Inductance MCQ Quiz - Objective Question with Answer for Inductance - Download Free PDF
Last updated on May 8, 2025
Latest Inductance MCQ Objective Questions
Inductance Question 1:
What is the function of the armature core in relation to the magnetic circuit?
Answer (Detailed Solution Below)
Inductance Question 1 Detailed Solution
The correct answer is option 4.
Function of the Armature Core in Magnetic Circuits
Magnetic Path Completion:
- The armature core is a key component in the magnetic circuit of machines like motors and generators.
- It provides a low-reluctance path for magnetic flux to travel between the field poles and yoke, completing the magnetic loop.
Efficient Magnetic Flux Conduction:
- Made of laminated soft iron or silicon steel to reduce eddy current losses.
- It efficiently conducts the alternating magnetic flux without significant energy losses.
Supports Induced EMF:
- As the armature rotates within the magnetic field, the armature windings (placed in slots on the core) cut through magnetic lines of force, inducing an electromotive force (EMF).
Mechanical Structure:
- It also serves as a mechanical support for the conductors/windings.
Inductance Question 2:
If the frequency of a series R-C circuit is increased, what happens to the capacitive reactance XC?
Answer (Detailed Solution Below)
Inductance Question 2 Detailed Solution
Concept
The capacitive reactance XC for a series RC circuit is given by:
\(X_C={1\over 2\pi fC}\)
From the above expression, we found that the frequency is inversely proportional to the capacitive reactance.
So, if the frequency of a series R-C circuit is increased, than the capacitive reactance will decrease.
Inductance Question 3:
In an inductive coil, the rate of change of current is maximum:
Answer (Detailed Solution Below)
Inductance Question 3 Detailed Solution
Explanation:
Inductive Coil and Rate of Change of Current
Definition: An inductive coil, often simply referred to as an inductor, is a passive electrical component that stores energy in its magnetic field when electrical current passes through it. It resists changes in the current flowing through it due to its property known as inductance.
Working Principle: When a voltage is applied across an inductor, it creates a time-varying magnetic field, resulting in an induced electromotive force (emf) that opposes the change in current. This phenomenon is described by Faraday's Law of Electromagnetic Induction and Lenz's Law.
Correct Option Analysis:
The correct option is:
Option 2: At the start of current flow.
This option accurately describes the behavior of an inductive coil when a voltage is first applied. At the initial moment of current flow, the rate of change of current is at its maximum. This can be understood by examining the fundamental principles of inductance and the behavior of an RL (resistor-inductor) circuit.
When a voltage \( V \) is applied to an inductive coil with inductance \( L \) and resistance \( R \), the current \( I(t) \) through the coil at any time \( t \) is given by the differential equation:
\( V = L \frac{dI(t)}{dt} + IR \)
At the moment the voltage is applied (\( t = 0 \)), the current \( I(0) \) is zero, and the rate of change of current \( \frac{dI(0)}{dt} \) is at its peak because the entire voltage is initially dropped across the inductor:
\( V = L \frac{dI(0)}{dt} \)
Thus, the maximum rate of change of current occurs at \( t = 0 \), immediately after the voltage is applied.
Additional Information
To further understand the analysis, let’s evaluate the other options:
Option 1: After one time constant.
This option is incorrect. The time constant (\( \tau = \frac{L}{R} \)) of an RL circuit represents the time required for the current to reach approximately 63.2% of its maximum value. After one time constant, the rate of change of current is not at its maximum; it has significantly decreased as the current approaches its steady-state value.
Option 3: Near the final maximum value of current.
This option is also incorrect. As the current approaches its final steady-state value, the rate of change of current diminishes and approaches zero. At the maximum current value, the inductor behaves almost like a short circuit, and there is no further change in current.
Option 4: At 36.8% of its maximum steady-state value.
This option is incorrect. At 36.8% of its maximum steady-state value (which corresponds to \( 1 - \frac{1}{e} \)), the rate of change of current is not at its maximum. The maximum rate of change occurs at the very start, not at this particular value of current.
Conclusion:
Understanding the behavior of inductive coils in electrical circuits is essential for analyzing how current changes over time. The rate of change of current in an inductor is highest at the start of current flow, immediately after a voltage is applied. This behavior is crucial for the design and analysis of circuits involving inductive components, including those used in power supplies, transformers, and various types of electronic equipment.
Inductance Question 4:
An iron cored coil is converted to an air cored coil by removing the iron core. The coil's inductance will ____________
Answer (Detailed Solution Below)
Inductance Question 4 Detailed Solution
- When the iron core is removed from an iron cored coil, it becomes an air cored coil. The inductance of the coil will decrease.
Explanation:
- Inductance is a property of an electrical circuit that opposes the change in current flowing through it. It is a measure of the ability of a coil to store electrical energy in the form of a magnetic field.
