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

Concept:

Heating coil:

The heating coil is a strip of wire that gives off heat much like a lamp filament.

→ when electric current flows through it glow red hot and convert electric energy into heat.

→ The most widely used wire is Nichrome wire due to its high resistance for heating coil.

→ Heat produced in heating coil \(\left( P \right) = \frac{{{V^2}}}{R}\) 

→ In series equivalent resistance is

R_eq = R₁ + R₂ + R₃ + R₄

→ In parallel equivalent resistance is

\(\frac{1}{{{R_{eq}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_2}}} + \frac{1}{{{R_3}}} + \frac{1}{{{R_4}}}\)

Analysis:

Since heat produce in heating coil is

\(P = \frac{{{V^2}}}{R}\)

So, when Resistance is maximum then produced heat is minimum.

Maximum resistance occurs when all resistance are connected in series i.e.

So heating coil will produce minimum heat when all resistance is connected in series.

R_eq = 4R

Option 2 is the correct choice.

More Information:

Properties of Heating coil Material.

1) High melting point.

2) Free from oxidation in open atmosphere

3) High tensile strength

4) High Resistivity

5) Low-temperature coefficient of resistance.

CONCEPT :

• Self-Induction: Whenever the electric current passing through a coil changes, the magnetic flux linked with it will also change.
• According to 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.
• When a current flows through a coil a flux linked with the circuit changes, the amount induced flux is given by

ϕ = L I 

Where I = Amount of current, L= Coefficient of self-induction

• Then the amount of induced EMF is given by

\(\Rightarrow e= -L\frac{dI}{dt}\)

Where ϕ = Flux linked with the circuit, \(\frac{dI}{dt}\) =rate of change of current, and L =  Coefficient of Self-inductance 

EXPLANATION:

• The amount EMF induced in a coil is given by

\(\Rightarrow e = - L\frac{dI}{dt} \)    

• Applying equation 1 from 0 to \(\frac{T}{4}\) the value of induced emf will be a negative value

∴ e = Negative value                    [ From 0 to \(\frac{T}{4}\)]

• Applying equation 1 from \(\frac{T}{4}\) to \(\frac{T}{2}\) the value of induced emf will be zero. Since the variation of current is zero in that region.

∴ e = 0              [ From \(\frac{T}{4}\) to \(\frac{T}{2}\)] 

• Applying equation 1 from \(\frac{T}{2}\) to T the value of induced emf will be a positive quantity. Since the variation of current is negative in that region.

∴ e = Positive Value           [\(\frac{T}{2}\) to T]

• From the discussion's it is clear that the value of EMF(ϵ) has to be

1. from 0 to \(\frac{T}{4}\) the value of induced emf will be a negative value
2.  from \(\frac{T}{4}\) to \(\frac{T}{2}\) the value of induced emf will be Zero
3. ​ from \(\frac{T}{2}\) to T, the value of induced emf will be a positive 

• Option 2  matches with the above discussion, Hence option 2 is the answer.

Given that, μ_r = 60

N₁ = 1200

N₂ = 1000

d₁ = 5 cm ⇒ r₁ = 2.5 cm

l = 100 cm

We know that:

\(M = \frac{{{\mu _0}{\mu _r}{N_1}{N_2}{A_1}}}{l}\)

\(= \frac{{60 \times 4\pi \times {{10}^{ - 7}} \times 1200 \times 1000 \times \pi \times {{\left( {2.5 \times {{10}^{ - 2}}} \right)}^2}}}{{100 \times {{10}^{ - 2}}}}\)

\( = 177.6\;mH\)

Concept:

The inductance of a solenoid is given by:

\(L = \frac{{{\mu _0}{N^2}A}}{l} \)

N = Number of turns

A = Area of the solenoid

l = length of the solenoid

Also, the energy stored by the inductor is given by:

\(E=\frac{1}{2}LI^2\)

Calculation:

Putting on the respective values, we get:

\(L= \frac{{4\pi \times {{10}^{ - 7}} \times {{1000}^2} \times \frac{\pi }{4} \times {{\left( {3 \times {{10}^{ - 2}}} \right)}^2}}}{{30 \times {{10}^{ - 2}}}} \)

\(L= 2.96 \times {10^{ - 3}}\)

Now, the energy stored by the inductor will be:

\(E = \frac{1}{2}L{I^2} = \frac{1}{2} \times 2.96 \times {10^{ - 3}} \times {10^2} \)

\(E= 0.148\;J\approx0.15~ J\)

\(L = \frac{{{\mu _o}{N^2}A}}{L} = \frac{{4\pi \times {{10}^{ - 7}} \times {{\left( {400} \right)}^2} \times \left( {16 \times {{10}^{ - 4}}} \right)}}{{1 \times {{10}^{ - 3}}}}\)

= 321.6 mH

\(\begin{array}{l} V{\rm{ }} = {\rm{ }}I{X_L}\\ \Rightarrow I = \frac{{230}}{{2\pi fL}} = \frac{{230}}{{2 \times 3.14 \times 50 \times 321.6 \times {{10}^{ - 3}}}} \end{array}\)

= 2.28 A

Concept:

The choke input filter, also known as the L-Section filter is a combination of an inductor filter and a capacitor filter. It consists of an inductor in series and a capacitor in shunt with the load R_L. It is used to smooth out the pulsating dc obtained from the rectifiers.

