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Combination Of Capacitors

Combination Of Capacitors :- When a capacitor of the desired capacitance is not available, two or more capacitors are connected together to obtain the required capacitance. The combination of capacitors is done in two ways :-

Series combination and
Parallel combination

(1). Series Combination Of Capacitors

In a series combination of capacitors, the capacitors are connected in such a way that the negative plate of the first capacitor is connected to the positive plate of the second capacitor. If there are more capacitors, their negative plates are connected to the positive plates of the next capacitor in the same sequence. Finally, the free ends of the first and the last capacitors are connected to the external voltage source.

Properties :

Decrease in Capacitance : In a series, the equivalent capacitance is less than the smallest individual capacitance.
Different Potential Differences: The voltage across each capacitor is divided in inverse proportion to their capacitances (the capacitor with smaller capacitance will have a larger voltage across it).
Same Charge: In a series combination, the charge on all capacitors is the same.

Suppose three capacitors with capacitances C₁, C₂ and C₃ are connected in series. The combination of capacitors is connected to a battery of volts.

The two central sections, which include the right plate of C₁ and the left plate of C₂, as well as the right plate of C₂ and the left plate of C₃, are electrically neutral because they are isolated bodies. The battery can neither supply charge to these sections nor draw charge from them.

Assume the battery transfers a charge of +Q from the right plate of C₃ to the left plate of C₁. As a result of this charge transfer, the central isolated sections become polarized, although their net charge remains zero (Q_net = 0). Consequently, the charge on each capacitor remains the same, but because their capacitances differ, the potential difference across each capacitor’s plates is also different.

Equivalent Capacitance Of Series Combination Of Capacitors ( C_s)

Potential difference across the ends of each capacitor :

$\displaystyle {{V}_{1}}=\frac{Q}{{{C}_{1}}},\text{ }{{V}_{2}}=\frac{Q}{{{C}_{2}}},\text{ }{{V}_{3}}=\frac{Q}{{{C}_{3}}}$ …..(1)

Total potential difference across the combination of capacitors :

$\displaystyle V={{V}_{1}}+{{V}_{2}}+{{V}_{3}}$ …..(2)

If a capacitor with capacitance C_s is connected in place of the series combination, and it stores the same charge Q when the same potential difference V is applied, then that single capacitor is called the equivalent capacitor of the series combination. If the capacitance of the equivalent capacitor is C_s, then

$\displaystyle V=\frac{Q}{{{C}_{s}}}$ …..(3)

From equations (1), (2) and (3),

$\displaystyle \frac{Q}{{{C}_{s}}}=\frac{Q}{{{C}_{1}}}+\frac{Q}{{{C}_{2}}}+\frac{Q}{{{C}_{3}}}$

$\displaystyle \Rightarrow \frac{1}{{{C}_{s}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}$ …..(4)

For n capacitors,

$\displaystyle \frac{1}{{{C}_{s}}}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}+..........+\frac{1}{{{C}_{n}}}$ …..(5)

Therefore :

In series combination, the charge on each capacitor is the same and is equal to Q = VC_S (From equation (3)).
In series combination, the potential difference across each capacitor is inversely proportional to its capacitance ( $\displaystyle V\propto \frac{1}{C}$ ).
The reciprocal of the equivalent capacitance of the combination is equal to the sum of the reciprocals of the individual capacitances of the capacitors in the combination.
The value of the equivalent capacitance is less than the capacitance of the capacitor with the smallest capacitance in the combination.

Note :-

(1). If n capacitors, each of capacitance , are connected in series, then the equivalent capacitance will be $\displaystyle {{C}_{s}}=\frac{C}{n}$ .

(2). Total energy stored in series combination of capacitors : Since in a series combination of capacitors the charge (Q) on each capacitor is the same, the total energy (U) is given by :-

$\displaystyle U=\frac{1}{2}\frac{{{Q}^{2}}}{{{C}_{s}}}=\frac{{{Q}^{2}}}{2}\left( \frac{1}{{{C}_{s}}} \right)=\frac{{{Q}^{2}}}{2}\left( \frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}+.......... \right)$

$\displaystyle \Rightarrow U=\frac{1}{2}\frac{{{Q}^{2}}}{{{C}_{1}}}+\frac{1}{2}\frac{{{Q}^{2}}}{{{C}_{2}}}+\frac{1}{2}\frac{{{Q}^{2}}}{{{C}_{3}}}+..........$

$\displaystyle \therefore U={{U}_{1}}+{{U}_{2}}+{{U}_{3}}+..........$

(3). If multiple parallel strips of dielectric with dielectric constants K₁ , K₂ , K₃ ……. are placed between the plates of a capacitor, then this arrangement is equivalent to a series combination of capacitors formed by placing the strips separately. This arrangement is called distance division.

