃ + …

## Capacitors in Parallel - Consider two parallel plate capacitors $C_1$ and $C_2$ connected in parallel, each with the same p.d. $V$

Capacitors in parallel have the same p.d. across them but store different charge

- Since the current splits at each junction in a parallel circuit, the charge stored on each capacitor is different - Therefore: - The charge on capacitor $C_1$ is $Q_1$ - The charge on capacitor $C_2$ is $Q_2$ - The total charge $Q$ is the sum of the individual charges: $$Q = Q_1 + Q_2$$ - Rearranging the capacitance equation for charge $Q$ means $Q_1$ and $Q_2$ can be written as: $$Q_1 = C_1V \quad \text{and} \quad Q_2 = C_2V$$ - The total charge $Q$ is defined by the total capacitance: $$Q = C_{total}V$$ - Substituting these into $Q = Q_1 + Q_2$ gives: $$C_{total}V = C_1V + C_2V = (C_1 + C_2)V$$ - Since the p.d. is the same through all components in each branch of a parallel circuit, the p.d. $V$ cancels out

For capacitors in parallel: C_total = C₁ + C₂ + C₃ + …

## Solving Problems with Capacitors in Series & Parallel - For capacitors in series: $$\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + ...$$ - For capacitors in parallel: $$C_{total} = C_1 + C_2 + C_3 + ...$$ ### Worked Example Three capacitors with capacitance of 15 μF, 25 μF and 50 μF are connected as shown below, where two capacitors are in parallel and one is in series with this combination. Calculate the total capacitance between the two terminals. **Answer:** **Step 1: Calculate the combined capacitance of the two capacitors in parallel** - Capacitors in parallel: $C_{total} = C_1 + C_2 + C_3 + ...$ - $C_{parallel} = 15 + 25 = 40 \text{ μF}$ **Step 2: Connect this combined capacitance with the final capacitor in series** - Capacitors in series: $\dfrac{1}{C_{total}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} + ...$ $$\frac{1}{C_{total}} = \frac{1}{40} + \frac{1}{50} = \frac{90}{2000} = \frac{9}{200}$$ **Step 3: Rearrange for the total capacitance** $$C_{total} = \frac{200}{9} = 22.2... = 22 \text{ μF (2 s.f.)}$$

Combined capacitance equations are opposite to those for resistors in series and parallel

## Key Comparisons | Configuration | Capacitors | What's the Same | What's Different | |---------------|------------|-----------------|------------------| | **Series** | $\dfrac{1}{C_{total}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + ...$ | Charge $Q$ is the same on each | P.d. $V$ is different across each | | **Parallel** | $C_{total} = C_1 + C_2 + ...$ | P.d. $V$ is the same across each | Charge $Q$ is different on each | ## Special Cases ### Two Identical Capacitors in Series - If two capacitors of equal capacitance $C$ are connected in series: $$\frac{1}{C_{series}} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$$ - Therefore: $C_{series} = \dfrac{C}{2}$ - The combined capacitance is **half** the capacitance of one capacitor ### Two Identical Capacitors in Parallel - If two capacitors of equal capacitance $C$ are connected in parallel: $$C_{parallel} = C + C = 2C$$ - The combined capacitance is **double** the capacitance of one capacitor ## Practical Tips - When solving complex circuits with both series and parallel combinations: 1. Identify which capacitors are in parallel and combine them first 2. Then identify which capacitors (or combinations) are in series 3. Work systematically through the circuit, simplifying step by step 4. Draw simplified circuit diagrams at each stage to track your progress - Always check your answer makes physical sense: - Series combinations should give a **smaller** total capacitance than the smallest individual capacitor - Parallel combinations should give a **larger** total capacitance than the largest individual capacitor ## Graph Analysis for Series vs Parallel - When investigating capacitors experimentally, plotting $Q$ against $V$ gives a straight line through the origin - The gradient equals the capacitance: $\text{gradient} = \dfrac{\Delta Q}{\Delta V} = C$ - For identical capacitors: - The gradient for the **parallel** combination is steeper (higher capacitance) - The gradient for the **series** combination is shallower (lower capacitance) - Specifically: $C_{parallel} > C_{series}$ | Configuration | Gradient | Comparison | |---------------|----------|------------| | Single capacitor $C$ | $C$ | Reference | | Two identical in parallel | $2C$ | Double the gradient | | Two identical in series | $C/2$ | Half the gradient | ### Graphical Comparison The following chart shows the relationship between charge and potential difference for two identical capacitors (each with $C = 10$ μF) in different configurations:

- Notice how the **parallel** line has the steepest gradient (highest capacitance = 20 μF) - The **series** line has the shallowest gradient (lowest capacitance = 5 μF) - The **single capacitor** line sits between them (capacitance = 10 μF) - All three lines pass through the origin, confirming $Q = 0$ when $V = 0$