Physics Topics can be challenging to grasp, but the rewards for understanding them are immense.
What are the Combination of Capacitors?
Series combination: In this type of combination of capacitors, the first plate of the first capacitor is joined to the electric source, its second plate is joined to the first plate of the second capacitor, the second plate of the second capacitor is joined to the first plate of the third capacitor and so on. The second plate of the last capacitor is grounded, the rest of the system being kept insulated [Fig.].
Calculation of equivalent capacitance: Let three capacitors of capacitances C1, C2, C3 be connected in series [Fig.]. Now Ifa charge +Q be given from a source to the first plate A of the first capacitor, this will induce a charge -Q on the other plate B of this capacitor and a charge + Q on the first plate C of the second capacitor and so on. All the capacitors will have the same charge Q. The final free positive charge from the last plate of the system moves to the earth.
If V1, V2 and V3 be the potential differences across the capacitors C1, C2 and C3 and V be the potential difference between the first plate A and the last plate F of the combination, then
V = V1 + V2 + V3
For the first capacitor, V1 = \(\frac{Q}{C_1}\)
For the second capacitor, V2 = \(\frac{Q}{C_2}\)
For the third capacitor, V3 = \(\frac{Q}{C_3}\)
∴ From equation (1) we have,
V = \(\frac{Q}{C_1}\) + \(\frac{Q}{C_2}\) + \(\frac{Q}{C_3}\)
or V = Q\(\left(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right)\) ……. (2)
Now suppose that instead of this combination, a single capacitor is used such that the same charge (Q) given to this new capacitor produces a same potential difference (V) between its two plates. This single capacitor is known as the equivalent capacitor of the combination and its capacitance is known as the equivalent capacitance.
If C be the equivalent capacitance of the series combination of the capacitors C1, C2, C3, then
V = \(\frac{Q}{C}\) ……. (3)
From equations (2) and (3) we get,
\(\frac{Q}{C}\) = \(Q\left(\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}\right)\)
or, \(\frac{1}{C}\) = \(\frac{1}{C_1}\) + \(\frac{1}{C_2}\) + \(\frac{1}{C_3}\) …….. (4)
For n number of capacitors connected in series, the equivalent capacitance C is given by,
\(\frac{1}{C}\) = \(\frac{1}{C_1}\) + \(\frac{1}{C_2}\) + …….. + \(\frac{1}{C_n}\) …… (5)
Thus if several capacitors are connected in series, the reciprocal of the capacitance of the equivalent capacitor is equal to the sum of the reciprocals of the capacitances of the individual capacitors.
Clearly, for series combination, the equivalent capacitance is always less than any individual capacitance in the series.
Parallel combination: In this type of combination the first plates, i.e., the insulated plates of all the capacitors are connected to a common point A and the second plates, i.e., the grounded plates to another common point B. The point A is connected to an electric source and the point B is earthed[Fig.]
Calculation of equivalent capacitance: Fig. shows three capacitors of capacitances C1, C2 and C3 connected in parallel. The insulated plates of the three capacitors are connected to an electric source having potential V and other plates are earthed. So the potential difference between the two plates
of each capacitor is V. A charge +Q drawn from the supply divides into Q1, Q2 and Q3 according to the capacity of the different capacitors. So,
Q = Q1 + Q2 + Q3 ……. (3)
For the first capacitor,
Q1 = C1V
For the second capacitor,
Q2 = C2V
For the third capacitor,
Q3 = C3V
∴ From the equation (6) we get,
Q = C1V + C2V + C3V
or, Q = V(C1 + C2 + C3) ……… (7)
If the capacitors connected in parallel be replaced by a single capacitor so that the same potential difference V is produced if charge + Q is given to its insulated plate, the single capacitor is known as the equivalent capacitor of the combination and its capacitance is known as the equivalent capacitance.
If C be the equivalent capacitance of the parallel combination of the capacitors C1, C2, C3, then
Q = CV …….. (8)
From equations (7) and (8) we have,
CV = V(C1 + C2 + C3)
or, C = C1 + C2 + C3 …… (9)
For n number of capacitors connected in parallel, the equivalent capacitance C is
c = c1 + c2+…….+ cn
Thus the equivalent capacitance of the capacitors joined in parallel is equal to the sum of their individual capacitances. Clearly, the equivalent capacitance of a number of capacitors in parallel is greater than any individual capacitance in the combination. Capacitors are connected in parallel when a large capacitance for a small potential is required.
