Inductance MCQ Quiz in मल्याळम - Objective Question with Answer for Inductance - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Energy stored in an inductor of inductance L and current I is given by

\(E = \frac{1}{2}L{I^2}\)

When current is being changed from \({I_1}\) to \({I_2}\) change in energy will be

ΔE = E2 – E1

\({\rm{\Delta }}E = \frac{1}{2}LI_2^2 - \frac{1}{2}LI_1^2\)     ----(1)

As and  are given, we need to find value of L.

Now, induced emf in a coil is

\(\epsilon = L\frac{{dI}}{{dt}}\)

Calculation:

Given,

Emf of the coil, ϵ = 26V

Time, t = 5s

Current I1 = 10 A

Current, I2 = 75 A

Here, ϵ = 25 V,

dI = I2 – I1 = (75 – 10)A = 65A

dt = 5 s

\(\epsilon = L\frac{{dI}}{{dt}}\)

\(\Rightarrow 26 = L \times \frac{{65}}{5}\)

\(L = \frac{{26}}{{13}} = 2 \ H\)

Putting values of L, \({I_1}\) and \({I_2}\) in equation (1), we get,

\({\rm{\Delta }}E = \frac{1}{2} \times 2\times \left[ {{{75}^2} - {{10}^2}} \right]\)

\({\rm{\Delta }}E = \frac{1}{2} \times 2 \times \left( {5625 - 100} \right)\)

\({\rm{\Delta }}E = \frac{1}{2} \times 2\times 5525\)

ΔE = 5525 J

In resistors, when two wires are wound side by side in opposite directions, it is done to reduce the coil inductance.

Concept:

When current flows through a conductor, it creates a magnetic field around the conductor. This magnetic field induces a self-inductance in the conductor itself and in any adjacent conductors or coils. Inductance is a property that opposes changes in the current flowing through the conductor. In a coil, inductance is primarily determined by the number of turns and the geometry of the coil.

Coil Winding Technique:

By winding two wires side by side in opposite directions, a resistor designer can reduce the coil inductance. This technique is known as "bifilar winding" or "counter-wound winding." The two wires are wound closely together, with one wire wound clockwise and the other wire wound counterclockwise.

How Bifilar Winding Reduces Coil Inductance:

• When the wires are wound in opposite directions, the magnetic fields created by the current flowing through each wire interact and cancel each other out to a certain extent. This cancellation effect reduces the overall magnetic field and, consequently, the inductance of the coil.
• The reduction in coil inductance is desirable in resistors because resistors are primarily designed to provide a specified resistance value without significant inductive or capacitive effects. By minimizing inductance, the resistor behaves more like an ideal resistance component, making it suitable for various applications in electrical and electronic circuits.
• It's worth noting that other factors, such as the wire spacing, wire diameter, and the presence of magnetic core material, can also affect the coil inductance. However, the specific technique of winding two wires side by side in opposite directions helps to reduce the coil inductance by utilizing the canceling effect of the magnetic fields generated by the wires.

Therefore, the purpose of winding two wires side by side in opposite directions in resistors is to reduce the coil inductance.

Current: Rate of flow of charge is known as current. It can be expressed as

\(I = \frac{dq}{dt}\)

Where, q = Charge in coulomb

t = time in seconds

The energy stored by an inductor carrying a current I is given by:

Energy, E = \(\frac{1}{2}L{I^2}\)  Joule

Where, L = Inductance of the coil

Now energy stored by inductor can also be expressed as

E = \(\frac{1}{2}L(\frac {dq}{dt})^2 \) J

Self-inductance is the property of the coil due to which it opposes any increase or decrease of current ( which produces flux) through it.

If self-inductance is reduced, this implies that now the opposition to the current is less i.e. in less time the current can become steady through the coil.

