Resonant Frequency MCQ Quiz in मराठी - Objective Question with Answer for Resonant Frequency - मोफत PDF डाउनलोड करा

\(\begin{array}{l} {\rm{Y}} = {{\rm{Y}}_{\rm{c}}} + {{\rm{Y}}_{{\rm{LR}}}}\\ {\rm{Y}} = {\rm{j\omega C}} + \frac{1}{{\left( {{\rm{j\omega L}} + {\rm{R}}} \right)}} = {\rm{j\omega C}} + \frac{{\left( {{\rm{R}} - {\rm{j\omega L}}} \right)}}{{\left( {{{\rm{R}}^2} + {{\rm{\omega }}^2}{{\rm{L}}^2}} \right)}} \end{array}\)

At resonance, we should have the Imaginary part to zero i.e.

\({l} {\rm{ω C}} = \frac{{{\rm{ω L}}}}{{{{\rm{R}}^2} + {{\rm{ω }}^2}{{\rm{L}}^2}}}\\ {{\rm{ω }}^2} = \frac{{{\rm{L}} - {{\rm{C}\rm{R}}^2}}}{{{{\rm{L}}^2}{\rm{C}}}}\\ {\rm{ω }} = \frac{1}{{\sqrt {{\rm{LC}}} }}\sqrt {1 - {{\rm{R}}^2}\frac{{\rm{C}}}{{\rm{L}}}}\\ {\rm{f }} = \frac{1}{{2\pi\sqrt {{\rm{LC}}} }}\sqrt {1 - {{\rm{R}}^2}\frac{{\rm{C}}}{{\rm{L}}}} \)

Explanation:

Force in a Magnetic Circuit with Coil Inductance L Dependent on x

Definition: The force in a magnetic circuit can be derived from the energy stored in the magnetic field. When a magnetic circuit with coil inductance \( L \) is dependent on a variable \( x \), the force can be calculated based on the rate of change of the inductance with respect to \( x \).

Working Principle: In electromagnetic systems, the force can be derived from the energy stored in the magnetic field. The inductance \( L \) of the coil in the magnetic circuit is a function of the variable \( x \), such as the position of a movable element. The energy stored in the inductor is given by:

Energy Stored in Inductor:

\[ W = \frac{1}{2} L i^2 \]

where \( i \) is the current through the coil.

The force \( F \) can be derived from the gradient of the stored energy with respect to the position \( x \):

\[ F = -\frac{dW}{dx} \]

Substituting the expression for the energy stored in the inductor, we get:

\[ F = -\frac{d}{dx} \left( \frac{1}{2} L i^2 \right) \]

Since \( L \) is a function of \( x \), we use the chain rule for differentiation:

\[ F = -\frac{1}{2} i^2 \frac{dL}{dx} \]

Advantages:

• Provides a direct relationship between the force and the rate of change of inductance with respect to the position.
• Simplifies the calculation of force in electromagnetic systems where the inductance varies with position.

Disadvantages:

• Requires knowledge of the exact relationship between inductance and position, which may be complex in certain systems.
• Assumes a linear relationship between energy and inductance, which may not hold in all practical scenarios.

Applications: This principle is commonly used in the design and analysis of electromagnetic actuators, solenoids, and other devices where the position-dependent inductance is a critical factor in determining the force generated.

Correct Option Analysis:

The correct option is:

Option 1: \(\rm -\frac{1}{2}i^2\frac{dL}{dx}\)

This option correctly represents the force derived from the energy stored in the magnetic field, considering the inductance dependent on the variable \( x \). The negative sign indicates that the force is in the direction of decreasing energy.

Additional Information

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

Option 2: \(\rm L^2\frac{dL}{dx}\)

This option is incorrect because it suggests a different relationship between the force and the inductance. The force should be proportional to the current squared and the rate of change of inductance, not the square of the inductance itself.

Option 3: \(\rm \frac{1}{2}L^2\frac{dL}{dx}\)

This option is incorrect as it misrepresents the dependence of the force on the inductance. The correct relationship involves the current squared, not the inductance squared.

Option 4: \(\rm -i^2\frac{dL}{dx}\)

This option is incorrect because it does not include the factor of \(\frac{1}{2}\) that arises from the expression for the energy stored in the inductor. The correct force expression should be \(\rm -\frac{1}{2}i^2\frac{dL}{dx}\).

Conclusion:

Understanding the relationship between the inductance and the position in a magnetic circuit is crucial for correctly determining the force. The correct expression for the force, considering the energy stored in the magnetic field and the position-dependent inductance, is given by \(\rm -\frac{1}{2}i^2\frac{dL}{dx}\). This principle is vital in the design and analysis of various electromagnetic devices where precise control of force and position is required.

