Resonant Frequency MCQ Quiz - Objective Question with Answer for Resonant Frequency - Download Free PDF

Parallel RLC circuit

The bandwidth in a parallel RLC circuit is given by:

\(BW={\omega_o\over QF}={1\over RC}\)

From the above observation, Bandwidth is inversely proportional to the resistance.

Hence, increasing the resistance R in a parallel RLC circuit will decrease the bandwidth.

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:

The characteristic equation of a series RLC circuit is given by,

\({\rm{C}}.{\rm{E}} = {{\rm{s}}^2} + \frac{{\rm{R}}}{{\rm{L}}}{\rm{s}} + \frac{1}{{{\rm{LC}}}}\)

We have \({{\rm{\omega }}_{\rm{n}}} = \frac{1}{{\sqrt {{\rm{LC}}} }}\)

And \(2{\rm{\zeta }}{{\rm{\omega }}_{\rm{n}}} = \frac{{\rm{R}}}{{\rm{L}}}\)

\(\begin{array}{l} \Rightarrow {{\rm{\zeta }}_{\rm{n}}} = \frac{{\rm{R}}}{{\rm{L}}} \times \frac{1}{2} \times \sqrt {{\rm{LC}}} \\ \Rightarrow {{\rm{\zeta }}_{\rm{n}}} = \frac{{\rm{R}}}{2}\sqrt {\frac{{\rm{C}}}{{\rm{L}}}} \end{array}\)

Hence the correct answer is option 2.

Concept:

The graph between impedance Z and the frequency of the parallel RLC circuit:

Here,

f1 is the lower cutoff frequency

f2 is the upper cutoff frequency

fr is the resonant frequency

BW is the bandwidth

Formula:

BW = f2 – f1

\({f_1} = {f_r} - \left( {\frac{{BW}}{2}} \right)\)

\({f_2} = {f_r} + \left( {\frac{{BW}}{2}} \right)\)

Calculation:

Given

Resonant frequency fr = 20 kHz

Bandwidth = 2 kHz

The upper cutoff frequency is given as:

\({f_2} = {f_r} + \left( {\frac{{BW}}{2}} \right)\)

\({f_2} = {20kHz} + \left( {\frac{{2kHz}}{2}} \right)\)

f₂ = 21 kHz

Concept:

The graph between impedance Z and the frequency of the parallel RLC circuit:

Here,

f1 is the lower cutoff frequency

f2 is the upper cutoff frequency

fr is the resonant frequency

BW is the bandwidth

Formula:

BW = f2 – f1

\({f_1} = {f_r} - \left( {\frac{{BW}}{2}} \right)\)

\({f_2} = {f_r} + \left( {\frac{{BW}}{2}} \right)\)

Calculation:

Given

Resonant frequency fr = 20 kHz

Bandwidth = 2 kHz

The upper cutoff frequency is given as:

\({f_2} = {f_r} + \left( {\frac{{BW}}{2}} \right)\)

\({f_2} = {20kHz} + \left( {\frac{{2kHz}}{2}} \right)\)

f₂ = 21 kHz

CONCEPT:

• LC Circuit: The circuit containing both an inductor (L) and a capacitor (C) can oscillate without a source of emf by shifting the energy stored in the circuit between the electric and magnetic fields is called LC circuit.
• The tuned circuit has a very high impedance at its resonant frequency. 

EXPLANATION:

• The frequency of oscillations generated by the LC circuit entirely depends on the values of the capacitor and inductor and their resonance condition.

It can be expressed as:

\(f = \frac{1}{{2\pi \sqrt {LC} }}\)

• In an LC oscillator, the frequency of the oscillator is Inversely proportional to the square root of L or C. So option 1 is correct.

Concept:

\({Y_{eq}} = \frac{1}{{R + j\omega L}} + j\omega C\)

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

\(= \frac{R}{{{R^2} + {\omega ^2}{L^2}}} + \left[ {\omega C - \frac{{\omega L}}{{{R^2} + {\omega ^2}{L^2}}}} \right]\)

At resonant frequency, imaginary part of equivalent admittance is zero.

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

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

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

\(\Rightarrow {\omega ^2} = \frac{1}{{LC}} - {\left( {\frac{R}{L}} \right)^2}\;\)

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

Application:

From the above expression, the resonant frequency decreases with an increase in resistance.

\({Y_{pq}} = \frac{1}{{10 + j100}} + j\omega C\)

\(= \frac{{10 - j100}}{{10,100}} + j\omega C\)

\(= \frac{{10}}{{10,100}} + j\left( {\omega C - \frac{1}{{101}}} \right)\)

At resonance, the imaginary part of admittance is equal to zero.

The admittance becomes pure resistive.

