Parallel Resonance MCQ Quiz - Objective Question with Answer for Parallel Resonance - Download Free PDF

Explanation:

Resonant Frequency in a Parallel RLC Circuit

Definition: Resonant frequency (f₀) in a parallel RLC circuit is the frequency at which the inductive reactance and capacitive reactance become equal in magnitude, causing the circuit to exhibit purely resistive behavior. At this frequency, the impedance of the circuit is maximized, and the circuit resonates.

Formula: The resonant frequency (f₀) for a parallel RLC circuit is given by the following equation:

f₀ = 1 / (2 × π × √(L × C))

Where:

• L: Inductance of the inductor (in henries, H)
• C: Capacitance of the capacitor (in farads, F)

Working Principle: Resonance occurs in a parallel RLC circuit when the energy alternates between the inductor and capacitor, creating oscillations at the resonant frequency. At this point, the inductive reactance (X_L) and capacitive reactance (X_C) cancel each other out, leading to the circuit behaving as though only the resistance is present.

Impact of Decreasing Capacitance (C):

From the formula for resonant frequency (f₀):

f₀ = 1 / (2 × π × √(L × C))

It is evident that the resonant frequency is inversely proportional to the square root of the product of inductance (L) and capacitance (C). Specifically:

• If C decreases, the denominator (√(L × C)) becomes smaller.
• This causes the value of f₀ to increase because the resonant frequency is inversely proportional to the square root of C.

Therefore, when the capacitance (C) in a parallel RLC circuit decreases, the resonant frequency (f₀) increases.

Correct Option Analysis:

The correct option is:

Option 2: Increases

This option correctly identifies the effect of decreasing capacitance on the resonant frequency in a parallel RLC circuit. As explained above, a reduction in capacitance leads to an increase in the resonant frequency due to their inverse relationship in the resonant frequency formula.

Additional Information

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

Option 1: Decreases

This option is incorrect because decreasing capacitance does not cause the resonant frequency to decrease. As shown in the formula, resonant frequency is inversely proportional to the square root of capacitance (C); hence, a reduction in C increases the value of f₀.

Option 3: Remains constant

This option is incorrect because the resonant frequency depends on the values of both inductance (L) and capacitance (C). Changes in either L or C will affect the resonant frequency. If capacitance decreases, f₀ will increase, as explained earlier.

Option 4: Becomes zero

This option is incorrect because the resonant frequency cannot become zero simply by decreasing capacitance. Even with very small values of C, the resonant frequency remains finite and increases. Resonant frequency becoming zero would imply the absence of oscillation, which is not the case here.

Conclusion:

Understanding the relationship between capacitance and resonant frequency is crucial for analyzing parallel RLC circuits. The resonant frequency is determined by the values of inductance and capacitance. When capacitance decreases, the resonant frequency increases due to their inverse proportionality. This principle is fundamental in electronics and telecommunications, where resonant circuits are used for signal filtering and frequency selection.

Concept

The resonant frequency of a parallel RLC circuit is given by:

$f_o=\frac{1}{2\pi \sqrt{LC}}$

where, f_o = Resonant frequency

L = Inductance

C = Capacitance

Calculation

Given, L = 1 H

C = 1 μF = 1 × 10^-6 F

$f_o=\frac{1}{2\pi \sqrt{1\times 1\times {10}^{-6}}}$

$f_o=\frac{1}{2\pi \times {10}^{-3}}\mathrm{H}\mathrm{z}$

Parallel RLC circuit

The bandwidth in a parallel RLC circuit is given by:

$BW=\frac{{\omega }_o}{QF}=\frac{1}{RC}$

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: $-\frac{1}{2}{\mathrm{i}}^2\frac{\mathrm{d}\mathrm{L}}{\mathrm{d}\mathrm{x}}$

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: ${\mathrm{L}}^2\frac{\mathrm{d}\mathrm{L}}{\mathrm{d}\mathrm{x}}$

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: $\frac{1}{2}{\mathrm{L}}^2\frac{\mathrm{d}\mathrm{L}}{\mathrm{d}\mathrm{x}}$

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: $-{\mathrm{i}}^2\frac{\mathrm{d}\mathrm{L}}{\mathrm{d}\mathrm{x}}$

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 $-\frac{1}{2}{\mathrm{i}}^2\frac{\mathrm{d}\mathrm{L}}{\mathrm{d}\mathrm{x}}$.

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 $-\frac{1}{2}{\mathrm{i}}^2\frac{\mathrm{d}\mathrm{L}}{\mathrm{d}\mathrm{x}}$. This principle is vital in the design and analysis of various electromagnetic devices where precise control of force and position is required.

The bandwidth of an RLC parallel circuit is defined as the range of frequencies over which the circuit can operate effectively. It is an important parameter in determining the performance of the circuit in filtering and resonance applications.

The bandwidth (BW) of an RLC parallel circuit is given by:

BW = R/L
where,

• R is the resistance
• L is the inductance

The Q-factor (quality factor) is given by:

$Q=\frac{1}{R}\sqrt{\frac{L}{C}}$

where,

• C is the capacitance

The relationship between bandwidth and Q-factor is given by:

BW  $=\frac{{\omega }_0}{Q}$

where,

• ω0" id="MathJax-Element-128-Frame" role="presentation" style="position: relative;" tabindex="0">ω0${\omega }_0$ $\omega_0$ is the resonant angular frequency

Solution

The given question asks for the relationship of bandwidth with other parameters in an RLC parallel circuit. Based on the formula:

BW = R/L

We can infer the following:

• Bandwidth is inversely proportional to inductance (L).
• Bandwidth is directly proportional to resistance (R).
• Bandwidth is inversely proportional to Q-factor.