- The presence of an iron core in a coil increases its inductance. This is because iron is a ferromagnetic material, which means it can easily magnetize and demagnetize. When a current flows through a coil with an iron core, the magnetic field produced by the current induces a strong magnetic field in the iron core. This increases the inductance of the coil.
- The inductance of a coil is given by the formula L = (μ₀μᵣN²A)/l, where L is the inductance, μ₀ is the permeability of free space, μᵣ is the relative permeability of the core material, N is the number of turns in the coil, A is the cross-sectional area of the coil, and l is the length of the coil.
- When the iron core is replaced by air, the relative permeability (μᵣ) becomes 1, as air has a relative permeability very close to 1. As a result, the inductance of the coil decreases, as the term μ₀μᵣ in the formula becomes μ₀ × 1 = μ₀.
- Therefore, the inductance of the coil will decrease when the iron core is removed and the coil becomes an air cored coil.
Inductance Question 5:
What will be the self-inductance of the coil if an EMF of 10 V is induced in it when the current flowing through it changes at the rate of 5 A/sec?
Answer (Detailed Solution Below)
Inductance Question 5 Detailed Solution
Concept:
Self-Induction
- Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
- As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
- This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
Induced e.m.f can be given as
\(⇒ V_{l}=-L\frac{dI}{dt}\) -----(1)
Where
VL = induced voltage in volts
N = self-inductance of the coil
\(\frac{dI}{dt}=\) rate of change of current in ampere/second
Calculation:
Given: di/dt = 5 A/sec, V = 10 volt
From equation 1,
\(\Rightarrow 10=-L\times 5\)
The negative sign indicates that induced emf (e) opposes the change of flux.
⇒ L = 2 H
Top Inductance MCQ Objective Questions
The inductance of a single layer coil of 50 turns is 5 mH. If the no. of turns is doubled, the inductance of the coil will become ______
Answer (Detailed Solution Below)
Inductance Question 6 Detailed Solution
Download Solution PDFConcept:
The formula for inductance of a coil is
\(L = \frac{{\mu {N^2}A}}{l}\)
N is no. of turns
A is the cross-sectional area
l length of the solenoid
\(L \propto \frac{{{N^2}}}{l}\)
\(\frac{{{L_2}}}{{{L_1}}} = {\left( {\frac{{{N_2}}}{{{N_1}}}} \right)^2}\frac{{{l_1}}}{{{l_2}}}\)
Calculation:
Given:
L1 = 5 mH, N1 = 50 turns
Now, The number of turns is doubled i.e.
N2 = 2N1 = 100
l2 = l1 = l
\(\frac{{{L_2}}}{{5 \ mH}} = {\left( {\frac{100}{50}} \right)^2} \)
L2 = 20 mH
An iron-cored coil has an inductance of 4 H. If the reluctance of the flux path is 100 AT/Wb, then the number of turns in the coil is:
Answer (Detailed Solution Below)
Inductance Question 7 Detailed Solution
Download Solution PDFConcept:
Self-inductance \(L = \frac{{{{\rm{μ }}_0}{{\rm{μ }}_{\rm{r}}}{\rm{A}}{{\rm{N}}^2}}}{l}\) ... (1)
Here, A = Area
N = Number of turns
l = Length of coil
Reluctance (S) = \(\frac{1}{\mu_{0}\mu_{r}}\times \frac{l}{A}\) ... (2)
From equation (1) and (2)
Reluctance \(s = \frac{{{N^2}}}{L}\)
Calculation:
L = 4 H, s = 100 AT / Wb
From equation (i)
\(N = \sqrt {4 \times 100} \)
= 20
If both the number of turns and core length of an inductive coil are doubled, then its self inductance will be
Answer (Detailed Solution Below)
Inductance Question 8 Detailed Solution
Download Solution PDFCONCEPT:
Self-Induction
- Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
- As a result of this, in accordance with Faraday’s laws of electromagnetic induction, an emf is induced in the coil which opposes the change that causes it.
- This phenomenon is called ‘self-induction’ and the emf induced is called back emf, current so produced in the coil is called induced current.
- Self-inductance of a solenoid is given by –
\(⇒ L=\frac{{{\mu }_{o}}{{N}^{2}}A}{l}\)
Where μo = Absolute permeability, N = Number of turns, l = length of the solenoid, and A = Area of the solenoid.
EXPLANATION:
Given - l2 = 2l1
N2 = 2 N1
Self-inductance of a solenoid is given by:
\(⇒ L=\frac{{{\mu }_{o}}{{N}^{2}}A}{l}\)
\(⇒ L\propto \frac{N^2}{l}\)
\(\Rightarrow \frac{L_2}{L_1}=\frac{N_2^2}{N_1^2}\times \frac{l_1}{l_2}=\frac{(2N_1)^2}{N_1^2}\times \frac{l_1}{2l_1}=2\)
L2 = 2 L1
In the figure shown below, if the current I decreases at a rate of β, then VP - VQ is:
Answer (Detailed Solution Below)
Inductance Question 9 Detailed Solution
Download Solution PDFThe correct answer is option 4):(- L β )
Concept:
For an inductor, The potential difference
ΔV = L \(di\over dt\)
where
ΔV is the change in voltage
\(di\over dt\) is the rate of change of current with respect to time
L is the self-inductance
Calculation:
\(di\over dt\) = - β
VP - VQ = L \(di\over dt\)
= - Lβ V
Basic unit of inductance is ______.
