__ Series combination:__

Capacitors connected as shown in the figure are said to be connected in series.

In series combination :

(a)the charges on individual capacitors are equal .

(b) and the total p.d. across the combination is to shared by the capacitors .

i.e. Q = C_{1} V_{1} = C_{2} V_{2} = C_{3} V_{3} – – – – –

and V = V_{1} + V_{2} + V_{3} + – – – – – –

$ \displaystyle \frac{Q}{C} = \frac{Q}{C_1} + \frac{Q}{C_2} + \frac{Q}{C_3} + —–+ \frac{Q}{C_n} $

Effective capacitance of the combination C can be found from the relation

$ \displaystyle \frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + —–+ \frac{1}{C_n} $

__ Parallel combination__

In this combination

(a)p.d. across each of the capacitors is the same but

(b)the charge supplied at points A and B is shared by the capacitors

$ \displaystyle V = \frac{Q_1}{C_1} = \frac{Q_2}{C_2} = …… $

and total charge

Q = Q_{1} + Q_{2} + Q_{3} + – – – – – –

C V = C_{1} V_{1} + C_{2}V_{2} + C_{3}V_{3} + – – – – –

Thus C = C_{1} + C_{2} + C_{3} + – – – – –

Solved Example : Three capacitors each of capacitance 9 pF are connected in series.

(a) What is the total capacitance of the combination ?

(b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply ?

Sol. C = 9 pF = 9 × 10^{-12} F , n = 3 , V = 120 volt

(a) As Three capacitors are in series

By Formul $ \displaystyle C_S = \frac{C}{n} $

$ \displaystyle C_S = \frac{9\times 10^{-12}}{3} $

C_{S} = 3 × 10^{-12} F

C_{S} = 3 pF

(b) Since in series, charge q is same on each capacitor,

C_{1}V_{1} = C_{2}V_{2} = C_{3}V_{3}

As C_{1} = C_{2}= C_{3} ,

We have, V_{1} = V_{2} = V_{3} = V_{c} (say)

Since V_{1} + V_{2} + V_{3} = V

3V_{c} = 120, V_{c} = 40V

Potential difference across each capacitor is 40V.

Solved Example : Three capacitors of capacitances 2 pF, 3 pF and 4 pF are connected in parallel.

(a) What is the total capacitance of the combination?

(b) Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Sol. (a) Total capacitance , C = 2 + 3 + 4 = 9 pF

(b)As all capacitors are connected in parallel, hence potential Difference across each capacitor is same .

V = 100 volt

q_{1} = C_{1} V = 2 × 100 = 200pC

q_{2} = C_{2} V = 3 × 100 = 300pC

q_{3} = C_{3} V = 4 × 100 = 400pC

Solved Example : Five identical capacitors, when connected in series, produce equivalent capacitance of 6μF. Now these capacitors are rearranged so as to produce the maximum value of equivalent capacitance. This maximum value will be

(A) 30 μF

(B) 36 μF

(C) 150 μF

(D) 6 μF

Solution: Let C be the capacitance of each capacitor

Then 6 = C/5 ⇒ C = 30 μF

If these are connected in parallel, then equivalent capacitance will be maximum.

C’ = 30 × 5 = 150 μF

Hence (C) is correct.

Solved Example : The expression for the equivalent capacitance of the system shown above is (A is the cross-sectional area of one of the plates)

(a) $\large \frac{\epsilon_0 A}{3 d}$

(b) $\large \frac{3\epsilon_0 A}{ d}$

(c) $\large \frac{\epsilon_0 A}{6 d}$

(d) None of these

Solution : The three capacitors are in parallel.

Equivalent capacitance , $\large C= C_1 + C_2 + C_3 $

$\large C= \frac{\epsilon_0 A}{d} + \frac{\epsilon_0 A}{2d} + \frac{\epsilon_0 A}{3 d} $

$\large = \frac{\epsilon_0 A}{d}(1+\frac{1}{2} + \frac{1}{3}) $

$\large \frac{11}{6} (\frac{\epsilon_0 A}{d} )$

Hence (D) is correct.

Solved Example : Find the equivalent capacitance between points A and B of the circuit shown, each capacitance = C.

Solution : By symmetry, the points D, E and F are at the same potentials, hence the given circuit is equivalent to figure (ii)

$\large C_{eq} = \frac{C}{2} + \frac{C}{2} + \frac{C}{2} + \frac{C}{2}$

The equivalent capacitance = 2C

Solved Example : Find the equivalent capacitance between points A and B. Capacitance of each capacitor is 2μF.

Solution: The circuit can be redrawn as

Solved Example : Four identical metal plates are placed in air parallel to each other with distances d from one another. The area of each plate is equal to A. The arrangement is shown in the figure. Find the capacitance of the system between plates P and R.

Solution : In order to find the equivalent capacitance between two given plates, we can connect a battery across them and then find the charge distribution on the surfaces of each plate using Gauss theorem or the concept :

” Net electric field in a conductor is always zero.”

Let the plates P and R have the potential V and 0 respectively and the plates 1 and 4 are at the potential V1. Charge distribution is shown in the figure.

q_{1} = C(V – V_{1}) from (4, 3) . . . (1)

q_{2} = CV from (3, 2) . . . (2)

q_{1} = CV_{1} from (2, 1) . . . (3)

From equation (1) and (2)

q_{1} = CV/2 . . . (4)

Charge supplied by the battery is

Q = q_{1} + q_{2} = (3/2)CV using (2) and (4)

=> CV = (2/3)Q . . . (5)

If equivalent capacitance be C’ then C’V = Q . . . (6)

Dividing (5) by (6)

C/C’ = 2/3

C’ = 3C/2

Exercise : Two concentric shells of radii R and 2R are shown in the figure. Initially a charge q is imparted to the inner shell. After the keys K_{1} & K_{2} are alternately closed n times each , find the potential difference between the shells.