Informative line

### Equivalent Capacitance

Learn steps to calculate determine equivalent capacitance for the given capacitor short circuit, practice series and parallel connection of capacitors.

# Series and Parallel Connection of Capacitors

## Series Combination

• Two capacitors connected without any other joint present between them are said to be in series.
• Capacitors C1 and C2 are in series combination.  • In series connection, charge across each capacitor is same.

$$Q_1=Q_2$$  ## Parallel connection

• Two capacitors connected with two joints present between them are said to be in parallel.
• Capacitors C1 and C2 are connected in parallel, as shown in figure.
• In parallel combination, charge across each capacitor may different, but potential difference must be same.  • Consider a circuit in which three capacitors are connected, as shown in figure.
• In the circuit C1, C2 and C3 are neither in series nor in parallel combination.  #### Choose the correct option regarding the following circuit.

A C1 and C2 are in series

B C1 and C3 are in parallel

C C2 and C3 are in series

D None

×

Capacitors C1, C2 and C3 are neither in series nor in parallel.

Hence, option (D) is correct.

### Choose the correct option regarding the following circuit. A

C1 and C2 are in series

.

B

C1 and C3 are in parallel

C

C2 and C3 are in series

D

None

Option D is Correct

# Series Combination of Capacitors

• Two electrodes connected in such a way that bottom electrode of C1 is connected to top electrode of C2, then they are said to be in series.
• In series combination, charge across both capacitors, C1 and C2 are same.
• Potential difference across capacitor C1

$$\Delta V_1=\dfrac{Q}{C_1}$$

• Potential difference across capacitor C2.

$$\Delta V_2=\dfrac{Q}{C_2}$$

• The total potential across capacitor will be

$$\Delta V_C=\Delta V_1+\Delta V_2$$  • Now the circuit reduces to single capacitance same as the equivalent capacitance of C1 and C2 connected in series and potential difference across this capacitance is given as

$$\Delta V_C=\Delta V_1+\Delta V_2$$.

• Thus, equivalent capacitance is given as

$$\dfrac{1}{C_{eq}}=\dfrac{\Delta V_C}{Q}$$

or,  $$\dfrac{1}{C_{eq}}=\dfrac{\Delta V_1}{Q}+\dfrac{\Delta V_2}{Q}$$

or,  $$\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}$$

Conclusion - The equivalent capacitance of two or more capacitors connected in series, is given as

$$\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+...$$

$$C_{eq}=\left (\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+\dfrac{1}{C_4}+...\right)^{-1}$$

#### Determine the equivalent capacitance of the circuit given. $$C_1=C_2=C_3=C_4=4\,\mu F$$

A $$8\,\mu F$$

B $$1\,\mu F$$

C $$2\,\mu F$$

D $$4\,\mu F$$

×

All capacitors C1, C2, C3 and C4 are connected in series.

The equivalent capacitance is given as

$$\dfrac{1}{C_{eq}}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}+\dfrac{1}{C_4}$$

or,  $$\dfrac{1}{C_{eq}}=\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}$$

or,  $$\dfrac{1}{C_{eq}}=\dfrac{4}{4}$$

or,  $$C_{eq}=1\,\mu F$$

### Determine the equivalent capacitance of the circuit given. $$C_1=C_2=C_3=C_4=4\,\mu F$$ A

$$8\,\mu F$$

.

B

$$1\,\mu F$$

C

$$2\,\mu F$$

D

$$4\,\mu F$$

Option B is Correct

# Combination of Series and Parallel Connection of More than Four Capacitors

• Consider a circuit with four capacitors connected, as shown in figure.
• In the circuit, C1 and C2 are connected in parallel and C3 and C4 are also connected in parallel.
• The equivalent of both are connected in series.
• To calculate equivalent capacitance,
• $$C_{eq_1}=C_1+C_2$$ [C1 and C2 connected in parallel]

$$C_{eq_2}=C_3+C_4$$ [C3 and C4 connected in parallel]

$$\dfrac{1}{C_{eq}}=\dfrac{1}{C_{eq_1}}+\dfrac{1}{C_{eq_2}}$$ [$$C_{eq_1}$$ and $$C_{eq_2}$$are in series]

or, $$\dfrac{1}{C_{eq}}=\dfrac{1}{(C_1+C_2)}+\dfrac{1}{(C_3+C_4)}$$  #### Find the equivalent capacitance of the given circuit.