Numerical Examples
Example 1.
A condenser is composed of 21 circular plates placed one after the other. The diameter of each plate is 10 cm. The consecutive plates are separated by 0.2 mm thick mica sheets of dielectric constant 6. If the alternate circular plates are connected, calculate the capacitance of the condenser in µF.
Solution:
The condenser is composed of 21 circular plates and alternate plates are connected. So here we get 20 identical capacitors connected in parallel whose capacitance is
C = \(\frac{20 \kappa \epsilon_0 \alpha}{d}\)
= \(\frac{20 \times 6 \times 8.854 \times 10^{-12} \times \pi \times 2.5 \times 10^{-3}}{2 \times 10^{-4}}\)
= 4.17 × 10-8 F = 0.0417 µF
[Here k = 6, α = π × (5)2 cm2 = π × 2.5 × 10-3 m2;
d = 2 × 10-4 m, ε0 = 8.854 × 10-12C2 ᐧ N-1 ᐧ m2]
Example 2.
A condenser is composed of 200 circular tin plates placed one after the other. The consecutive plates are separated by 0.5 mm thick mica sheets of dielectric constant 6. If the alternate tin plates are connected and the capacitance of the entire condenser is 0.4µF, what is the radius of each tin plate?
Solution:
The condenser is composed of 200 circular plates and alternate plates are connected. So, here we get 199 identical capacitors connected in parallel. Nów capacitance of each capacitor is
C = \(\frac{\kappa \alpha}{4 \pi d}\) = \(\frac{\kappa \pi r^2}{4 \pi d}\) = \(\frac{k r^2}{4 d}\)
Example 3.
The equivalent capacitances of the parallel and the series combinations of two capacitors are 5 µF and 1.2 µF, respectively. Calculate the capacitances of each capacitor.
Solution:
Let the capacitances of the two capacitors be C1 µF and C2 µ F. According to the question,
C1 + C2 = 5 ……. (1)
and \(\frac{C_1 C_2}{C_1+C_2}\) = 1.2
or, C1C2 = 1.2 × 5 = 6 ….. (2)
or, C1(5 – C1) – 6 = 0 [with the help of equation (1)]
or, \(C_1^2\) – 5C1 + 6 = 0 or, (C1 – 3)(C1 – 2) = 0
∴ C1 = 3 or,2
If C1 = 3; C2 = 5 – 3 = 2
and if C1 = 2; C2 = 5 – 2 = 3
So, the capacitances of the two capacitors are 3 µF and 2 µF.
Example 4.
Two capacitors of capacitances 20 µF and 60 µF are connected in series. If the potential difference between the two ends of the combination is 40 V, calculate the terminal potential differences of each capacitor.
Solution:
If C be the equivalent capacitance of the combination, then
C = \(\frac{20 \times 60}{20+60}\) = 15 µF = 15 × 10-6F
∴ Total charge of the combination,
Q = CV = 15 × 10-6 × 40 = 6 × 10-4C
Since the two capacitors are connected in series, charge on each capacitor is equal to the total charge of the combination, i.e., 6 × 10-4C.
∴ Potential difference between the two plates of the capacitor having capacitance C1 is,
V1 = \(\frac{Q}{C_1}\) = \(\frac{6 \times 10^{-4}}{20 \times 10^{-6}}\) = 30V
Again, potential difference between the two plates of the capacitor having capacitance C2 is,
V1 = \(\frac{Q}{C_2}\) = \(\frac{6 \times 10^{-4}}{60 \times 10^{-6}}\) = 10V
Example 5.
A charged condenser is made to share its charge with an uncharged condenser of twice its capacitance. Find the sum of the energy of the two condensers.
Solution:
Let the capacitance of the charged condenser be C and let its charge be Q.
Before sharing of charge, energy of the charged condenser is,
E1 = \(\frac{Q^2}{2 C}\)
The capacitance of the other condenser = 2C; since the charged condenser is made to share its charge with the second condenser, it is clear that the condensers are connected in parallel
with each other.
∴ Equivalent capacitance of them = C+ 2C = 3C
∴ Energy of the combination = \(\frac{1}{2} \cdot \frac{Q^2}{3 C}\) = \(\frac{1}{3} \cdot \frac{1}{2} \frac{Q^2}{C}\) = \(\frac{1}{3} E_1\)
Example 6.
