Capacitors in Series and Parallel: Formula, Derivation and Examples

Capacitors in Series and in Parallel – Explanation, Formula, and Examples

18.5 CAPACITORS IN SERIES AND IN PARALLEL

18.5.1 Capacitors Connected in Series

A dc battery of emf V volt is connected to three capacitors of capacitance C₁, C₂ and C₃ connected in series, as shown in Figure 18.10(a). We wish to replace these three capacitors with a single capacitor of capacitance C_T so that this circuit with the single capacitor (Figure 18.10(b)) is equivalent to the circuit with the three capacitors in series.

Figure 18.10 (a) Three capacitors $C_1$, $C_2$, and $C_3$ connected in series across a dc battery of emf V.(b) An equivalent circuit where the three capacitors are replaced by a single capacitor of capacitance $C_T$, producing the same effect as the series combination.

The three capacitor circuit

1. The charge stored in each capacitor is the same.
   
   Explanation:

   (a) Refer to the circuit shown in Figure 18.10(a). Suppose that all the capacitors are initially uncharged. Plate J is then connected to the positive terminal of the battery and plate P is connected to the negative terminal.

   (b) Free electrons in J begin to move out of the plate and head towards the positive terminal of the battery. The plate becomes positively charged. The electron flow will continue until that capacitor has become fully charged. Plate J will have acquired a positive charge +Q.

   (c) Plate K is near to J. Since J is positively charged, free electrons from plate L are drawn into K by means of electrostatic induction. The electron flow will stop when the charge acquired by K is -Q.

   (d) Since free electrons flow out of plate L, the plate becomes positively charged. When K has acquired charge -Q from L, it means that finally L will hold a charge of +Q.

   (e) This process is repeated by plates M and N.

   (f) Plate P is connected to the negative terminal of the battery. When plate N becomes positively charged, electrons will be forced to move out of the negative terminal and into plate P.

   (g) When all three capacitors become fully charged at the same time, the charge stored in each plate is of magnitude Q. This means that the charge stored in each capacitor is the same.

2. After all the capacitors have become fully charged, the total amount of charge that has flowed in the external circuit and through the battery is Q (not 3Q). This is because of the fact that the amount of charge that has been 'sucked' out of plate J is +Q and the amount of charge that has been 'pumped' into plate P is -Q. Those electrons transferred from plates L to K and from N to M are not found in the external circuit containing the battery.
3. The p.d. across each fully charged capacitor may not be the same. If the p.d. across C₁, C₂, C₃ are V₁, V₂, V₃ respectively, then we have

   $\boxed{V = V_1 + V_2 + V_3}$

where V is the emf of the battery.

The single capacitor equivalent circuit:

When the single capacitor becomes fully charged,

1. it has to store charge of Q coulomb
2. the p.d. across the capacitor has to be equal to V, the emf of the battery

To determine the total capacitance C_T, use the expression

$\boxed{V = V_1 + V_2 + V_3}$

But: $V = \frac{Q}{C_T}$; $V_1 = \frac{Q}{C_1}$; $V_2 = \frac{Q}{C_2}$; $V_3 = \frac{Q}{C_3}$

Hence:

$\boxed{\frac{1}{C_T}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}}$

Example 18.5

Figure 18.11 A constant voltage of 20.0 V is applied to two capacitors connected in series

A constant voltage of 20.0 V is applied to two capacitors connected in series, as shown in Figure 18.11. Determine

(a) the total capacitance in the circuit
(b) the charge stored in each capacitor
(c) the p.d. across each capacitor

(a) Total capacitance

1/C_T = 1/6 + 1/10 = 4/15

C_T = 3.75 μF ≈ 3.8 μF

(b) The charge stored in the single equivalent capacitor is given by

Q = C_T V = (3.75)(20) = 75 μC

charged stored in $C_1$ = charged stored in $C_2$ = charged stored in $C_T$ = 75 μC

Charge is same in each capacitor: Q = 75 μC

(c) The potential difference $V_1$ across $C_1$ is given by

$V_1=\frac{Q}{C_1}=\frac{75 \ \mu C}{6 \ \mu F}=12.5 \ V$

The potential difference $V_2$ across $C_2$ is given by

$V_2=\frac{Q}{C_2}=\frac{75 \ \mu C}{10 \ \mu F}=7.5 \ V$

Exercise 18.3

Three capacitors are connected in series.

Charge in each capacitor = 20 μC

Voltages:

• 2.0 V
• 3.0 V
• 7.0 V

Find total capacitance.