Concept:

Natural Frequency of an LC Circuit:

• The natural frequency (f) of an LC circuit is given by the formula:
  f = 1 / (2π√(LC)),
  where, L is the inductance, C is the capacitance, and f is the frequency of oscillation.
• When a dielectric material with dielectric constant (K) is placed in the capacitor, the capacitance increases by a factor of K.Thus, the new capacitance becomes C' = K × C.
• When the dielectric is inserted, the frequency becomes: f' = 1 / (2π√(L × K × C)).
• The ratio of the original frequency to the new frequency is:
• f / f' = √K
   

Calculation:

Given, Original frequency, f = 120 kHz
New frequency with dielectric, f' = 120 kHz - 20 kHz = 100 kHz

Using the relation:

f / f' = √K

Substitute the values:

120 / 100 = √K

1.2 = √K

Now, square both sides:

1.44 = K

∴ The dielectric constant of the material is 1.44. Option 2) is correct.

Concept:

Mutually aiding inductors in series

The effective inductance is:

L_eq = L1 + L2 + 2M

Mutually opposing inductors in series

Leq = L1 + L2 - 2M

Mutually aiding inductors in parallel

\(L_{eq}=\frac{L_1 L_2 - M^2}{L_1+L_2-2M}\)

Mutually opposing inductors in parallel

\(L_{eq}=\frac{L_1 L_2 - M^2}{L_1+L_2+2M}\)

Calculation:

The given circuit has the inductors in aiding nature.

The coupling between the conductors is increased from zero.

The equivalent inductance also increases.

As equivalent inductance increases the ω₀ decreases.

\({\omega _0} = \frac{1}{{\sqrt {{L_{eq}}C} }}\)

And Q also decreases

\(Q = \frac{R}{{{\omega _0}{L_{eq}}}}\)

The resonant frequency for the above circuit is,

\({\omega _0} - \frac{1}{{\sqrt {LC} }}\sqrt {\frac{{R_L^2 - \frac{L}{C}}}{{R_c^2 - \frac{L}{C}}}} \)

For the given circuit diagram,

Resonant frequency (ω₀) = 2 rad/sec

R_L = 6 Ω
R_C = 4 Ω

L = 4 H

By substituting all the values, we get

\( \Rightarrow 2 = \frac{1}{{\sqrt {4C} }}\sqrt {\frac{{36 - \frac{4}{C}}}{{16 - \frac{4}{C}}}} \)

⇒ 64C²-25C + 1 = 0

⇒ C = 345.4 mF (or) 45.2 mF

Minimum Value of C = 45.2 mF

\({Z_{ab}} = j\omega L + \left( {R + ||\frac{1}{{j\omega L}}} \right)\)

\(= j\omega L + \left( {\frac{{R.\;\frac{1}{{j\omega C}}}}{{R + \frac{1}{{j\omega C}}}}} \right)\)

\(= j\omega L + \frac{R}{{1 + j\omega RC}}\;\left( {\frac{{1 - j\omega RC}}{{1 - j\omega RC}}} \right)\)

\({Z_{ab}} = j\omega L + \frac{{R\left( {1 - j\omega RC} \right)}}{{1 + {\omega ^2}{R^2}{C^2}}}\)

\(= \frac{R}{{1 + {\omega ^2}{R^2}{c^2}}} + j\omega L - \frac{{j\omega {R^2}C}}{{1 + {\omega ^2}{R^2}{C^2}}}\)

Given that Z_ab is purely resistive

Imaginary part of Z_ab = 0

\(\Rightarrow \omega L - \frac{{\omega {R^2}C}}{{1 + {\omega ^2}{R^2}{C^2}}} = 0\)

\(\Rightarrow L = \frac{{{R^2}C}}{{1 + {\omega ^2}{R^2}{c^2}}}\)

\(\Rightarrow {\omega ^2}{R^2}{c^2} + 1 = \frac{{{R^2}C}}{L}\)

\(\Rightarrow {{\rm{\omega }}^2}{\left( {100} \right)^2}{\left( {25 \times {{10}^{ - 9}}} \right)^2} = \frac{{{{\left( {100} \right)}^2} \times 25 \times {{10}^{ - 9}}}}{{160 \times {{10}^{ - 6}}}} - 1\)

⇒ ω = 300 k rad/s.