Hence capacitance :-

$\displaystyle C=\frac{{{\varepsilon }_{0}}A}{\left( \frac{{{t}_{1}}}{{{K}_{1}}}+\frac{{{t}_{2}}}{{{K}_{2}}}+\frac{{{t}_{3}}}{{{K}_{3}}} \right)}$

$\displaystyle \Rightarrow \frac{1}{C}=\frac{\left( \frac{{{t}_{1}}}{{{K}_{1}}}+\frac{{{t}_{2}}}{{{K}_{2}}}+\frac{{{t}_{3}}}{{{K}_{3}}} \right)}{{{\varepsilon }_{0}}A}$

$\displaystyle \Rightarrow \frac{1}{C}=\frac{{{t}_{1}}}{{{\varepsilon }_{0}}A{{K}_{1}}}+\frac{{{t}_{2}}}{{{\varepsilon }_{0}}A{{K}_{2}}}+\frac{{{t}_{3}}}{{{\varepsilon }_{0}}A{{K}_{3}}}$

$\displaystyle \Rightarrow \frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}$

(2). Parallel Combination Of Capacitors

In a parallel combination of capacitors, two or more capacitors are connected in such a way that their first plates are joined at one point and their second plates are joined at another point. That is, all the positive plates of the capacitors are connected together, and all the negative plates are also connected together.

Properties :

Increased capacitance : In a parallel combination, the total capacitance is equal to the sum of the capacitances of each capacitor, resulting in an increase in total capacitance.
Equal potential difference : The potential difference across all capacitors is the same because they are connected to the same power source.
Different charges : The charge on each capacitor is proportional to its capacitance (a capacitor with higher capacitance will have more charge).

Consider three capacitors with capacitances C₁, C₂, and C₃ connected in parallel. The combination of capacitors is connected to a battery of voltage .

Equivalent Capacitance Of Parallel Combination Of Capacitors ( C_p)

Charge stored on each capacitor :

$\displaystyle {{Q}_{1}}={{C}_{1}}V,\text{ }{{Q}_{2}}={{C}_{2}}V,\text{ }{{Q}_{3}}={{C}_{3}}V$ …..(6)

The total charge stored on the combination of capacitors :

$\displaystyle Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}$ …..(7)

If a capacitor with capacitance C_p is connected in place of a parallel combination such that it stores the same charge when the same potential difference is applied, then it is called the equivalent capacitor of the parallel combination. If the capacitance of the equivalent capacitor is C_p, then

$\displaystyle Q={{C}_{p}}V$ …..(8)

From equations (6), (7) and (8),

$\displaystyle {{C}_{p}}V={{C}_{1}}V+{{C}_{2}}V+{{C}_{3}}V$

$\displaystyle \Rightarrow {{C}_{p}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}$ …..(9)

For n capacitors,

$\displaystyle {{C}_{p}}={{C}_{1}}+{{C}_{2}}+{{C}_{3}}+..........+{{C}_{n}}$ …..(10)

Therefore :

In a parallel combination, the potential difference across each capacitor is the same.
The equivalent capacitance of the combination is equal to the sum of the individual capacitances of the capacitors in the combination.
The value of the equivalent capacitance is greater than the capacitance of the capacitor with the highest capacitance in the combination.

Note :-

(1). If n capacitors of equal capacitance are connected in parallel, then the equivalent capacitance will be $\displaystyle {{C}_{p}}=nC$ .

(2). Total energy stored in parallel combination of capacitors : Since in a parallel combination of capacitors, the potential difference (V) across each capacitor is the same, the total energy (U) is given by :-

$\displaystyle U=\frac{\text{1}}{2}{{C}_{p}}{{V}^{2}}=\frac{\text{1}}{2}\left( {{C}_{1}}+{{C}_{2}}+{{C}_{3}}+.......... \right){{V}^{2}}$

$\displaystyle \Rightarrow U=\frac{\text{1}}{2}{{C}_{1}}{{V}^{2}}+\frac{\text{1}}{2}{{C}_{2}}{{V}^{2}}+\frac{\text{1}}{2}{{C}_{3}}{{V}^{2}}+..........$

$\displaystyle \therefore U={{U}_{1}}+{{U}_{2}}+{{U}_{3}}+..........$

(3). If a capacitor has several parallel slabs with different dielectric constants K₁ , K₂ , K₃ ……. placed between its plates, each having different areas but the same thickness (equal to the distance between the capacitor plates), then this arrangement is equivalent to a parallel combination of capacitors formed by keeping these slabs separately. This arrangement is called area division.