The correct answer is Option 4, which states that the bandwidth of an RLC parallel circuit is inversely proportional to the Q-factor.

The quality factor is defined as the ratio of the maximum energy stored to maximum energy dissipated in a cycle

$Q=2\pi \frac{Maximumenergystored}{Totalenergylostperperiod}$

In a series RLC, Quality factor $Q=\frac{\omega L}{R}=\frac{1}{\omega RC}=\frac{X_L}{R}=\frac{X_C}{R}$

In a parallel RLC, $Q=\frac{R}{\omega \mathrm{L}}=\omega RC=\frac{R}{X_L}=\frac{R}{X_C}$

It is defined as, resistance to the reactance of reactive element.

The quality factor Q is also defined as the ratio of the resonant frequency to the bandwidth.

$Q=\frac{f_r}{BW}$

For parallel RLC Circuit ${\mathrm{Q}}_1=\mathrm{R}\sqrt{\frac{\mathrm{C}}{\mathrm{L}}}$

For Series RLC Circuit ${\mathrm{Q}}_2=\frac{1}{\mathrm{R}}\sqrt{\frac{\mathrm{L}}{\mathrm{C}}}$

Parallel Resonance:

The input admittance is given by:

$Y_{in}=\frac{1}{R}+j(\omega C-\frac{1}{\omega L})$

$Y_{in}=G+j(B_C-B_L)$

where, Y_in = Admittance

G = Conductance

B_C = Capacitive susceptance

B_L = Inductive susceptance

At resonance, BC = BL 

Hence, Yin = G and is minimum.

Additional Information

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

The correct answer is option 3):(Q-factor provides the voltage magnification.)

Concept:

• In series resonant circuit Q-factor gives the voltage magnification of the circuit, whereas in a parallel circuit it gives the current magnification
• Q-factor of parallel resonance is the inverse  as that of series resonance.
• The quality factor Q is also defined as the ratio of the resonant frequency to the bandwidth

$Q=\frac{f_r}{BW}$

• Q-factor of parallel resonance is reciprocal to that of series resonance.

The correct answer is 'option 1'

Concept:

• Parallel resonance occurs when the supply frequency creates zero phase difference between the supply voltage and current producing a resistive circuit

• If the impedance of the parallel circuit is at its maximum at resonance then consequently, the admittance of the circuit must be at its minimum and one of the characteristics of a parallel resonance circuit is that admittance is very low limiting the circuit's current.
• Unlike the series resonance circuit, the resistor in a parallel resonance circuit has a damping effect on the bandwidth of the circuit making the circuit less selective.
• Also, since the circuit current is constant for any value of impedance, Z, the voltage across a parallel resonance circuit will have the same shape as the total impedance and for a parallel circuit, the voltage waveform is generally taken from across the capacitor.
• We now know that at the resonant frequency, ƒ_r the admittance of the circuit is at its minimum and is equal to the conductance, G is given by 1/R because in a parallel resonance circuit the imaginary part of admittance, i.e. the susceptance, B is zero because B_L = B_C as shown.
• The shape of impedance versus frequency is shown below

Concept:

The quality factor is given by:

$Q=\frac{1}{R}\sqrt{\frac{L}{C}}$

where Q = Quality factor

R = Resistance

L = Inductor

C = Capacitance

Calculation:

Given R= 5Ω, L= 1H, C= 2mF

$Q=\frac{1}{5}\sqrt{\frac{1}{2\times {10}^{-3}}}$

$Q=\sqrt{\frac{1000}{2\times 25}}$

$Q=\sqrt{20}$

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.

The admittance for the given circuit is

$Y_{in}=j\omega C+\frac{1}{R+j\omega L}$

$=j\omega C+\frac{1}{R+j\omega L}\times \frac{R-j\omega L}{R-j\omega L}$

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

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

At resonance, the imaginary part is zero.

$\Rightarrow \omega C-\frac{\omega L}{R^2+{\omega }^2{L}^2}=0$

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

The admittance at resonance is,

$Y_{in}=\frac{R}{R^2+{\omega }^2{L}^2}=\frac{R}{\frac{L}{C}}=\frac{RC}{L}$

The impedance at resonance is

A capacitor and coil or LC circuit in parallel resonant circuit is called a tuned or tank circuit.

They are used in tuner circuit to generate signals of a particular frequency or picking up signals of a particular frequency.

At Resonance:

X_c = X_L

$\frac{1}{2\pi fC}=2\pi fL$

The resonant frequency of a parallel tuned circuit is given by:

$f=\frac{1}{2\pi \sqrt{LC}}$

If an LC tank circuit consists of an ideal capacitor C connected in parallel with a coil of inductance L having an internal resistance R.

The resonant frequency of the tank circuit is

$f=\frac{1}{2\pi \sqrt{\mathrm{L}\mathrm{C}}}\sqrt{1-{\mathrm{R}}^2\frac{\mathrm{C}}{\mathrm{L}}}$

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.