Calculation:

By applying KVL in the first loop,

\( - V + {L_1}\frac{{d{i_1}}}{{dt}} + M\frac{{d{i_2}}}{{dt}} = 0\)

By applying KVL in the second loop,

\({L_2}\frac{{d{i_2}}}{{dt}} + M\frac{{d{i_1}}}{{dt}} = 0\)

\( \Rightarrow \frac{{d{i_2}}}{{dt}} = - \frac{M}{{{L_2}}}\frac{{d{i_1}}}{{dt}}\)

From the above equations,

\( \Rightarrow - V + {L_1}\frac{{d{i_1}}}{{dt}} + M\left( { - \frac{M}{{{L_2}}}\frac{{d{i_1}}}{{dt}}} \right) = 0\)

\( \Rightarrow V = \left( {{L_1} - \frac{{{M^2}}}{{{L_2}}}} \right)\frac{{d{i_1}}}{{dt}}\)

Derivation:

By applying KVL in the first loop:

\( - V + {L_1}\frac{{d{i_1}}}{{dt}} + M\frac{{d{i_2}}}{{dt}} = 0\)

By applying KVL in the second loop,

\({L_2}\frac{{d{i_2}}}{{dt}} + M\frac{{d{i_1}}}{{dt}} = 0\)

\( \frac{{d{i_2}}}{{dt}} = - \frac{M}{{{L_2}}}\frac{{d{i_1}}}{{dt}}\)

From the above equations,

\( - V + {L_1}\frac{{d{i_1}}}{{dt}} + M\left( { - \frac{M}{{{L_2}}}\frac{{d{i_1}}}{{dt}}} \right) = 0\)

\( V = \left( {{L_1} - \frac{{{M^2}}}{{{L_2}}}} \right)\frac{{d{i_1}}}{{dt}}\)

The inductance of an inductor is primarily determined by four factors:

• The type of core material
• The number of turns of wire
• The spacing between turns of wire
• The diameter of the coil (core)

The type of core material:

• The core of an inductor is the material that occupies the space enclosed by the turns of the inductor.
• The amount of current in an iron-core inductor also influences its inductance. This is because of the magnetic properties of the iron core change as the current changes.
• The amount of inductance is determined by the amount of emf produced by a specified current change. The amount of emf depends on how much flux interacts with the conductors of the coil.
• If all the other factors are equal, an iron-core inductor has more inductance than an air-core inductor. This is because the iron has a higher permeability i.e. it can carry more flux.
• The brass core has more reluctance which opposes the flux compared to air core or an iron core. Therefore, the brass slug decreases inductance when it is centered in the coil.
• Therefore, if a brass core of an inductor is replaced by an iron core, the inductance of coil increases.

The correct answer is option 2):(1 V)

Concept:

The induced emf is related to the change in current as follows:

e=- L \(di \over dt\)

Where

e is the induced emf,

 L is the self-inductance of the coil,

and

di/dt is the rate of change of current in the coil.

Calculation:

L = 20 H

\(di \over dt\) = 50 × 10^-3 A\sec

e =  -20 ×  50 × 10-3 

e = 1 V

Caclulation:

from the figure given in the question,

L1 = 12 H and L₂ = 12 H

L₃ = 6 H and L₄ = 12 H

L₅ = 12 H and L₆ =  12 H.

L₇ = 6 H

L₁ and L₂ are parallel to one another and their combination is in series with L₃ , 

= (12/2) + 6 

= 12 H

Then in parallel with L4

= 12/2

= 6 H

the equivalent 6 H inductance is in series with L₇

= 6 + 6

= 12 H ----(1)

L₆ and L₅ are parallel to each other.

= 12/2

= 6 H ----(2)

The (1) and (2) are parallel to each other

then the required L_eq would be

Leq = \(\frac{12\times6}{12+6} = 4\)

Leq = 4 H

The correct answer is option (2)

Concept:

Parallel combination:

When two or more inductors are connected in such a way that their ends are connected at the same two points and have an equal potential difference for all inductors is called a parallel combination of the inductor.

Equivalent Inductance(Leq) for parallel combination:

\(\frac{1}{{{L_{eq}}}} = \frac{1}{{{L_1}}} + \frac{1}{{{L_2}}} + \frac{1}{{{L_3}}}\)

Where

L1 is the inductance of the first inductor,

L2  is the inductance of the second inductor,

L3  is the inductance of the third inductor,

Series combination:

When two or more inductors are connected end to end and have the same electric current on each is called the series combination of the inductors.​

Equivalent inductance (Leq) in series combination:

Leq = L1 + L2 + L3

Calculation:

Here, 20 H, 12 H & 10 H are connected in series.

Hence, the circuit can be drawn as,

Here, 7 H & 42 H are connected in parallel [7||42 = \(\frac{7\times 42}{49}=6\ H\)]

Hence, the circuit can be drawn as,

∴ L_eq = 4 + 6 + 8 = 18 H