Circuit of Choke Input Filter:

Solution:

The inductor (choke) allows the dc component to pass and restricts the ac component of the input. For this, XL >> XC

If any component of the ac is still present after passing through the inductor, it is bypassed through the capacitor in the shunt. For that, the impedance of the capacitance must be very less than the resistance R_L. Hence, X_C << R_L. 

For the well-designed choke input filter with a high load, 

\(R_L>>X_L>>X_C\)

Hence, X_L is much greater than X_C at the input frequency.

Additional Information

For a half-wave rectifier, \(R_L>>\omega L>>{1 \over \omega C}\)

For a full-wave rectifier, \(R_L>>2\omega L>>{1 \over 2\omega C}\)

Inductor

An inductor is an electrical element that stores energy in the form of a magnetic field.

An inductor opposes the change in current passing through it.

The S.I. unit of inductance is henry (H) and is denoted by the symbol L.

If current 'I' flows through the inductor, then the induced voltage is given by:

\(V_L=L{di\over dt}\)

Additional Information A capacitor stores energy in the form of electrical energy.

Concept:

In steady-state inductor behave as short circuit

→ In steady-state capacitor behave as an open circuit

Calculation:

At t = 0^- → the switch will be open, the inductor will be short-circuited.

Using current division Rule

\({i_L} = \frac{{10\; \times \;1}}{{3 \;+\; 1}} = \frac{{10}}{4} = 2.5\;A\)

Note: None of the options is correct, the most appropriate option is 4. 

Concept:

Inductor:

• A passive two-terminal electrical component that stores energy in a magnetic field when electric current flows through it.
• When current flows through inductor then time-varying magnetic field induces an EMF in the conductor (Faraday's law of induction).
• According to Lenz's law, the induced voltage opposes the change in current that created it. So inductor opposes any changes in current through them.
• It is also known as coil, choke or reactor.
• The energy stored in an inductor is given by

E = (½)LI²

Where E = Energy stored in Inductor (Joule)

L = Inductance of the inductor (H)

I = current trough inductor (A)

Calculation:

Given:

L = 1 H

I = 2 A

E = (½)LI² = (½) × 1 × 2² = 2 J

Explanation:

Energy Stored in the Magnetic Field of a Solenoid

Definition: The energy stored in the magnetic field of a solenoid is the energy associated with the magnetic field created by the current flowing through the solenoid. This energy can be calculated using the formula for magnetic energy density and the volume of the solenoid.

Given Data:

• Length of the solenoid (L) = 40 cm = 0.4 m
• Diameter of the solenoid = 4 cm = 0.04 m
• Radius of the solenoid (r) = Diameter / 2 = 0.04 / 2 = 0.02 m
• Number of turns (N) = 100
• Current through the solenoid (I) = 20 A

Formula for Energy Stored in the Magnetic Field:

The energy stored in the magnetic field of a solenoid is given by:

U = (1/2) × L × I²

where:

• U = Energy stored (in Joules)
• L = Inductance of the solenoid (in Henry)
• I = Current through the solenoid (in Amperes)

Step 1: Calculate the Inductance (L) of the Solenoid

The inductance of a solenoid is given by the formula:

L = (μ₀ × N² × A) / l

where:

• L = Inductance (in Henry)
• μ₀ = Permeability of free space = 4π × 10^-7 H/m
• N = Number of turns
• A = Cross-sectional area of the solenoid = π × r² (in m²)
• l = Length of the solenoid (in meters)

Substitute the values:

A = π × r² = π × (0.02)² = 3.14 × 0.0004 = 0.001256 m²

L = (4π × 10^-7 × 100² × 0.001256) / 0.4

L = (4π × 10^-7 × 10000 × 0.001256) / 0.4

L = (1.577 × 10^-2) / 0.4

L = 3.9425 × 10^-2 H

Step 2: Calculate the Energy Stored (U)

U = (1/2) × L × I²

Substitute the values:

U = (1/2) × 3.9425 × 10^-2 × (20)²

U = 0.5 × 3.9425 × 10^-2 × 400

U = 0.5 × 15.77

U = 7.885 Joules

Therefore, the energy stored in the magnetic field of the solenoid is approximately 0.789 Joules.

Correct Option Analysis:

The correct option is:

Option 2: 0.789 Joule

This matches the calculated result, confirming that the energy stored in the magnetic field of the solenoid is approximately 0.789 Joules.

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: 0.705 Joule

This value is incorrect because it results from either an error in the calculation of the inductance or the energy formula. The actual energy stored is higher than this value.

Option 3: 0.587 Joule

This value is also incorrect and is significantly lower than the actual calculated energy stored. This could result from an incorrect assumption about the cross-sectional area or current value.

Option 4: 0.658 Joule

This value is close to the correct value but still incorrect. It may result from a rounding or calculation error in one of the intermediate steps.

Conclusion:

The correct energy stored in the magnetic field of the solenoid, based on the given data and formula, is 0.789 Joules. This matches Option 2. The other options deviate due to calculation errors, incorrect assumptions, or approximations.