18.5.2 Capacitors Connected in Parallel

Three capacitors C₁, C₂, C₃ are connected in parallel, and a dc battery of emf V volt is applied to the parellel combination, as shown in Fugure 18.12(a). We wish to replace these three capacitors with a single capacitor of capacitanci $C_T$ so that this circuit with the single capacitor (Figure 18.12(b)) is equivalent to the circuit with the three capacitors in parallel.

Figure 18.12 (a) Three capacitors $C_1$, $C_2$, and $C_3$ connected in parallel across a dc source of emf V.(b) An equivalent circuit where the three capacitors are replaced by a single capacitor of capacitance $C_T$, producing the same effect as the parallel combination.

The three capacitor circuit

• The p.d. across each capacitor is the same and is equal to the applied voltage V
• The charges stored in the capacitors may be different.
• After all the capacitors have become fully charged, the total amount of charge Q that has flowed in the external circuit and through the battery is given by

$\boxed{Q = Q_1 + Q_2 + Q_3}$

where, $Q_1$, $Q_2$, $Q_3$ are the charges stored in $C_1$, $C_2$ and $C_3$ respectively.

The single capacitor equivalent circuit:

1. The charge Q stored by the single capacitor is given by

2. The p.d. across the capacitor is equal to the applied voltage V.

To determine the total capacitance $C_T$, use the expression

$\boxed{Q = Q_1 + Q_2 + Q_3}$

We have $Q_1=C_1V$, $Q_2=C_2V$, $Q_3=C_3V$, $Q=C_TV$

Hence, $C_TV=C_1V+C_2V+C_3V$

$\boxed{C_T=C_1+C_2+C_3}$

Example 18.6

Figure 18.13 Three capacitors are connected as shown in Figure 18.13. A dc battery of 12.0 V is applied across XY

Three capacitors are connected as shown in Figure 18.13. A dc battery of 12.0 V is applied across XY. Determine

(a) the total capacitance across XY
(b) the p.d. across the 4 μF capacitor
(c) the charge stored in each fully charged capacitor

Answer

(a) The total capacitance $C_\parallel$ of the parallel capacitors 2 μF and 3 μF is given by

$C_\parallel= 2 + 3 = 5 \ \mu F$

Notoce that $C_\parallel$ is in series with the 4 μF capacitor. Hence the total capacitance $C_{xy}$ across XY is given by

Series with 4 μF:

$\frac{1}{C_{xy}} = \frac{1}{4} + \frac{1}{5} =\frac{9}{20}$

$C_{xy} = 2.22 \ \mu F \approx  2.2 \ \mu F$

(b)  

Figure 18.14 An equivalent circuit

An equivalent circuit is shown in Figure 18.14. Since the tow capacitors  are connected in series we have 

Charge stored in 4 μC capacitor = charge stored in $C_T$

= ($C_T$)(p.d. across $C_T$)

= ($2.22 \ \mu F$)(12 V)

$\cong 26.7 \ \mu F$

p.d across 4 μC capacitor $V=\frac{Q}{C}$

$=\frac{26.67 \ \mu F}{4 \ \mu F}=6.67 \ V$

(c) 4 μC capacitor charged stored = 26.7 μC

p.d. across capacitor $C_\parallel$ = 12 - V

= 12.0 - 6.67 = 5.33 V

2 μC capacitor: charged stored = (2 μF)(5.33 V)

= 10.66 μC $\cong 10.7 \ \mu F$

3 μC capacitor: charged stored = (3 μF)(5.33 V)

= 15.99 μC $\cong 16.0 \ \mu F$

Note: Notice that

charge in 4 μF capacitor (26.7 μC) = charge in 2 μF capacitor (10.7 μF) + charge in 3 μF (16.0 4 μC)

Example 18.7

A capacitor $C_1$ of capacitance 20 μF is charged fully by a 6.0 V voltage supply and after that tje voltage supply is removed. An uncharged capacitor $C_2$ of capacitance 5.0 μF is then connected directly across the 20 μF capacitor. Determine

(a) the p.d. across each capacitor
(b) the charge stored in each capacitor

Figure 18.15 A capacitor $C_1$ of capacitance 20 μF is charged fully by a 6.0 V voltage supply and after that tje voltage supply is removed

Answer

(a) Refer to Figure 18.15(a). The initial charge $Q_0$ stored in $C_1$ is given by

$Q_0=C_1V$

$=(20)(6)=120 \ \mu C$

Refer to Figure 18.15(b). After connecting $C_2$ to $C_1$, let

$Q_1=$ charge stored in $C_1$

$Q_2=$ charge stored in $C_2$

$V_1=$ p.d. across the combination

Since the p.d. across each capacitor is the same, the combination can be cinsidered to be a parellel combination. We can substitute this parallel combination with a single capacitor C whose capacitance given by

$C=C_1+C_2$

$C=20+5=25 \ \mu F$

The charge stored in the single capacitor has to be equal to the initial amount of charged $Q_0$.