\({Z_{in}} = {R_1} + \frac{1}{{j\omega C}} + \frac{{\left( {{R_2}} \right)\left( {j\omega L} \right)}}{{{R_2} + j\omega L}}\)

\(= {R_1} - \frac{j}{{\omega C}} + \frac{{j{R_2}\omega L}}{{\left( {{R_2} + j\omega L} \right)}}\frac{{\left( {{R_2} - j\omega L} \right)}}{{\left( {{R_2} - j\omega L} \right)}}\)

\(= {R_1} - \frac{j}{{\omega C}} + \frac{{j\;R_2^2\;\omega L}}{{R_2^2 + {\omega ^2}{L^2}}} + \frac{{{R_2}{\omega ^2}{L^2}}}{{R_2^2 + {\omega ^2}{L^2}}}\)

Imaginary should be equal to zero.

\(\Rightarrow \frac{{R_2^2\omega L}}{{R_2^2 + {\omega ^2}{L^2}}} = \frac{1}{{\omega C}}\)

\(\Rightarrow R_2^2{\omega ^2}LC = R_2^2 + {\omega ^2}{L^2}\)

ω = 50 × 10³, R₁ = 2 × 10³, R₂ = 10 × 10³, C = 5 × 10^-9

⇒ (10 × 10³)³ × (50 × 10³)² × (L) × (5 × 10^-9) = (10 × 10³)² + (50 + 10³)² L²

⇒ 12.5 × 10⁸ × L = 10⁸ + (25 × 10⁸) L²

⇒ 25 L² - 12.5 L + 1 = 0

\({Z_{in}} = 500 + \frac{1}{{j\omega C}} + \frac{{R + j\omega L}}{{\left( {R + j\omega L} \right)}}\)

\(= 500 - \frac{j}{{\omega C}} + \frac{{jR\omega L\left( {R - j\omega L} \right)}}{{{R^2} + {\omega ^2}{L^2}}}\)

\(= 500 - \frac{j}{{\omega C}} + \frac{{j\;{R^2}\omega L}}{{{R^2} + {\omega ^2}{L^2}}} + \frac{{{\omega ^2}R{L^2}}}{{{R^2} + {\omega ^2}{L^2}}}\)

Z_in should be purely real

\(\Rightarrow \frac{1}{{\omega C}} = \frac{{{R^2}\omega L}}{{{R^2} + {\omega ^2}{L^2}}}\)

⇒ ω²R²LC = R² + ω²L²

⇒ ω²(1 × 10³)² (500 × 10^-3) (1 × 10^-6) = (1 × 10³)² + ω² (500 × 10-³)²

 ⇒ ω² (0.5) = 10⁶ + ω² (0.25)

For maximum power transfer

Theorem

R_s = R_L.

Hence:

Hence Z_in should be 100 Ω.

Since Z_in is purely resistance.

The LC is in resonance

The resonant frequency here is the source frequency 1 MHZ

The resonant frequency of the given circuit is given by

\({\omega _r} = \sqrt {\frac{1}{{LC}} - \frac{{{R^2}}}{{{L^2}}}} \) 

\( \Rightarrow {\left( {2\pi \times {{10}^6}} \right)^2} = \frac{1}{{LC}} - \frac{{{{20}^2}}}{{{L^2}}}\)        _____(1)

The impedance at resonance

\(Z\left( {j{\omega _r}} \right) = \frac{L}{{RC}} = 100\)      ____(2)

\( = \frac{L}{{20\left( C \right)}} = 100\) 

Solving equation (1) & (2)

L = 6.37 μH

C = 3.18 nF

The circuit consist 2 Parallel LC circuits connected in series

V₁ = sin (2π × 10000 t)

V₂ = sin (2π × 30,000 t)

f₁ = 10,000 Hz

f₂ = 30,000 Hz = 3f₁

For LC – 2

\({\omega _0} = \frac{1}{{\sqrt {LC} }} \Rightarrow {f_0} = \frac{1}{{2\pi \sqrt {LC} }}\)

\({f_0} = \frac{1}{{2\pi \sqrt {100 \times {{10}^{ - 6}} \times 2.53 \times {{10}^{ - 6}}} }}\)

f₀ = 10,000 Hz

f₁ = 10,000 Hz

Hence for V₁ source LC₂ circuit is in Resonance.

A parallel LC circuit behaves as open circuit at resonance.

The current drawn will be minimum when circuit 1 is also in Resonance.

for LC-1 to be open circuit i.e. in Resonance

\(\left( {3{f_1}} \right)\left( {30,000} \right) = \frac{1}{{2\pi \sqrt {LC} }}\)

C = 0.28 μF

When Both LC-1 and LC-2 is in Resonance, current drawn is minimum = 0 and independent of value of Resistance R.