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Charging and Discharging of Capacitor

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Last updated date: 28th May 2024
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Charging and Discharging of Capacitor: A JEE Main Essential

The capacitor is a device used to store energy in the form of electrical charge which can be later utilised to supply charge or energy once the power source is disconnected from it. It is used in the electric circuits of radios, computers, etc. along with these capacitors. It also provides temporary storage of energy in circuits which can be supplied when required. The property of the capacitor to store energy is known as capacitance. But how do they work? How do they fill up with electric power, and how do they release it? Buckle up, JEE Main aspirants, as we delve into the fascinating world of charging and discharging capacitors!


Before we get into the details, let's understand three key concepts:


  1. Capacitance: Imagine a bucket. The bigger the bucket, the more water it can hold. Similarly, a capacitor's capacitance determines how much electric charge (Q) it can store. Measured in farads (F), the higher the capacitance, the more charge it can handle.

  2. Voltage: Think of voltage as the pressure pushing the water in the bucket. In electrical terms, voltage (V) represents the potential difference between the two plates of a capacitor. The higher the voltage, the stronger the electric field pushing the charges.

  3. Current: Imagine water flowing into the bucket as current (I). It's the rate at which charge flows through the circuit, measured in amperes (A).


Charging and Discharging of Capacitor Derivation

Charging and discharging of capacitors holds importance because it is the ability to control as well as predict the rate at which a capacitor charges and discharges that makes capacitors useful in electronic timing circuits. It happens when the voltage is placed across the capacitor and the potential cannot rise to the applied value instantaneously. As the charge on the terminals gets accumulated to its final value, it tends to repel the addition of further charge accumulation.

Thus following are the factors on which rate at which a capacitor can be charged or discharged depends on:

  1. The capacitance of the capacitor to store electric charge.

  2. The resistance of the circuit through which it is being charged or is discharged.


Capacitance

Imagine pouring liquid into a container – as you pour more, the level of the liquid keeps rising. Similarly, when you add electrical charge to a conductor (like a metal object), its electrical potential increases. This relationship is expressed as:


Amount of charge (Q) $\propto$ to potential (V)

Or

Q = CV — (1)


In this equation, C is a constant called capacitance. It's like a measure of how much charge a conductor can hold for a given increase in potential.


We can express capacitance as:

$C = \dfrac{Q }{V}$ — (2)


This tells us that capacitance is the ratio of the charge on a conductor to its potential. The value of C depends on factors like the size and shape of the conductor, the type of material around it, and the arrangement of nearby charges. Interestingly, it doesn't depend on the material of the conductor itself.


If we set the potential (V) to 1, then from Eqn. (1):

Q = C or C = Q


So, capacitance is essentially the amount of charge needed to raise the potential of a conductor by one unit.


The cgs unit of capacitance is called a statfarad (st F). 

$1 \text{ statfarad} = \dfrac{\text{1 statcoulomb}}{\text{1 statvolt}}$


A conductor is said to have one statfarad of capacitance if its potential increases by one statvolt when a charge of one statcoulomb is added to it. In the SI unit system, capacitance is measured in farads (F).


From eqn. (2)

1 farad (F) = $\dfrac{txt{1 coulomb(C)}}{\text{1 volt(V)}}$


Charging and Discharging of a Capacitor through a Resistor

Let's talk about what happens when a capacitor gets charged up and then loses its charge through a resistor. Imagine there's a setup with a capacitor (let's call it C), a resistor (let's call it R), and a battery with some power (let's call it ε). They're all connected in a line using a switch (Morse key K), as you can see in the picture.


[Image will be Uploaded Soon]


Charging of a Capacitor

When you press the key, the capacitor starts to store electric charge. If we use "I" to represent the current flowing through the circuit and "Q" for the charge on the capacitor during charging, we can express the potential difference across the resistor as IR and the potential difference between the capacitor plates as Q/C. These potentials add up to the total electromotive force (ε) in the circuit.


$RI + \dfrac{Q}{C} = \epsilon$ — (1)

When the capacitor is fully charged and no more current flows (I = 0), the equation becomes $\dfrac{Q_0}{C} = \epsilon$ — (2), 

where $Q_0$ is the maximum charge the capacitor can hold.

From equations ( 1 ) and (2),

$R I+\frac{Q}{C}=\frac{Q_0}{C}$

$\frac{Q_0}{C}-\frac{Q}{C}=R I$

$\frac{Q_0-Q}{C R}=I \ldots(3)$

Since $\mathrm{I}=\mathrm{dQ} / \mathrm{dt}$, from equation ...(3),

$\frac{Q_0-Q}{C R}=\frac{d Q}{d t} \text { or } \frac{d Q}{Q_0-Q}=\frac{d t}{C R}$

When $\mathrm{t}=0, \mathrm{Q}=0$, and $\mathrm{t}=\mathrm{t}, \mathrm{Q}=\mathrm{Q}$.