A spherical drop of water carries a charge of 10 × 10-12 C and has a potential of 100 V at its surface.
(i) What is the radius of the drop?
(ii) If eight such charged drops combine to form a single drop, what will be the potential at the surface of the new drop? [HS ’02]
Solution:
i) Charge of a spherical water drop,
Q = 10 × 10-12C
= 10 × 10-12 × 3 × 109 esu of charge
= 3 × 10-2 esu of charge
Potential of the drop, V = \(\frac{100}{300}\) = \(\frac{1}{3}\) statV
Capacitance of the drop,
C = \(\frac{Q}{V}\) = \(\frac{3 \times 10^{-2}}{\frac{1}{3}}\) = 0.09 statF
In CGS units, the radius of a spherical conductor = its capacitance.
∴ Radius of the drop, r = 0.09 cm.
ii) If R be the radius of the large drop, then
\(\frac{4}{3} \pi R^3\) = 8 × \(\frac{4}{3} \pi R^3\)
or, R = 2r = 2 × 0.09 = 0.18 cm
Total charge, Q = 8 × 3 × 10-2 = 24 × 10-2 statC
∴ Potential at the surface of the new drop,
V = \(\stackrel{Q}{C}\) = \(\stackrel{Q}{R}\) = \(\frac{24 \times 10^{-2}}{0.18}\) = \(\frac{4}{3}\) = 1.33 statV
= 1.33 × 300 V = 400 V
Example 7.
Three plates of the same size form a system of capacitors. Each plate has an area α. The intermediate distances between the plates are d1 and d2, respectively. The space between the first two plates is occupied by a dielectric of constant k1 and that between the second and third plates by a dielectric of constant k2. Calculate the capacitance of the system.
Solution:
The three plates form two capacitors connected in series (Fig.).
Capacitance of the first capacitor,
C1 = \(\frac{\kappa_1 \epsilon_0 \alpha}{d_1}\)
Capacitance of the second capacitor, .
C2 = \(\frac{\kappa_2 \epsilon_0 \alpha}{d_2}\)
If C be the equivalent capacitance of the whole system, then
Example 8.
The capacitance of a parallel plate air capacitor is 9 pF. The separation between the plates is d. The Intermediate space is filled up by two dielectric media. The widths of them are \(\frac{d}{3}\) and \(\frac{2 d}{3}\), and their dielectric constants are 3 and 6, respectively. Find the capacitance of the parallel plate capacitor. (AIEEE ‘08)
Solution:
As the capacitors are connected in series the equivalent capacitance,
Ceq = \(\frac{C_1 C_2}{C_1+C_2}\) … (1)
We know that,
Example 9.
Three capacitors having capacitances 1 µF, 2 µF and 3, µF are joined in series. A potential difference of 1100 V is applied to the combination. Find the charge and potential difference across each capacitor. [HS ‘05]
Solution:
If C be the equivalent capacitance of the combination, then,
\(\frac{1}{C}\) = \(\frac{1}{1}\) + \(\frac{1}{2}\) + \(\frac{1}{3}\) = \(\frac{11}{6}\) or, C = \(\frac{6}{11}\) µF = \(\frac{6}{11}\) × 10-6F
∴ Total charge of the combination,
Q = CV = \(\frac{6}{11}\) × 10-6 × 11oo = 6 × 10-4C
Since the capacitors are connected in series, charge on each capacitor is equal to the total charge, i.e., 6 × 10-4 C.
Potential difference across the plates of the first capacitor,
V1 = \(\frac{Q}{C_1}\) = \(\frac{6 \times 10^{-4}}{1 \times 10^{-6}}\) = 600V
Similarly, for the other two capacitors, respectively,
V2 = \(\frac{Q}{C_2}\) = \(\frac{6 \times 10^{-4}}{2 \times 10^{-6}}\) = 300V
and V3 = \(\frac{Q}{C_3}\) = \(\frac{6 \times 10^{-4}}{3 \times 10^{-6}}\) = 200V
Example 10.
Two capacitors having capacitances 0.1 µF and 0.01 µF are joined in series. A potential difference of 22 V is applied to the combination. If the capacitors are now joined in parallel, what will be the change in stored energy?
Solution:
If the capacitors are joined in series, their equivalent capacitance,
The change in stored energy = E2 – E1 = 266.2 – 22 = 2442 erg
∴ Stored energy increases by 244.2 erg.