Combination of Capacitors

Introduction

In many electrical circuits, we need a specific value of capacitance that may not be available as a single component. To achieve the desired value, we combine multiple capacitors. There are two fundamental ways to connect them: Series and Parallel.

Capacitors in Series

In a series combination, capacitors are connected end-to-end like a chain so that the same charge $Q$ flows through each of them when connected to a battery.

1. The Logic:

• The charge $Q$ on each capacitor is the same.
• The total potential difference $V$ across the combination is the sum of potential differences across individual capacitors: $V = V_1 + V_2 + V_3$.

2. Derivation: We know that for any capacitor, $V = Q/C$. Substituting this for each capacitor: $$V_1 = \frac{Q}{C_1}, \quad V_2 = \frac{Q}{C_2}, \quad V_3 = \frac{Q}{C_3}$$ If $C_{eq}$ is the equivalent capacitance of the series combination: $$V = \frac{Q}{C_{eq}}$$ Substituting these into the total voltage equation: $$\frac{Q}{C_{eq}} = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3}$$ Dividing both sides by $Q$, we get the Series Formula: $$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots$$

Key Characteristics:

• The equivalent capacitance is smaller than the smallest individual capacitance.
• Voltage divides in the inverse ratio of capacitance: $V_1 : V_2 : V_3 = \frac{1}{C_1} : \frac{1}{C_2} : \frac{1}{C_3}$.

Capacitors in Parallel

In a parallel combination, all the positive plates are connected to one point and all negative plates to another. This ensures that every capacitor experiences the same potential difference $V$.

1. The Logic:

• The potential difference $V$ across each capacitor is the same.
• The total charge $Q$ supplied by the battery is the sum of charges on individual capacitors: $Q = Q_1 + Q_2 + Q_3$.

2. Derivation: Using the relation $Q = CV$: $$Q_1 = C_1V, \quad Q_2 = C_2V, \quad Q_3 = C_3V$$ If $C_{eq}$ is the equivalent capacitance: $$Q = C_{eq}V$$ Substituting into the total charge equation: $$C_{eq}V = C_1V + C_2V + C_3V$$ Dividing both sides by $V$, we get the Parallel Formula: $$C_{eq} = C_1 + C_2 + C_3 + \dots$$

Key Characteristics:

• The equivalent capacitance is larger than the largest individual capacitance.
• Charge divides in the direct ratio of capacitance: $Q_1 : Q_2 : Q_3 = C_1 : C_2 : C_3$.

🧠 Memory Capsule

• Series = Sum of Reciprocals: Good for reducing capacitance or sharing high voltage.
• Parallel = Direct Sum: Good for increasing total charge storage capacity.
• Quick Rule for 2 Capacitors in Series: $C_{eq} = \frac{C_1 C_2}{C_1 + C_2}$ (Product over Sum).
• Identical Capacitors: $n$ identical capacitors in series give $C/n$; in parallel they give $nC$.

Example 1: Basic Series Calculation Three capacitors of $2 \, \mu F, 3 \, \mu F$, and $6 \, \mu F$ are connected in series. Find the equivalent capacitance.

Solution:

1. Formula: $$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3}$$
2. Substitute: $$\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$$
3. Calculation: $$\frac{1}{C_{eq}} = \frac{3+2+1}{6} = \frac{6}{6} = 1 \; \mu F^{-1}$$
4. Final Answer: $$C_{eq} = 1 \, \mu F$$

Example 2: Basic Parallel Calculation Three capacitors of $10 \, pF, 20 \, pF$, and $30 \, pF$ are connected in parallel. Find the equivalent capacitance.

Solution:

1. Formula: $$C_{eq} = C_1 + C_2 + C_3$$
2. Substitute: $$C_{eq} = 10 + 20 + 30 = 60 \, pF$$
3. Final Answer: $$C_{eq} = 60 \, pF$$

Example 3: Mixed Combination Two $4 \, \mu F$ capacitors are in parallel, and this combination is in series with a $12 \, \mu F$ capacitor. Find the equivalent capacitance.

Solution:

1. Parallel part: $$C_p = 4 + 4 = 8 \, \mu F$$
2. Series with $12 \, \mu F$: $$C_{eq} = \frac{8 \times 12}{8 + 12} = \frac{96}{20} = 4.8 \, \mu F$$
3. Final Answer: $$C_{eq} = 4.8 \, \mu F$$

Example 4: Voltage Division in Series Two capacitors $3 \, \mu F$ and $6 \, \mu F$ are in series across a $12 \, V$ battery. Find the voltage across each.

Solution:

1. Equivalent capacitance: $$C_{eq} = \frac{3 \times 6}{3 + 6} = 2 \, \mu F$$
2. Common charge in series: $$Q = C_{eq}V = 2 \, \mu F \times 12 \, V = 24 \, \mu C$$
3. Voltage across $3 \, \mu F$ capacitor: $$V_1 = \frac{Q}{C_1} = \frac{24}{3} = 8 \, V$$
4. Voltage across $6 \, \mu F$ capacitor: $$V_2 = \frac{Q}{C_2} = \frac{24}{6} = 4 \, V$$
5. Check: $$V_1 + V_2 = 8 + 4 = 12 \, V$$
6. Final Answer: $$V_1 = 8 \, V, \quad V_2 = 4 \, V$$

Example 5: Charge Distribution in Parallel Two capacitors $2 \, \mu F$ and $4 \, \mu F$ are in parallel across a $10 \, V$ battery. Find the charge on each.

Solution:

1. Same voltage in parallel: $$V = 10 \, V$$
2. Charge on first capacitor: $$Q_1 = C_1V = 2 \, \mu F \times 10 \, V = 20 \, \mu C$$
3. Charge on second capacitor: $$Q_2 = C_2V = 4 \, \mu F \times 10 \, V = 40 \, \mu C$$
4. Final Answer: $$Q_1 = 20 \, \mu C, \quad Q_2 = 40 \, \mu C$$

Example 6: Finding an Unknown Capacitance in Series What capacitance must be connected in series with a $100 \, pF$ capacitor to give an equivalent capacitance of $75 \, pF$?

Solution:

1. Series formula: $$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_x}$$
2. Substitute values: $$\frac{1}{75} = \frac{1}{100} + \frac{1}{C_x}$$
3. Solve: $$\frac{1}{C_x} = \frac{1}{75} - \frac{1}{100} = \frac{4-3}{300} = \frac{1}{300}$$
4. Final Answer: $$C_x = 300 \, pF$$

Example 7: Identical Capacitors in Series and Parallel If $n$ identical capacitors each of capacitance $C$ are first connected in series and then in parallel, find the ratio $C_p/C_s$.

Solution:

1. Series equivalent capacitance: $$C_s = \frac{C}{n}$$
2. Parallel equivalent capacitance: $$C_p = nC$$
3. Ratio: $$\frac{C_p}{C_s} = \frac{nC}{C/n} = n^2$$
4. Final Answer: $$\frac{C_p}{C_s} = n^2$$