By integrating both sides, we get

$\int_0^Q \frac{d Q}{\left(Q_0-Q\right)}=\int_0^t \frac{d t}{C R}=\frac{1}{C R} \int_0^t d t$

$\left|-\ln \left(Q_0-Q\right)\right|_0^Q=\frac{1}{C R}|t|_0^t$

$-\ln \left(Q_0-Q\right)+\ln Q_0=\frac{t}{C R}$

$\ln \left(Q_0-Q\right)-\ln Q_0=-\frac{t}{C R}$

$\ln \frac{Q_0-Q}{Q_0}=-\frac{t}{C R}$

$\frac{Q_0-Q}{Q_0}=e^{-t / C R}$

$Q_0-Q=Q_0 e^{-t / C R}$

$Q=Q_0\left(1-e^{-t / C R}\right)$

$Q=Q_0\left(1-e^{-t / \tau}\right) \ldots$

Here, $\tau=C R$


This formula helps us understand how the charge on the capacitor changes over time during the charging process.


Transient Period

After a time period equivalent to 4-time Constants (4T), the capacitor in this RC charging circuit is virtually fully charged and the voltage across the capacitor now becomes approx 98% of its maximum value, 0.98Vs. This time taken for the capacitor to reach this 4T point is known as the Transient Period.


Steady-State Period

After a time of 5T, the capacitor is said to be fully charged with the voltage across the capacitor (Vc ) being equal to the supply voltage( Vs ). As the capacitor becomes fully charged, no more current flows in the circuit. The time period after 5T  is called the Steady-State Period.


Discharging of a Capacitor

When you let go of the key K [Figure], the circuit breaks, and there's no extra resistance added. The battery is now not part of the circuit, and the capacitor releases its stored energy through R. If I represents the current during this release, and we set ε (voltage) to zero in the equation RI + Q/C = ε, we get:


$R I+\frac{Q}{C}=0 \text { or } R \frac{d Q}{d t}+\frac{Q}{C}=0$

$R \frac{d Q}{d t}=-\frac{Q}{C} \text { or } \frac{d Q}{Q}=-\frac{d t}{C R}$

When $\mathrm{t}=0, \mathrm{Q}=\mathrm{Q}_0$, and $\mathrm{t}=\mathrm{t}, \mathrm{Q}=\mathrm{Q}$.

On integrating both sides, we get

$\int_{Q_0}^Q \frac{d Q}{Q}=-\int_0^t \frac{d t}{C R}=-\frac{1}{C R} \int_0^t d t$

$|\ln Q|_{Q_0}^Q=-\frac{1}{C R}|t|_0^t$

$\ln Q-\ln Q_0=-\frac{t}{C R}$

$\ln \frac{Q}{Q_0}=-\frac{t}{C R}$

$Q=Q_0 e^{-t / C R}=Q_0 e^{-t / \tau} \ldots(5)$

where

$\tau=C R$

Hence Equation (5) tells us the charge on the capacitor at any time during the discharge.

If $\mathrm{t}=\mathrm{CR}$, then from equation (5),

$Q=Q_0 e^{-1}=Q_0 / e=0.368 Q=36.8 \% \text { of } Q_0$


The time constant of a CR circuit is also the time it takes for the capacitor's charge to drop from its maximum value to about 0.368 (approximately 1/3) of its maximum value.

So, the charge on the capacitor will only become zero after an infinite amount of time.


Time Constant

The Time Constant is like the timer for a capacitor in a circuit. It's represented by the symbol CR. If CR is much smaller than 1, the capacitor quickly reaches its final charge. If CR is much larger than 1, it takes a long time for the capacitor to reach its final charge. 


If $\mathrm{t}=\mathrm{T}$, then, from equation (4),

$Q=Q_0\left(1-e^{-1}\right)=Q_0\left(1-\frac{1}{e}\right)$

$ Q=Q_0\left(1-\frac{1}{2.718}\right) $

$ =Q_0(1-0.368)=0.632 Q_0 $

$ =63.2 \% \text { of } \mathrm{Q}_0$


So, CR decides how fast the capacitor charges or discharges through a resistor.


When the time t equals the time constant τ, a specific equation shows that the charge on the capacitor is about 63.2% of its maximum value (Q0). In simpler terms, it takes one time constant for the capacitor to reach around two-thirds of its full charge.


