Uses of Inductor

Inductors are widely used in electrical and electronic circuits because they can store energy as a magnetic field. One of their main uses is in filters—like low-pass, high-pass, and band-pass filters—where they help control which frequencies can pass through a circuit. They're also key components in transformers, allowing for efficient voltage conversion and power transfer. In power supplies, inductors act as chokes, blocking unwanted high-frequency AC while letting DC or low-frequency AC pass. In tuned circuits, inductors help pick out specific frequencies, which is why they’re essential in radios, TVs, and other communication devices. They’re also used in inductive sensors, such as metal detectors and traffic sensors, to detect changes in magnetic fields. In switching power supplies, inductors help regulate current, while in devices like energy meters and relays, they improve accuracy and performance.

1.0Definition of  Inductors

• An inductor is a component typically made of a coil of conducting wire. When current flows through it, the inductor resists changes in the current due to its magnetic field.
• Inductors are passive electronic components that play a critical role in electrical circuits. While resistors resist current and capacitors store energy in electric fields, inductors store energy in magnetic fields. They are commonly used in power electronics, signal processing, and radio frequency applications.

2.0Diagram of Inductor

3.0Key Point of Inductor

1. A circuit element having a fixed value of inductance is known as an inductor.
2. It is represented by

3. Function of an inductor is to oppose the change in current in the circuit.
4. Potential difference across an inductor in the direction of current is

$V_A-V_B=L\frac{di}{dt}$

1. If $\frac{di}{dt}$ is positive, potential drops from A to B.
2. If $\frac{di}{dt}$ is negative, potential drops from B to A.
3. If $\frac{di}{dt}$ is zero, the potential between  A and B is zero.

4.0Principle of Operation

• An inductor works on Faraday’s Law: a changing magnetic flux through the coil induces an EMF in the circuit.
• When current through an inductor changes, a back EMF is produced that opposes the change in current, in accordance with Lenz's Law.

5.0Self Induction(L)

When current through the coil changes, with respect to time then magnetic flux linked with the coil also changes with respect to time. Due to this an emf and a current is induced in the coil. According to Lenz law, induced current opposes the change in magnetic flux. This phenomenon is called self-induction and a factor by virtue of which the coil shows opposition to change in magnetic flux called self-inductance of the coil. 

Case 1. Current through  the coil is constant

$L=\frac{N\varphi }{I}=\frac{NBA}{I}=\frac{{\varphi }_{\text{Total }}}{I}$

Case 2.Induced EMF in Self Induction

$e_s=\frac{-LdI}{dt}$

• Self Inductance is a scalar quantity
• S I Unit- Henry
• Dimensional Formula-[ML²T^-2A^-2]
• L depends on 
• Geometry of inductor 
• Medium($\mu ={\mu }_0{\mu }_r$)

Self Inductance of Solenoid

Let a current is streaming in the solenoid, magnetic induction in the solenoid is given as, $B={\mu }_{o}nI$

Flux Linkage, $N\varphi =N[BA]=N\left[{\mu }_{o}nIA\right]=N{\mu }_o\left(\frac{N}{l}\right)IA$

$N\varphi =\left(\frac{{\mu }_o{N}^{2}A}{l}\right)I$

Self Induction, $L=\frac{N\varphi }{I}=\frac{{\mu }_o{N}^{2}A}{l}$

$L={\mu }_0\left(\frac{N^2}{l^2}\right)(lA)={\mu }_0{n}^{2}Al={\mu }_0{n}^{2}V$

6.0Power in an Inductor

A battery that drives current through an inductor works against the back EMF, with part of the supplied energy getting stored as magnetic energy in the inductor.

$E-L\frac{dI}{dt}-IR=0$

$E=L\frac{dI}{dt}+IR$

Instantaneous power supplied by battery,

$P=VI=EI$

$P=L\frac{dI}{dt}+I^{2}R$

$L\frac{dI}{dt}\to \text{ Power supplied to the inductor }$

$I^{2}R\to \to \text{ Power dissipated in the resistor }$

7.0Energy Stored in an Inductor

Power supplied to the inductor, $P=L\frac{dI}{dt}$

$\frac{dU}{dt}=LI\frac{dI}{dt}$

$dU=LIdI$

$\int dU=\int LIdI$

$U=\frac{1}{2}L{I}^2$

8.0Energy Density of Inductor

Energy stored in the solenoid is, $U=\frac{1}{2}L{I}^2$

For Solenoid, $L={\mu }_o{n}^{2}V\text{ and }B={\mu }_{o}nI\text{ {here }V\text{ is the volume of inductor }\}$

