Problems On Energy Stored In A Capacitor

According to experts, energy is the ability to perform work. Studies have shown that it is possible to transfer energy from one medium to another and use it to perform tasks that allow the way of life to exist. Walking, cooking, driving, and other daily activities require a certain amount of energy.

Energy is required for all living things, and no process in the universe can occur without energy. The Sun is the most important energy source for living things on Earth. Solar energy is available actively in the form of photosynthesis and passively in the form of carbon energy. 

Energy Stored In A Capacitor

The energy stored in a capacitor can be viewed as the quantum of work that a battery can perform. It is because the voltage represents stored energy when compared with every additional unit of charge induced. 

A certain amount of work is associated with dispersing charge from the negative side of the capacitor to the positive portion, and the same is expressed in terms of charge and Voltage. A direct relationship exists between the quantum of charge in the capacitor and Voltage.

There is an interesting problem pertaining to the amount of energy stored in the capacitor. It is the result of a few points that cause some confusion. When we calculate the energy generated, the capacitor is charged at a certain level of voltage.

The interesting thing here is that half of the energy generated is released in the form of heat in the passage of charge, and only half is stored in the capacitor.

Even though the capacitor is charged faster, lowering the charging path tolerance to provide more energy upon that capacitance doesn’t perform since the energy dissipation rate in the resistor increases dramatically.

The fact that you’ll always lose 50% of the energy in the form of heat in this incremental charge attempt is counterproductive. Therefore, this typical issue is a good example of the versatility of mathematics and its comprehensive nature. Because the large percentage of the voltage level starts dropping throughout the resistance and just a little portion of the energy is incurred, getting the very first component of charging onto the plates of a capacitor requires much less effort.

Mathematical Expression:    

                                              dU = qdV

Where, U = energy

V = Voltage

q = Charge

The total energy on the circuit is obtained by persisting with the essential component that has an algebraic form, which can be represented by “q”.

Mathematical Expression:

                                  U =  Q2/2C = CV2/2 = QV/2, 

Here Vb is the battery voltage.

So, to put one unit of energy in the circuit, one has to incur twice the same; the other unit incurred gets dissipated in the form of heat.

Another method is to decrease the charging resistor to the juncture in which the preliminary charge is exceptionally high, enabling a substantial chunk of the energy to be emitted as electromagnetic waves.

This exceeds the approach of antenna theory.

Following the same logic, the energy imparted by the battery can be expressed as: 

E = CV2,

However, just 50% of the above calculated gets incurred on the capacitor; the rest is either dissipated in the form of heat or in the case of relatively lower charge resistance, heat, and EM waves.

Questions On Energy Stored In A Capacitor 

Question 1: Find the charge stored in the capacitor between terminals A and B of the combination shown below. The capacitance of each capacitor is given as: C = 1μF

Solution:

The circuit shown above can be simplified as follows:

As both the capacitors in both the rows are in a series, the equivalent capacitance is calculated as:

First row =  1/C +1/C = 2/C →C/2

Second row =  C/2

Therefore,

Total capacitance = C/2 +C/2 = C

Charge (Q) = CV

The charge on the capacitor between the two terminals A and B → Q/2 =CV/2

Hence, as per the formula, the energy stored in the capacitor is as follows:

(U) =  (Q2)2/2C =Q2/4 × 2C = C2V2/8C = CV2 /8

C = 1μF; V = 9V (Given)

Substituting the values of C and V:

U = (1 × 10-6) × 8 × 9/8

= 9 × 10-6 = 9μJ

Question 2: (a) What is the energy stored in the 10.0 μF capacitor charged to 9.00 × 103 V?

         (b) Find the amount of stored charge.

According to the capacitor formula:

U = 1/ 2 (CV2)

On putting the given values:

U = ½ x 10 x (9.00 × 103 )2 = 405 J

Therefore, the energy stored in the capacitor is 405 J.

As per the formula, the amount of charge stored is calculated by the following formula:

Q= C/V 

Q =  10/(9.00*10^3)

Q = 90.00 mC

Conclusion 

The energy stored in a capacitor is the electrostatic potential energy, which is proportional to the charge Q and voltage V between the capacitor plates. The electric field between the plates of a charged capacitor stores energy; it increases as the capacitor charges.