The time constant is crucial because it tells us how long it takes for the capacitor to get close to its maximum charge. 


$e^{-t / C R}=0 \text { or } t=\infty$


In theory, it takes an infinite amount of time for the capacitor to reach its absolute maximum charge.



Discharging of Circuit 

At time equals to 5-time constants i.e 5T, the capacitor remains fully charged as long as there is a constant supply applied to it. Now when this fully charged capacitor is disconnected from its DC battery supply, the stored energy accumulated during the charging process will stay indefinitely on its plates, keeping the voltage across its connecting terminals at a constant value.

Now if the battery is replaced by a short circuit when the switch is closed, the capacitor would discharge itself back through the resistor, R as we now have an RC discharging circuit. As the capacitor keeps on discharging, its current through the series resistor the stored energy inside the capacitor is extracted with the voltage Vc across the capacitor that decays to zero finally.

In the RC Circuit Discharging, the time constant (τ) is still equal to the value of 63%. Thus for the RC discharging circuit that is initially fully charged, the voltage across the capacitor after one time constant, 1T, has dropped by 63% of its initial value thus is 1 – 0.63 = 0.37 or 37% of the final value.

[Image will be Uploaded Soon]

As shown in the graph when the switch is first closed, the capacitor starts to discharge. The rate of decay of the RC discharging curve can be seen to be steeper at the beginning because the discharging rate is fastest at the start and then decreases exponentially as the capacitor loses charge at a slower rate. As the discharging continues, the value of VC reduces resulting in a less discharging current.

We know that from the previous RC charging circuit that the voltage across the capacitor, C is equal to 0.5Vc at 0.7T with the steady-state fully discharged value being finally reached at 5T.

Now For the RC discharging circuit, the voltage across the capacitor ( VC ) is the function of time during the discharge period and is defined as 

Vc=Vse-t/RC

Where:

  • VC is the voltage across the capacitor

  • VS is the supply voltage

  • t is the elapsed time since the removal of the supply voltage

  • RC is the time constant


Current During Charging and Discharging of a Capacitor

We know that

$I=\frac{d Q}{d t}$


When the capacitor is charging:

As it charges, 

$=Q_0\left(1-e^{-t / \tau}\right) $

$ I=\frac{d}{d t}(Q)=\frac{d}{d t}\left[Q_0\left(1-e^{-t / \tau}\right)\right]$

$I_{c h}=\frac{Q_0}{\tau} e^{-t / \tau}=I_0 e^{-t / \tau} \ldots(6)$

Where

$I_0=\frac{Q_0}{\tau}$ - Maximum value of the current flowing through the circuit. 

When,

$t=0, I_{c h}=I_0$


When the capacitor is discharging:

As it discharges,

$Q=Q_0 e^{-t / \tau}$

$I=\frac{d}{d t}(Q)=\frac{d}{d t}\left(Q_0 e^{-t / \tau}\right)$

$I_{d i s}=-\frac{Q_0}{\tau} e^{-t / \tau}=-I_0 e^{-t / \tau} \ldots$

When,

$t=0, I_{d i s}=-I_0=I_0$


In terms of strength (magnitude), equations (6) and (7) show that the maximum current values (Ich and Idis) are the same during both charging and discharging.


So, whether the capacitor is charging or discharging through a resistor, the current always decreases from its maximum to zero.


Additionally, at the start (t = 0), $I_{charging} = I_0$ and $I_{discharging} = -I_0$, meaning the directions of the currents in both cases are opposite.


Conclusion

Understanding the charging and discharging of capacitors is crucial for JEE Main aspirants. When a capacitor charges, it stores electrical energy, gradually reaching its maximum capacity. This process is governed by the time constant, which depends on the resistance and capacitance in the circuit. On the other hand, discharging involves the release of stored energy, creating a flow of current. The time taken for the capacitor to discharge is also determined by the time constant. Mastery of these concepts not only aids in problem-solving but also provides a foundation for grasping more advanced electrical principles, making it a fundamental topic for JEE Main preparation.

FAQs on Charging and Discharging of Capacitor

Q1. Explain the Polarity of the Capacitor During Charging and Discharging?

Ans: During the process of charging the capacitor, the current flows towards the positive plate (and positive charge gets added to that plate) and away from the negative plate. While during the discharging of the capacitor, current flows away from the positive and towards the negative plate, in the opposite direction.

Q2. Is the Time for Charging and Discharging of the Capacitor is Equal?

Ans: Process of charging (storage) and discharging (release) of the energy of a capacitor is never instantaneous but it takes a certain amount of time to occur with the time taken for the capacitor to charge or discharge within a certain percentage of its maximum supply value is called as the time Constant (τ).