$Q_0=CV_1$

$V_1=\frac{Q_0)}{C}$

$V_1=\frac{120}{25}=4.8 \ V$

(b) $Q_1=C_1V$

$Q_1=(20)(4.8)=96 \ \mu C$

$Q_2=Q_0-C_1V$

$Q_1=120 -96 =24 \ \mu C$

---

Example 18.8

Find the equivalent capacitance between a and b for the combination of capacitors shown in Figure 18.16(a). All capacitances are in microfarads.

Solution

We reduce the combination step by step as indicated in the figure. The $1.0 \ \mu F$ and $1.0 \ \mu F$ capacitors are in parallel and combine according to the expression $C_{eq}=C_1+C_2=4 \ \mu F$. The $2.0 \mu F$ and $6.0 \mu F$ capacitors also are in parallel and have an equivalent capacitance of $8.0 \mu F$. Thus, the upper branch in Figure 4b consists of two $4.0 \mu F$ capacitors in series, which combine as follows:

$\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$

$\frac{1}{C_{eq}}=\frac{1}{4.0 \mu F}+\frac{1}{4.0 \mu F}=\frac{1}{2.0 \mu F}$

$C_{eq}=2 \ \mu F$

Fig. 18.16 A network of six capacitors arranged in a combination of series and parallel connections between points aand b, used to find the equivalent capacitance.

The lower branch in Figure 18.16(b) consists of two $8.0 \mu F$ capacitors in series, which combine to yield an equivalent capacitance of $4.0 \mu F$. Finally, the $2.0 \mu F$ and $4.0 \mu F$ capacitors in Figure 18.16(c) are in parallel and thus have an equivalent capacitance of $6.0 \mu F$.

---

Example 18.9

Two capacitors of $5 \mu F$ and $25 \mu F$ are connected in parallel and their terminals are connected to a 6 V voltage source. Calculate:

• (a) the equivalent capacitance
• (b) the total charge
• (c) the charge and potential difference of each capacitor

Answer

(a) Equivalent capacitance

$C_{eq}=C_1+C_2$

$C_{eq}=5 \ \mu F+25 \ \mu F$

$C_{eq}=30 \ \mu F$

(b) Total charge

$q_{eq}=C_{eq} V$

$q_{eq}=(30 \ \mu F)(6 \ V)$

$q_{eq}=180 \ \mu C$

(c) Charge and potential difference of each capacitor

For capacitors connected in parallel, the potential difference across each capacitor is the same:

$V_1=V_2=V=6 \ V$

Charge on each capacitor:

$q_1=C_1 V_1$

$q_1=5 \ \mu F \times 6 \ V=30 \ \mu C$

$q_2=C_2 V_2$

$q_2=25 \ \mu F \times 6 \ V=150 \ \mu C$

---

Example 18.10

Determine the potential difference (voltage) and the charge stored on each capacitor in the capacitor networks shown in the circuit diagrams.

Figure 18.17 A network of four capacitors arranged in a combination of series and parallel connections between points aand b, used to find the equivalent capacitance.

Answer

Key Concepts of Capacitor Circuits

1. Capacitors in Series

The charge on each capacitor is the same

$Q_1=Q_2=Q_3$

The total voltage is the sum of individual voltages

$V=V_1+V_2+V_3$

2. Capacitors in Parallel

The voltage across each capacitor is the same

$V_1=V_2=V_3$

The total charge is the sum of charges

$Q=Q_1+Q_2+Q_3$

Step 1: Combine Capacitors $C_1$ and $C_2$

Capacitors $C_1$ and $C_2$ are connected in series.