\(\Rightarrow Y = \frac{1}{{101}}\)

Impedance (Z) will be:

\(Z= \frac{1}{Y} = 101\;{\rm{\Omega }}\)

The power factor is unity when the imaginary part of impedance is unity

\(\begin{array}{l} Z = R + jω L||\left( {R + \frac{1}{{jω c}}} \right)\\ = \frac{{\left( {R + jω L} \right)\left( {R + \frac{1}{{jω c}}} \right)}}{{2R + jω L + \frac{1}{{jω c}}}}\\ = \frac{{\left( {R + jω L} \right)\left( {jRω C + 1} \right)}}{{j2Rω C - {ω ^2}LC + 1}}\\ \Rightarrow Z = \frac{{R + jω {R^2}C - {ω ^2}RLC + jω L}}{{1 + j2ω RC - {ω ^2}LC}} \end{array}\)

Equating imaginary part of z = 0, we get

L- ω² L² C - R² C = 0

for this resonating frequency is

\(ω = \sqrt {\frac{1}{{LC}} - {{\left( {\frac{R}{L}} \right)}^2}} \)

 Following are the conditions for UPF (pure resistive circuit)

ω = 0

Now

\(R = \sqrt {\frac{L}{C}}\)

Parallel RLC circuit

The bandwidth in a parallel RLC circuit is given by:

\(BW={\omega_o\over QF}={1\over RC}\)

From the above observation, Bandwidth is inversely proportional to the resistance.

Hence, increasing the resistance R in a parallel RLC circuit will decrease the bandwidth.

The capacitance is proportional to the permittivity,

i.e. C ∝ ϵ

\(C=\frac{A\epsilon }{d}\)

If the capacitance changes with temperature it can be defined as:

C = C₀(1 + α_ϵT)      ----(1)

Where α_ϵ is the temperature coefficient of the dielectric

Differentiating equation (1) with respect to T (temperature), we get;

\(\Rightarrow \frac{dC}{dT}={{C}_{0}}{{\alpha }_{\epsilon }}\)      ----(2)

The resonant frequency of an LC tank circuit is calculated.

Since the capacitance is a function of temperature, differentiating the above with respect to T we get:

\(\Rightarrow \frac{d{{f}_{r}}}{dT}=\frac{d}{dT}\left( \frac{1}{2\pi \sqrt{LC}} \right)=\frac{1}{2\pi \sqrt{L}}.\frac{d}{dT}\left( \frac{1}{{{\left( C \right)}^{\frac{1}{2}}}} \right)\)

\(\Rightarrow \frac{d{{f}_{r}}}{dT}=\frac{1}{2\pi \sqrt{L}}.\left( \frac{-1}{2} \right).{{\left( C \right)}^{\frac{-3}{2}}}.\frac{dC}{dT}=\frac{1}{2\pi \sqrt{L}}\left( \frac{-1}{2C\sqrt{C}} \right)\frac{dC}{dT}\)

Putting \(\frac{dC}{dT}\) from equation (II) to the above equation we get;

\(\Rightarrow \frac{d{{f}_{r}}}{dT}=\frac{1}{2\pi \sqrt{L}}.\left( \frac{-1}{2C\sqrt{C}} \right).C{{\alpha }_{\epsilon }}\)

\(\Rightarrow \frac{d{{f}_{r}}}{dT}=\frac{-1}{2\pi \sqrt{LC}}.\frac{1}{2}{{\alpha }_{\epsilon }}\)

\(\Rightarrow \frac{d{{f}_{r}}}{dT}=-{{f}_{r}}.\frac{1}{2}{{\alpha }_{\epsilon }}\)

\(\Rightarrow {{f}_{r}}={{f}_{r}}+\frac{d{{f}_{r}}}{dt}.\text{ }\!\!\Delta\!\!\text{ }t\)

\(\Rightarrow {{f}_{r}}={{f}_{r}}-\frac{{{f}_{r}}}{2}.{{\alpha }_{\epsilon }}\text{ }\!\!\Delta\!\!\text{ }t={{f}_{r}}\left[ 1-\frac{{{\alpha }_{\epsilon }}}{2}.\text{ }\!\!\Delta\!\!\text{ }t \right]\)

Comparing the above with a standard equation, we can write \({{\alpha }_{fr}}=\frac{-{{\alpha }_{\epsilon }}}{2}\)

Given, α_ϵ = 6 ppm/°C

So, \({{\alpha }_{fr}}=\frac{-{{\alpha }_{\epsilon }}}{2}=-3~ppm/{}^\circ C\) 

For the given parallel RLC circuit, current will be in phase with the voltage when circuit operate at resonant frequency.

Resonant frequency in parallel RLC circuit,

\(\omega = \frac{1}{{\sqrt {LC} }} = \frac{1}{{\sqrt {5 \times {{10}^{ - 3}} \times 500 \times {{10}^{ - 9}}} }}\)

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

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

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

The frequency of oscillation is equal to the resonance frequency of the tank circuit is:

\({f_1} = \frac{1}{{2\pi \sqrt {L{C_1}} }}\)   ---(1)

The capacitance is increased by 10%, i.e.

⇒ C₂ = C₁ + 0.1 C₁

C₂ = 1.1 C₁

New resonant frequency will be:

\({f_2} = \frac{1}{{2\pi \sqrt {L{C_2}} }}\)

\(f_2= \frac{1}{{2\pi \sqrt {1.1L{C_1}} }}\)  ---(2)

Dividing Equation (2) with (1), we get:

\(\frac{f_2}{f_1}=\frac{1}{\sqrt{1.1}} \)

With f₁ = 100 kHz, f₂ is:

f₂ = 95.346 KHz