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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.

Illustration Questions

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.

image
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

 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

\(C_{eq}=C_1+C_2+C_3+...\)

 

 

Illustration Questions

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 \(C_3=4\,\mu F\).

image
A

\(7\,\mu F\)

.

B

\(9\,\mu F\)

C

\(8\,\mu F\)

D

\(4\,\mu F\)

Option B 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

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

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

 

 

Illustration Questions

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. \(C_1=C_2=C_3=C_4=4\,\mu F\)

image
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 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, \(\dfrac{1}{C_{eq}}=\dfrac{1}{(C_1+C_2)}+\dfrac{1}{(C_3+C_4)}\)

     

Illustration Questions

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.

image
A

\(1\,\mu F\)

.

B

\(2\,\mu F\)

C

\(4\,\mu F\)

D

\(8\,\mu F\)

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, \(\dfrac{1}{C_{eq}}=\dfrac{1}{(C_1+C_2)}+\dfrac{1}{(C_3+C_4)}\)

     

Illustration Questions

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, \(C_{eq_1}=1\,\mu F\)

image

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, \(C_{eq_2}=1\,\mu F\)

image

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, \(C_{eq_3}=2\,\mu F\)

image

Equivalent capacitance of  \(1\,\mu F\) and  \(1\,\mu F\), connected in parallel,

or, \(C_{eq_4}=1\,\mu F+1\,\mu F\)

or, \(C_{eq_4}=2\,\mu F\)

image

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, \(C_{eq}=1\,\mu F\)

image

Find the equivalent capacitance of the given circuit.

image
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.

Illustration Questions

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 \(C_3=3\,\mu F\).

image
A

\(6\,\mu F\)

.

B

\(5\,\mu F\)

C

\(1.5\,\mu F\)

D

\(zero\)

Option A is Correct

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