Answer (Detailed Solution Below)
Inductance Question 10 Detailed Solution
Download Solution PDFThe correct answer is Henry.
- Henry is the SI unit of inductance.
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.
Additional Information
- Weber is the SI unit of Magnetic flux.
- Farad is the SI unit of capacitance.
- Coulomb is the SI unit of Electric charge .
When a dc voltage of 60 V is applied across a solenoid, an electric current of 5 A flows through it. When the dc voltage is replaced by the ac voltage of the angular frequency 400 rad/s, the electric current is reduced to 3A. What is the inductance of solenoid?
Answer (Detailed Solution Below)
Inductance Question 11 Detailed Solution
Download Solution PDFConcept:
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: Vdc = 60 V, Idc = 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.
Vdc = Idc × R = 5 × R = 60
R = 12 Ω
Now, when AC supply is given, Current Iac = 3 A at ω = 400 rad/sec
Equivalent impedance(Z) = V/I = 60/3 = 20Ω
Therefore, \(Z=\sqrt{(R)^2+(ω L)^2}\)
\(20=\sqrt{(12)^2+(ω L)^2}\)
ωL = 16 Ω
L = 16/400 = 0.04 H
Inductance has the dimension of
Answer (Detailed Solution Below)
Inductance Question 12 Detailed Solution
Download Solution PDFExpressions for Self-inductance:
- The property of a coil that opposes any change in the amount of current flowing through it is called its self-inductance or inductance.
Factors affecting inductance
- The greater the self-induced voltage, the greater the self-inductance of the coil, and hence larger is the opposition to the changing current.
- According to Faraday’s laws of electromagnetic induction, induced voltage in a coil depends upon the number of turns (N) and the rate of change of flux (dφ/dt) linking the coil.
Hence, the inductance of a coil depends upon these factors:
- Shape and number of turns.
- Relative permeability of the material surrounding the coil.
- The speed with which the magnetic field changes.
Consider an iron-cored solenoid of dimensions as shown in Fig.
The inductance of the solenoid (L) is given by,
\(L=N\frac{dϕ}{dI}\)
Now, flux (ϕ) = \(\frac{NI}{S}=\frac{NI}{l/a\mu_0 \mu_r}\)
Where, NI is MMF and S is reluctance,
Differentiating ϕ wrt I,
\(\frac{d\phi}{dI}=\frac{Na\mu_0\mu_r}{l}\)
Now, \(L=N\frac{dϕ}{dI}=N\frac{Na\mu_0\mu_r}{l}=\frac{N^2}{S}\)
Now, inductive reactance (XL) = ωL = \(\dfrac{\omega N^2}{S}\)
Conclusion:
From the above concept,
Inductance has the dimension of Flux/current.
A 500 W discharge lamp takes a current of 4 A at unity p.f. Find the inductance of a choke required to enable the lamp to work on 250 V, 50 Hz main.
Answer (Detailed Solution Below)
Inductance Question 13 Detailed Solution
Download Solution PDFConcept :
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 XL = 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
\(Z = \sqrt {{R^2} + X_L^2} \)
XL = 2 π f L
\(L = \frac{{{X_L}}}{{2\;\pi \;f}}\)
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
XL = 54.125 Ω
L = XL / (2π f) = 0.1732 H
Calculate the inductance of a coil that induces 150 V, when a current changes at the rate of 3 A/s?
Answer (Detailed Solution Below)
Inductance Question 14 Detailed Solution
Download Solution PDFConcept
The voltage across an inductor in a coil is given by:
\(V_L=L{di\over dt}\)
where, VL = Voltage across inductor
L = Inductance
\({di\over dt}=\) Rate of change of current
Calculation
Given, VL = 150 V
\({di\over dt}= 3\space A/s\)
\(150=L\times 3\)
L = 50 H
Self-induction is sometimes analogously called:
Answer (Detailed Solution Below)
Inductance Question 15 Detailed Solution
Download Solution PDFSelf induction:
- The self inductance is defined as the property by virtue of which coil opposes any change in the current flowing through it.
- Due to this self inductance, when the current in the coil is increased, the self induced EMF in it will oppose this increase by acting in the direction opposite to that of the applied EMF.
- Similarly, if the current in the coil is decreased, he self induced EMF will tend to keep the current at its original value by acting in the same direction as the applied EMF.
- Thus, any change in the current through the coil is opposed due to its self inductance.
- Hence, self-inductance is sometimes analogously called electrical inertia or electromagnetic inertia.