Test: Inductance - 2 - Electrical Engineering (EE) MCQ

Concept:

• Induction: It is the magnetic field that is proportional to the rate of change of the magnetic field.
• This definition of induction holds for a conductor. Induction is also known as inductance.
• L is used to represent the inductance and Henry is the SI unit of inductance.
• 1 Henry is defined as the amount of inductance required to produce an emf of 1 volt in a conductor when the current change in the conductor is at the rate of 1 Ampere per second.
• Following are the factors that affect the inductance:
  (i) The number of turns of the wire used in the inductor.
  (ii) The material used in the core.
  (iii) The shape of the core.
• Electromagnetic Induction: The law was given by Faraday which states that by varying the magnetic flux electromotive force is induced in the circuit.
• Inductance: It can be defined as the electromotive force generated to oppose the change in current at a particular time duration.

According to Faraday’s Law:

Concept:

Inductive reactance:

• The inductive reactance is the opposition offered by the inductor in an AC circuit to the flow of ac current.
• Its SI unit is ohm(Ω).
• The inductive reactance is given as,

⇒ X_L = 2πfL

Where f = frequency of ac current and L = self - inductance of the coil

Impedance:

• Impedance is essentially everything that obstructs the flow of electrons within an electrical circuit.
• For a pure inductor, the inductive reactance is equal to the impedance.

AC voltage applied to an inductor:

• When an AC voltage is applied to an inductor, the current in the circuit is given as,

⇒ I = V / X_L

• In a pure inductor circuit, the current reaches its maximum value later than the voltage by one-fourth of a period.

Calculation:

Given Vrms = 220 V 

L = 20 mH = 20 × 10-3 H

f = 50 Hz

Inductive reactance is given as,

⇒ X_L = 2πfL

⇒ X_L = 2π × 50 × 20 × 10-3 

⇒ X_L = 6.28 Ω   

When an AC voltage is applied to an inductor, the current in the circuit is given as,

Concept:

Self-inductance 

Here, A = Area

N = Number of turns

l = Length of coil

μ₀ = absolute permeability of material 

μ_r = relative permeability of material 

Calculation:

N = 100, l = 1 m,

A = 100 cm² = 100 cm × cm = 100 × 10^-2 m × 10^-2 m = 100 × 10^-4 m²

μ_∘ = 4 π × 10^-7, μ_r = 1 (Air)

= 40 π μH 

Concept:

• Inductor: The coil which stores magnetic energy in a magnetic field is called an inductor.
• Inductance: The property of an inductor that causes the emf to generate by a change in electric current is called the inductance of the inductor.
• The SI unit of inductance is Henry (H).

Concept:

The current in an inductive circuit is given by:

I = V/X_L

where I = Current

V = Voltage

X_L = Inductive reactance

Explanation:

• XL is inversely proportional to the current.
• Hence, with an increase in the inductive reactance, the current decreases. 

Key Points

• Mutual Inductance is the interaction of one coil's magnetic field on another coil as it induces a voltage in the adjacent coil.
• It is the basic operating principle of the transformer, motors, generators and many other electrical components that interact with another magnetic field. 
• The mutual inductance between the two coils can be greatly increased by positioning them on a common soft iron core or by increasing the number of turns.

Concept:

Solenoid: 

A type of electromagnet that generates a controlled magnetic field through a coil wound into a tightly packed helix.

• The uniform magnetic field is produced when an electric current is passed through it.
• The magnetic field inside a solenoid does not depend on the diameter of the solenoid.
• The field inside is constant.
   

Calculation:

Given: V_dc = 60 V, I_dc = 5 A

Solenoid equivalent is given by series RL circuit.

Now by DC supply, the inductor behaves as a short circuit at a steady state.

V_dc = I_dc × R = 5 × R = 60

R = 12 Ω

Now, when AC supply is given, Current I_ac = 3 A at ω = 400 rad/sec

Equivalent impedance(Z) = V/I = 60/3 = 20Ω

Therefore, 

ωL = 16 Ω

L = 16/400 = 0.04 H

Concept:

The equivalent inductance of series aiding connection is

 L = L₁ + L₂ + 2M

The equivalent inductance of series opposing connection is

L = L₁ + L₂ – 2M

Equivalent inductance of parallel aiding connection is

Equivalent inductance of parallel opposing connection is

Calculation:

Given,

L₁ = 100 mH

L₂ = 200 mH

M = 50 mH

Inductance of parallel aiding connection is

Concept:

The formula for inductance of a coil is

L = μN²A / l

N is no. of turns

A is the cross-sectional area

l length of the solenoid

L ∝ N² / l

L₂L₁ = (N₂ / N₁)² x  l₁l₂

Calculation:

Given:

L₁ = 5 mH, N₁ = 50 turns

Now, The number of turns is doubled i.e.