A $$2\,\mu F$$

B $$1\,\mu F$$

C $$1.5\,\mu F$$

D $$3\,\mu F$$

×

Equivalent capacitance of  $$2\,\mu F$$ and  $$2\,\mu F$$, connected in series,

$$\dfrac{1}{C_{eq_1}}=\dfrac{1}{2\,\mu F}+\dfrac{1}{2\,\mu F}$$

or, $$C_{eq_1}=1\,\mu F$$ Equivalent capacitance of  $$2\,\mu F$$ and  $$2\,\mu F$$, connected in series,

$$\dfrac{1}{C_{eq_2}}=\dfrac{1}{2\,\mu F}+\dfrac{1}{2\,\mu F}$$

or, $$C_{eq_2}=1\,\mu F$$ Equivalent capacitance of  $$C_{eq_1}$$ and  $$C_{eq_2}$$, connected in parallel,

$$C_{eq_3}=C_{eq_1}+C_{eq_2}$$

or, $$C_{eq_3}=1\,\mu F+1\,\mu F$$

or, $$C_{eq_3}=2\,\mu F$$ Equivalent capacitance of  $$1\,\mu F$$ and  $$1\,\mu F$$, connected in parallel,

or, $$C_{eq_4}=1\,\mu F+1\,\mu F$$

or, $$C_{eq_4}=2\,\mu F$$ Equivalent capacitance of  $$C_{eq_3}$$ and  $$C_{eq_4}$$, connected in series,

$$\dfrac{1}{C_{eq}}=\dfrac{1}{C_{eq_3}}+\dfrac{1}{C_{eq_4}}$$

or, $$\dfrac{1}{C_{eq}}=\dfrac{1}{2\,\mu F}+\dfrac{1}{2\,\mu F}$$

or, $$\dfrac{1}{C_{eq}}=\dfrac{2}{2\,\mu F}$$

or, $$C_{eq}=1\,\mu F$$ ### Find the equivalent capacitance of the given circuit. A

$$2\,\mu F$$

.

B

$$1\,\mu F$$

C

$$1.5\,\mu F$$

D

$$3\,\mu F$$

Option B is Correct

# Short Circuit  • For series combination, same charge flows in both the capacitors.  • A connecting wire is connected across C2, as shown in figure.  • Capacitor C2 is short-circuited, as connecting wire is connected across it. Hence, potential difference across C2 is zero as both plates of capacitor are at same potential.
• Thus, C2 will behave as conducting wire.
• The equivalent capacitance for the circuit is

$$C_{eq}=C_1$$

Note - The shorted components are not damaged, they will function normally when short circuit is removed.

#### Determine the equivalent capacitance of the circuit when C4 is short circuited given $$C_1=1\,\mu F,\,\,C_2=2\,\mu F$$ and $$C_3=3\,\mu F$$.

A $$6\,\mu F$$

B $$5\,\mu F$$

C $$1.5\,\mu F$$

D $$zero$$

×

Since, capacitance C4 is short circuited.

Hence, capacitor C4 will behave as conducting wire.

Capacitor C1, C2 and C3 are connected in parallel.

Thus, equivalent capacitance is given as

$$C_{eq}=C_1+C_2+C_3$$

or,  $$C_{eq}=1\,\mu F+2\,\mu F+3\,\mu F=6\,\mu F$$

### Determine the equivalent capacitance of the circuit when C4 is short circuited given $$C_1=1\,\mu F,\,\,C_2=2\,\mu F$$ and $$C_3=3\,\mu F$$. A

$$6\,\mu F$$

.

