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Inductance Formula

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  • Last Updated : 04 Apr, 2022

Inductance is a term that every physics student should be familiar with. It has its own formula and is frequently combined with resistance and capacitance. Oliver Heaviside first coined the phrase in 1886. In addition, we use the L to represent inductors on circuit diagrams and inductance in equations, in honor of the eminent physicist Heinrich Let’s learn about the Inductance formula and how to use it to determine the inductance of any item.

Inductance

Inductance is a property of an electrical conductor that causes it to resist changes in the electric current passing through it. The flow of electric current generates a magnetic field around a conductor. The field strength is proportional to the current magnitude and is unaffected by current fluctuations. According to Faraday’s law of induction, any change in the magnetic field via a circuit generates an electromotive force (EMF) (voltage) in the conductors, a process is known as electromagnetic induction.

Inductance can be found in many electrical and electronic systems, as well as circuits. The gears are available in a range of shapes and sizes, as well as a variety of names. Examples include coils, chokes, transformers, inductors, and other parts.

The SI unit of inductance is the henry (H), which can be represented in the current and voltage rate of change.

Formula of Inductance

  • Following is the inductance formula,

L = μN2A/l

Where,

  • L = Inductance (H),
  • μ = Permeability (Wb/Am),
  • N = The coil’s number of turns,
  • A = The coil’s circumference,
  • l = Length of coil (m).

Derivation 

Given: 

E = N(dϕ/dt)

The number of turns in the coil is N, and the induced EMF across the coil is E.

Using Lenz’s law, rewrite the above equation,

E = -N(dϕ/dt)

The previous equation is modified to compute the value of inductance.

E = -N(dϕ/dt)

∴ E = -L(di/dt)

N = dΦ = L di

NΦ = Li

Hence,

The flux density is denoted by B, and the coil area is denoted by A.

Li = NΦ = NBA

Hl = Ni

The magnetizing force of the magnetic flux is denoted by H.

B = μH

Li = NBA

L = NBA/i = N2BA/Ni

N2BA/Hl = N2μHA/Hl

L = μN2A/l 

  • With an inductance of L, the voltage induced in a coil (V) is equal to,

V = L × (di/dt)

Where,

  • V = Voltage (volts),
  • L = Value of Inductance (H),
  • i = Current (A),
  • t = Time taken (s).
  • The inductance reactance is calculated as follows,

X = 2πfL

Where,

  • X = Reactance (ohm),
  • f = frequency (Hz),
  • L = Inductance (H).
  • If Inductance is in series 

L = L1 + L2 + L3 . . . . + Ln

  • If Inductance is in parallel

1/L = 1/L1 + 1/L2 + 1/L3 . . . . + 1/Ln

Sample Questions

Question 1: Define Inductance.

Answer:

Inductance is a property of an electrical conductor that causes it to resist changes in the electric current passing through it. Since inductance has N in the formula, it means that the number of turns in the conductor are directly proportional to the inductance present. However, an interesting fact is that even straight conductors carry inductance just very little in amount to be considerable.

Question 2: What is the SI unit of Inductance?

Answer:

SI unit of Inductance is the henry (H). The discovery of inductance is credited to Faraday, however, the introduction of self inductance for a single circuit was first introduced by Henry. Therefore, the unit of inductance is dedicated to the scientist’s name.

Question 3: Determine the self-inductance of a 210-turn solenoid with a cross-sectional area of 17 cm2 and a length of 66.2 cm.

Solution:

Given: μ = 4π × 10-7, N = 210 turns, A = 17 × 10-4, l = 66.2 × 10-2 

Since,

L = μN2A/l

∴ L = ((4π × 10-7) × (210)2 × (17 × 10-4))/(66.2 × 10-2)

∴ L = 0.0001422

L = 14.22 × 10-5 H

Question 4: What is the corresponding resistance when 16H, 10H, and 21H inductors are joined in series?

Solution:

Given: L1 = 16 H, L2 = 10 H, L3 = 21 H

Since,

L = L1 + L2 + L

∴ L = 16 + 10 + 21

L = 47 H

Question 5: A circuit is connected to a 61 H inductor, and a frequency of 240 Hz is provided. calculate the reactance?

Solution:

Given: f = 240 Hz, L = 61 H

Since,

X = 2πfL

∴ X = 2 × 3.14 × 240 × 61

X = 91939 ohm

Question 6: What is the corresponding resistance when 26H, 16H, 21H, and 30H inductors are connected in parallel?

Solution:

Given: L1 = 26 H, L2 = 16 H, L3 = 21 H, L4 = 30 H

Since,

1/L = 1/L1 + 1/L2 + 1/L3 + 1/L4 

∴ 1/L = 1/26 + 1/16 + 1/21 + 1/30

∴ 1/L = 0.03 + 0.06 + 0.04 + 0.03

∴ 1/L = 0.16 H

L = 6.25 H

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