Question

prove that the total energy stored in capacitor is sum of energy of each capacitor with they are connected in series or parallel​

Answer 1

Description:

Whenever a combination of capacitors is connected with a battery, some amount of energy is stored in it. In order to find the value of different energy stored in different combination, Connect three capacitors (C1, C2 and C3) first in series and then in parallel combination.

Series combination

Let there are three capacitors (C1, C2 and C3) having capacitance (C1, C2 and C3) connected in series. Capacitor C1 consists of two plates (say plate A and B), Capacitor C2 consists of two plates (say plate C and D) whereas Capacitor C3 consists of two plates (say plate E and F).

Plate B of C1 is connected with plate C of C2 whereas plate D of C2 is connected with plate E of C3. Plates A and F are connected to point P and Q. These points are connected to a battery carrying potential difference (V). Figure is shown below.

Series

Let the energy due to charge q in capacitors (C1, C2 and C3) be (U1, U2 and U3) respectively.

We know that energy due to charge q in a capacitor = (1/2)(q2/C)

So, Energy U1 = 1/2 q2/c1

Energy U2 = 1/2 q2/c2

Energy U3 = 1/2 q2/c3

But, total energy stored in a system is the sum of energy stored within each individual capacitors in a series.

If U is the total energy stored within a system,

Then, U = U1 + U2 + U3

So, U = 1/2 q2 (1/c1 + 1/c2 + 1/c3)

In series combination 1/cequivalent = 1/c1 + 1/c2 + 1/c3

Where Cequivalent is the net capacitance of the system

U = 1/2 q2 (1/Cequivalent)

We know that, q = CequivalentV

So, U = 1/2 V2 Cequivalent

This is the net energy contained by the system in series combination of capacitors.

Parallel Combination

Let there are three capacitors (C1, C2 and C3) having capacitance (C1, C2 and C3) connected in series. Capacitor C1 consists of two plates ( say plate A and B), Capacitor C2 consists of two plates ( say plate C and D) whereas Capacitor C3 consists of two plates ( say plate E and F).

First Plate each capacitor is connected with point P whereas another plate of each is connected with Q. The two points P and Q are connected with the two terminals of battery so, so the potential difference (V) is developed across point P and Q.

Plate

Let the energy due to charge q in capacitors (C1, C2 and C3) be (U1, U2 and U3) respectively.

We know that energy due to potential V in a capacitor = (1/2)CV2

So, Energy U1 = C1V2/2

So, Energy U2 = C2V2/2

So, Energy U3 = C3V2/2

But, total energy stored in a system is the sum of energy stored within each individual capacitors in a parallel.

If U is the total energy stored within a system,

Then, U = U1 + U2 + U3

So, U = 1/2 V2 (C1 + C2 + C3)

Where Cequivalent is the net capacitance of the system

So, U = 1/2 V2 Cequivalent

This is the net energy contained by the system in parallel combination of capacitors.

Note

Energy of a system is equal to the sum of all the energies stored within each individual capacitor.

As equivalent capacitance of parallel combination is higher than that of series combination, so, Energy of a system in series combination is always less than in parallel combination.