N₂ = 2N₁ = 100

l₂ = l₁ = l 

L₂ = 20 mH

Concept:

Inductance is the property of a component that opposes the change of current flowing through it and even a straight piece of wire will have some inductance.

Therefore, the below equation will give inductance L as being proportional to the number of turns squared N²

The inductance of a coil of wire is given by,

Where N is the number of turns

A is the cross-sectional area

L is the length of the solenoid.

μ_r is the relative permeability of the core material

From the above relation, we observe that the inductance is directly proportional to the permeability of the material, i.e. higher the permeability of the core material, higher will be the inductance.

Concept:
Self-Inductance 

Here, A = Area

N = Number of turns

l = Length of coil

magnetic flux 

Reluctance s = N² / L.......(i)

Calculation:

L = 4 H, s = 100 AT / Wb

From equation (i)

N = √4×100

= 20

Concept :

Choke coil (or ballast) is a device having high inductance and negligible resistance. It is used to control current in ac circuits and is used in fluorescent tubes. The power loss in a circuit containing choke coil is least.

(1) It consist of a Cu coil wound over a soft iron laminated core

(2) Thick Cu wire is used to reduce the resistance (R) of the circuit

 (3) Soft iron is used to improve inductance (L) of the circuit

(4) The inductive reactance or effective opposition of the choke cost is given by X_L = 2πfL

(5) For an ideal choke coil r= 0, no electric energy is wasted i.e. average power P=0

 6) In actual practice choke coil is equivalent to a R-L circuit.

(7) Choke coil for different frequencies are made by using different substances in their core.

 For low frequency L should be large thus iron core choke coil is used. For high frequency ac circuit, L should be small, so air cored choke coil is used.

Calculations:

P = I² R

X_L = 2 π f L

L= X_L / 2πf

Given

P = 500 W

I = 4 A

V = 250 V

f = 50 Hz

pf  = unity

R = P / I²  = 500 / 16 = 31.25 Ω

Z =  V / I = 250 / 4 = 62.5 Ω          

X_L = 54.125 Ω

L = X_L  / (2π f) = 0.1732 H

Concept:

• When inductors are connected in series so that the magnetic field of one links with the other, the effect of mutual inductance either increases or decreases the total inductance depending upon the amount of magnetic coupling. 
• The effect of this mutual inductance depends upon the distance apart of the coils and their orientation to each other.
• Mutually connected series inductors can be classed as either “Aiding” or “Opposing” the total inductance.

Mutually Aiding:

Leq = L₁ + L₂ + 2M

L : self-inductance

M: Mutual inductance

Mutually Opposing:

Leq = L₁ + L₂ – 2M

Concept:

Self-inductance

Here, A = Area

N = Number of turns

l = Length of coil

magnetic flux 

Calculation:

N= 1000, ϕ = 500 μwb, I = 50 A

R_{l =} reluctance

ϕ = (1000 × 50 ) / Rl

R_l = 10⁸

L = 0.01 H

Concept: 

The impedance Z of an RL circuit is given by,

Z = √(R² + (2πfL)²)

Where:

• R = resistance of the coil
• L = inductance of the coil
• f = frequency of the oscillator

At 50 Hz, the current is 1 A, and the voltage is 5 V, so the impedance is,

1 = 5 / Z₁

This gives

Z₁ = 5 Ω

Now, using the impedance formula,

5 = √(R² + (100πL)²)

25 = R² + (100πL)²

At 100 Hz, the current is 0.625 A, and the voltage is still 5 V, so the impedance is,

0.625 = 5 / Z₂

This gives,

Z₂ = 8 Ω

Now, using the impedance formula,

8 = √(R² + (200πL)²)

Squaring both sides,

64 = R² + (200πL)²

Now we have two equations,

25 = R² + (100πL)²                        ____(1)

64 = R² + (200πL)²                      ______(2)

Subtract the first equation from the second,

39 = (200πL)² - (100πL)²

Factoring out,

39 = 3(100πL)²

(100πL)² = 13

L = √13 / (100π)

L ≈ 0.0115 H

∴ The correct option is 2