$U=\frac{1}{2}\left({\mu }_0{n}^{2}V\right){\left(\frac{B}{{\mu }_{0}n}\right)}^2$

Energy stored per unit volume, $\frac{U}{V}=\frac{B^2}{2{\mu }_o}$

Note :

1. The above result is derived for an inductor but is true in general for any system.
2. Magnetic energy density in free space is, $\frac{U}{V}=\frac{B^2}{2{\mu }_o}$
3. Electric Energy density in free space is, $\frac{U}{V}=\frac{1}{2}{ϵ}_o{E}^2$

9.0Inductor Behavior as a Battery

Case 1: When current is increasing

Case 2: When current is decreasing

Case 3: When current is constant

Note: If the inductor has resistance in its coil, it not only opposes changes in current but also causes energy loss as heat. This internal resistance can reduce efficiency and affect circuit performance. 

10.0Inductors in Series and Parallel

1. Series Combination

$L_{eq}=L_1+L_2$

2. Parallel Combination

$\frac{1}{L_{eq}}=\frac{1}{L_1}+\frac{1}{L_2}$

Note-If an inductor is cut into two parts, its time constant remains the same.

11.0Properties of a Coil (Inductor)

1. Inductance (L) is the ability of an inductor to resist changes in current. It increases with more coil turns and better core materials.
2. Self-inductance means a coil can generate a voltage within itself when the current changes.
3. Mutual inductance happens when one coil induces voltage in a nearby coil.
4. The Q-factor or quality factor measures how efficient an inductor is—higher Q means less energy is lost.
5. The core material—like ferrite, air, or iron—affects how well the magnetic field forms and, therefore, the inductance.
6. Although inductors ideally have no resistance, real ones do have some because of the wire used.
7. Finally, saturation occurs when the core can’t handle more magnetic flux at high currents, causing the inductance to drop.

12.0Practical Uses of Inductors

1.  Power Supply Filters: Inductors are used in LC filters to smooth out the ripples in power supplies. They block unwanted high-frequency AC signals while letting the steady DC flow through.
2. Energy Storage in Switch Mode Power Supplies (SMPS): In DC-DC converters, inductors temporarily store energy and release it as needed to keep the voltage stable.
3. RF and Communication Circuits: Inductors work with capacitors in LC circuits to pick out or block specific frequencies in devices like radios, tuners, and oscillators.
4. Transformers (Mutual Inductance): When two or more inductors are wound on the same core, they form a transformer, which is used to change voltage levels and provide electrical isolation.
5. Inductive Sensors: These inductors help detect changes in magnetic fields, making them useful in devices like proximity sensors, metal detectors, and traffic sensors.
6. Audio Crossovers: In speaker systems, inductors prevent high frequencies from reaching the woofers, helping deliver clearer sound.
7. Induction Heating: Inductors generate eddy currents in metal objects, which produces heat for cooking and industrial heating applications.
8. Relay and Motor Control: Inductors are part of relay coils and motors, creating magnetic fields that enable mechanical movement and control.

13.0Illustrations on Inductors

Illustration-1.Find $V_A-V_B$ in the given circuit if-Current is decreasing at the rate $10^3\mathrm{A}/\mathrm{s}$

Solution: Given, $\frac{di}{dt}=10^3$

KVL equation from A to B

$V_A-iR-E-L\frac{di}{dt}=V_B$

$V_A-(5)(1)-15-\left(5\times 10^{-3}\right)\left(-10^{-3}\right)=V_B$

$V_A-V_B=15V$

Illustration-2.Find $V_A-V_B$ in the given circuit if-Current is increasing at the rate $10^3\mathrm{A}/\mathrm{s}$

Solution: Given, $\frac{di}{dt}=10^3$

KVL equation from A to B

$V_A-iR-E-L\frac{di}{dt}=V_B$

$V_A-(5)(1)-15-\left(5\times 10^{-3}\right)\left(10^{-3}\right)=V_B$

$V_A-V_B=25V$

Illustration-3.Find $V_A-V_B$ in the given circuit if-Current is constant.

Solution: Given, $\frac{di}{dt}=0$

KVL equation from A to B

$V_A-iR-E-L\frac{di}{dt}=V_B$

$V_A-(5)(1)-15-\left(5\times 10^{-3}\right)(0)=V_B$

$V_A-V_B=20V$