B

$$5\,\mu F$$

C

$$1.5\,\mu F$$

D

$$zero$$

Option A is Correct

# Parallel Combination of Capacitors

• Two capacitors connected in parallel are joined from top to top and from bottom to bottom.
• Top electrode of both the capacitors are connected by a conducting wire so that they are at same potential.
• Similarly, bottom electrode of both the capacitors are connected by a conducting wire so that they are at same potential.
• Thus, it means two capacitors connected in parallel, have same potential difference.
• Consider a circuit in which two capacitors (C1 and C2) are connected in parallel with a battery.
• The potential difference across each capacitor is assumed to be $$\Delta V_C$$.
• Charge stored by capacitor C1

?$$Q_1=C_1\,\Delta V_C$$

• Charge stored by capacitor C2

$$Q_2=C_2\,\Delta V_C$$  • Now, the above circuit is replaced with new circuit having single capacitor in place of two capacitors having total charge  $$Q=Q_1+Q_2$$  • The voltage across capacitor is $$\Delta V_C$$.
• Thus, the capacitance of capacitor is given as,

$$C_{eq}=\dfrac{Q}{\,\Delta V_C}$$

or,  $$C_{eq}=\dfrac{Q_1+Q_2}{\,\Delta V_C}$$

or,  $$C_{eq}=\dfrac{Q_1}{\,\Delta V_C}+\dfrac{Q_2}{\,\Delta V_C}$$

or,  $$C_{eq}=C_1+C_2$$

Conclusion - If two or more capacitors are connected in parallel then their equivalent capacitance is given as

$$C_{eq}=C_1+C_2+C_3+...$$

#### Determine equivalent capacitance for the circuit given $$C_1=2\,\mu F$$, $$C_2=3\,\mu F$$ and $$C_3=4\,\mu F$$.

A $$7\,\mu F$$

B $$9\,\mu F$$

C $$8\,\mu F$$

D $$4\,\mu F$$

×

Since, capacitors C1, C2 and C3 are connected in parallel.

So, equivalent capacitance is given as

$$C_{eq}=C_1+C_2+C_3$$

or,  $$C_{eq}=2\,\mu F+3\,\mu F+4\,\mu F$$

or,  $$C_{eq}=9\,\mu F$$

### Determine equivalent capacitance for the circuit given $$C_1=2\,\mu F$$, $$C_2=3\,\mu F$$ and $$C_3=4\,\mu F$$. A

$$7\,\mu F$$

.

B

$$9\,\mu F$$

C

$$8\,\mu F$$

D

$$4\,\mu F$$

Option B is Correct

# Combination of Series and Parallel Connection of Capacitors

• Consider a circuit with four capacitors connected, as shown in figure.
• In the circuit, C1 and C2 are connected in parallel and C3 and C4 are also connected in parallel.
• The equivalent of both are connected in series.
• To calculate equivalent capacitance,
• $$C_{eq_1}=C_1+C_2$$ [C1 and C2 connected in parallel]

$$C_{eq_2}=C_3+C_4$$ [C3 and C4 connected in parallel]

$$\dfrac{1}{C_{eq}}=\dfrac{1}{C_{eq_1}}+\dfrac{1}{C_{eq_2}}$$ [$$C_{eq_1}$$ and $$C_{eq_2}$$are in series]

or, $$\dfrac{1}{C_{eq}}=\dfrac{1}{(C_1+C_2)}+\dfrac{1}{(C_3+C_4)}$$  #### Determine the equivalent capacitance for the given circuit.

A $$1\,\mu F$$

B $$2\,\mu F$$

C $$4\,\mu F$$

D $$8\,\mu F$$

×

Equivalent capacitance of $$2\,\mu F$$ and $$2\,\mu F$$, connected in series,

$$\dfrac{1}{C_{eq_1}}=\dfrac{1}{2\,\mu F}+\dfrac{1}{2\,\mu F}=\dfrac{2}{2\,\mu F}$$

or,$$C_{eq_1}=1\,\mu F$$

Equivalent capacitance of $$2\,\mu F$$ and $$2\,\mu F$$, connected in series,

$$\dfrac{1}{C_{eq_2}}=\dfrac{1}{2\,\mu F}+\dfrac{1}{2\,\mu F}=\dfrac{2}{2\,\mu F}$$

or,$$C_{eq_2}=1\,\mu F$$

Since, $$C_{eq_1}$$and $$C_{eq_2}$$are connected in parallel. So, equivalent capacitance of circuit is

$$C_{eq}=C_{eq_1}+C_{eq_2}=1\,\mu F+1\,\mu F$$

or,  $$C_{eq}=2\,\mu F$$

### Determine the equivalent capacitance for the given circuit. A

$$1\,\mu F$$

.

B

$$2\,\mu F$$

C

$$4\,\mu F$$

D

$$8\,\mu F$$

Option B is Correct