$C_{12}=\frac{C_1 C_2}{C_1+C_2 }$

$C_{12}=\frac{(10 \ \mu F)(2 \ \mu F)}{10 \ \mu F+2 \ \mu F}$

$C_{12}=\frac{5}{3} \mu F$

Step 2: Total Capacitance

Since $C_{12}$ is parallel with $C_3$:

$C_{eq}=C_{12}+C_3$

$C_{eq}= \frac{5}{3} \ \mu F+6 \ \mu F$

$C_{eq}=\frac{23}{3} \ \mu F$

Step 3: Total Charge

$Q_{eq}=C_{eq} V$

$Q_{eq}=\left(\frac{23}{3}\right) \ \mu F(6V)$

$Q_{eq}=46 \ \mu C$

Step 4: Voltage Across Parallel Branch

$V_{12}=V_3=6 \ V$

Step 5: Charge on $C_3$

$Q_3=C_3 V_3$

$Q_3=(6 \ \mu F)(6V)$

$Q_3=36 \ \mu C$

Step 6: Charge on $C_1$ and $C_2$

$Q_1=Q_2=Q_{12}$

$Q_{12}=Q_{eq}-Q_3$

$Q_{12}=46 \ \mu C-36 \ \mu C$

$Q_{12}=10 \ \mu C$

Step 7: Voltage on Each Capacitor

For $C_1$

$V_1=\frac{Q_1}{C_1}$

$V_1=\frac{10 \ \mu C}{10 \ \mu F}=1 \ V$

For $C_2$

$V_2=\frac{Q_2}{C_2}$

$V_2=\frac{10 \ \mu C}{2 \ \mu F}=5 \ V$

Additional example problems can be studied at the link below.

Additional Example Problems.

Applications of Capacitors in Series and in Parallel

Capacitors in Series – Applications

Increasing Voltage Rating

Capacitors connected in series are used when a higher voltage rating is required. The total voltage is shared among the capacitors, allowing the circuit to handle higher voltages.

Reducing Total Capacitance

Series connection is used when a smaller equivalent capacitance is needed in a circuit.

High-Voltage Power Systems

Used in power transmission and high-voltage equipment where components must withstand large voltages safely.

Tuning and Filtering Circuits

Used in electronic circuits to adjust capacitance values for tuning and filtering purposes.

Capacitors in Parallel – Applications

Increasing Total Capacitance

Capacitors in parallel are used to increase the total capacitance, allowing more charge to be stored.

Energy Storage Systems

Used in power supplies and backup systems to store larger amounts of energy.

Smoothing Circuits

Parallel capacitors help smooth voltage fluctuations in rectifier circuits, providing a steady DC output.

Decoupling and Bypass Applications

Used in electronic circuits to reduce noise and stabilize voltage levels.

Audio and Signal Processing

Used to improve signal quality by filtering unwanted frequencies.

Conclusion

Capacitors can be connected in series or in parallel to achieve different electrical properties depending on the requirements of a circuit. In a series connection, the total capacitance decreases while the overall voltage rating increases, making it suitable for high-voltage applications. On the other hand, in a parallel connection, the total capacitance increases, allowing the system to store more charge and energy.

Understanding these configurations is essential in designing efficient electronic circuits. Series connections are useful when controlling voltage distribution, while parallel connections are ideal for energy storage and voltage stabilization. By choosing the appropriate arrangement, capacitors can be effectively used in a wide range of electrical and electronic applications.

Frequently Asked Questions (FAQ)

1. What is the difference between capacitors in series and in parallel?

Capacitors in series have a lower total capacitance but a higher voltage rating, while capacitors in parallel have a higher total capacitance and the same voltage across each capacitor.

2. What is the formula for capacitors in series?

$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_1} + ⋯$

3. What is the formula for capacitors in parallel?

$C_{eq} = C_1 + C_2 + ⋯$

4. Why does capacitance decrease in series?

In a series connection, the effective distance between plates increases, which reduces the overall capacitance.

5. Why does capacitance increase in parallel?

In parallel, the effective plate area increases, allowing more charge to be stored.

6. Where are capacitors in series used?

They are used in high-voltage applications where the voltage needs to be distributed across multiple capacitors.

7. Where are capacitors in parallel used?

They are used in power supplies, energy storage systems, and circuits that require stable voltage.

8. Do capacitors in series have the same charge?

Yes, all capacitors in series carry the same charge.

9. Do capacitors in parallel have the same voltage?

Yes, all capacitors in parallel have the same voltage across them.

10. Which configuration stores more energy?

Parallel capacitors store more energy because the total capacitance is higher.

Related Topics

• Coulomb's Law
• Electric Field
• Electric Potential
• Gauss Law
• Relation Between Electric Field and Electric Potential
• Motion of Charged Particle
• Equipotential Surfaces
• Capacitance

Share :