Hence capacitance :-

$\displaystyle C={{C}_{1}}+{{C}_{2}}+{{C}_{3}}$

$\displaystyle C={{K}_{1}}\left( \frac{{{\varepsilon }_{0}}{{A}_{1}}}{d} \right)+{{K}_{2}}\left( \frac{{{\varepsilon }_{0}}{{A}_{2}}}{d} \right)+{{K}_{3}}\left( \frac{{{\varepsilon }_{0}}{{A}_{3}}}{d} \right)$

$\displaystyle C=\frac{{{\varepsilon }_{0}}}{d}\left( {{K}_{1}}{{A}_{1}}+{{K}_{2}}{{A}_{2}}+{{K}_{3}}{{A}_{3}} \right)$

Note (For JEE) :-

If a combination of N capacitors (series, parallel, or mixed) is being charged by a battery, the energy loss in the form of heat (H) during charging is :

$\displaystyle H=\sum\limits_{k=1}^{N}{\frac{{{\left( {{Q}_{fk}}-{{Q}_{ik}} \right)}^{2}}}{2{{C}_{k}}}}=\sum\limits_{k=1}^{N}{\frac{{{\left( \Delta {{Q}_{k}} \right)}^{2}}}{2{{C}_{k}}}}$

Where is the change in charge of the k^th capacitor and is the capacitance of the k^th capacitor.

If initially all the capacitors are uncharged (Q_ik= 0), then :

$\displaystyle H=\sum\limits_{k=1}^{N}{\frac{{{\left( {{Q}_{fk}} \right)}^{2}}}{2{{C}_{k}}}}$

To find the equivalent capacitance of a combination of capacitors made of plates with equal area and placed at equal distances

Example :- If the area of each plate is and the separation between successive plates is d, find the equivalent capacitance between points A and B.

Solution :

(i).

The steps to solve such questions are as follows :

Step 1 : Number the plates.

Step 2 : If some plates (for example, plate numbers 1 and 4) are connected at a single point, mark that point as well and give it an appropriate name.

Step 3 : Now, mark on the sheet the two points (A and B) at appropriate distances between which the equivalent capacitance is to be determined. If there is any additional point (for example, point C in Step 2), mark it as well. Along with the points, also write the number of plates connected to each of them.

Step 4 : Now, looking at Step 2, draw the capacitors connected between all the points one by one in Step 3.

In step 2, there is a capacitor between plate number 1 and plate number 2, so a capacitor will appear between points A and C.
Similarly, in step 2, there is a capacitor between plate number 2 and plate number 3, so a capacitor will appear between points A and B.
In the same way, in step 2, there is a capacitor between plate number 3 and plate number 4, so a capacitor will appear between points B and C.

Step 5 : Now you have obtained a simple circuit. Remove the extra points (point C) and identify all the capacitors in this circuit, determining which capacitors are connected in series and which are connected in parallel. Finally, calculate the desired equivalent capacitance. Since all the capacitors have plate area $A$ (the same) and the distance between their plates is d, the capacitance of each capacitor will be $\displaystyle C=\frac{{{\varepsilon }_{0}}A}{d}$ .

Therefore, equivalent capacitance between points A and B :

$\displaystyle {{C}_{AB}}=\frac{3}{2}C=\frac{3}{2}\frac{{{\varepsilon }_{0}}A}{d}$

(ii). There will be no capacitor connected between plates number 2 and 3 because both these plates are connected to the same point B.

Equivalent capacitance between points A and B :

$\displaystyle {{C}_{AB}}=2C=\frac{2{{\varepsilon }_{0}}A}{d}$

(iii).

Equivalent capacitance between points A and B :

$\displaystyle {{C}_{AB}}=3C=\frac{3{{\varepsilon }_{0}}A}{d}$

(iv).

Therefore, equivalent capacitance between points A and B :

$\displaystyle {{C}_{AB}}=\frac{2}{3}C=\frac{2}{3}\frac{{{\varepsilon }_{0}}A}{d}$

Note :- With N plates, a maximum of (N-1) capacitors can be made.

Example 1.

If the area of each plate is A and the separation between successive plates is d, then the equivalent capacitance between A and B will be –

(A) $\displaystyle \frac{{{\varepsilon }_{0}}A}{d}$

(B) $\displaystyle \frac{{{\varepsilon }_{0}}A}{2d}$

(D) $\displaystyle \frac{3{{\varepsilon }_{0}}A}{d}$

Solution :

This is a balanced Wheatstone bridge circuit. Therefore, it can be solved as follows :

Therefore, equivalent capacitance between points A and B :

$\displaystyle {{C}_{AB}}=C=\frac{{{\varepsilon }_{0}}A}{d}$

Example 2.

A parallel plate capacitor is made by stacking N equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C, then what will be the resultant capacitance ?

Solution :

This is a parallel combination of capacitors. Now, since a maximum of (N-1) capacitors can be formed from N plates, and the capacitance between any two plates is C, therefore the resultant capacitance between X and Y is :

C_XY = (N-1)C

Example 3.

A segment AB of a circuit is shown in the figure. The electromotive force (emf) of the source is E = 10 V, the capacitances of the capacitors are C₁ = 1 μF and C₂ = 2 μF, and the potential difference V_A – V_B = 5 V. Determine the voltage across each capacitor.

Solution :

Here, both capacitors C₁ and C₂ are connected in series, so the magnitude of charge on both will be the same. Let the charge on them be Q, and the polarity of the plates is as follows :-