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

Explanation:

Resonance in a Series Circuit

Definition: Resonance in an electrical series circuit occurs when the inductive reactance (XL) is equal to the capacitive reactance (XC), resulting in their effects canceling each other out. This leads to a purely resistive circuit where the impedance is minimized and is equal to the resistance of the circuit. At resonance, the circuit oscillates at its natural frequency, which is determined by the inductance (L) and capacitance (C) in the circuit.

Correct Option Analysis:

The correct option is:

Option 4: The phase angle of the circuit is 90°.

This option is NOT correct because, at resonance in a series circuit, the phase angle between the voltage and current is 0°, not 90°. At resonance, the inductive and capacitive reactances cancel each other out, and the circuit behaves as a purely resistive circuit. In a purely resistive circuit, the voltage and current are in phase, meaning the phase angle is 0°. A phase angle of 90° would indicate a completely reactive circuit (either purely inductive or purely capacitive), which is not the case at resonance in a series circuit.

Important Information: At resonance, the following conditions are observed:

• Inductive reactance (XL) = Capacitive reactance (XC).
• The impedance of the circuit is equal to the resistance (Z = R).
• The circuit behaves as a purely resistive circuit, with no reactive component.
• The current is at its maximum value, as the impedance is minimized.
• The voltage across the inductor and capacitor is equal in magnitude but opposite in phase, effectively canceling each other out.

Additional Information

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

Option 1: The voltage across the capacitor and inductor is equal and opposite.

This statement is correct. At resonance, the voltage across the inductor and capacitor is equal in magnitude but opposite in phase. This means that the inductive and capacitive voltages cancel each other out, leaving only the resistive voltage in the circuit.

Option 2: The circuit behaves as a purely resistive circuit.

This statement is also correct. At resonance, the inductive and capacitive reactances cancel out, leaving only the resistive component. The impedance of the circuit is purely resistive, and the current and voltage are in phase.

Option 3: The net reactance of the circuit is zero.

This statement is correct as well. At resonance, the inductive reactance (XL) is equal to the capacitive reactance (XC), resulting in a net reactance of zero. This is why the circuit behaves as a purely resistive circuit at resonance.

Option 5: [This option is not provided in the question, so no analysis is required for it.]

Conclusion:

Understanding resonance in a series circuit is crucial for analyzing the behavior of electrical circuits at specific frequencies. At resonance, the circuit achieves unique characteristics such as maximum current flow, zero net reactance, and purely resistive behavior. While most statements provided in the options align with the principles of resonance, the incorrect statement is Option 4, as the phase angle of a resonant series circuit is 0° and not 90°.

Explanation:

Power Factor of Series R-L Circuit

Definition: In an R-L series circuit, the power factor is a measure of how effectively electrical power is converted into useful work. It represents the cosine of the angle ($\theta$) between the voltage and current vectors in the circuit. The power factor is crucial in AC circuits as it determines the efficiency of power usage.

The power factor of a series R-L circuit is given by:

$\cos \theta =\frac{\mathrm{R}}{\mathrm{Z}}$

Where:

• $R$: Resistance of the circuit
• $Z$: Impedance of the circuit

Impedance ($Z$): In an R-L series circuit, the impedance is the vector sum of the resistance ($R$) and the reactive inductance ($X_L$). It is given by:

$\mathrm{Z}=\sqrt{{\mathrm{R}}^2+{\mathrm{X}}_{\mathrm{L}}^2}$

Here:

• $X_L$: Reactive inductance, which is calculated as $X_L=\omega L$, where $\omega$ is the angular frequency and $L$ is the inductance.

Detailed Explanation:

In an R-L series circuit, the total impedance ($Z$) is a combination of the resistive and inductive components. The resistance ($R$) represents the real part of the impedance and is associated with energy dissipation as heat. The inductive reactance ($X_L$) represents the imaginary part of the impedance and is associated with energy storage in the magnetic field of the inductor.

The phase angle ($\theta$) between the voltage and current in the circuit is determined by the relationship between the resistance ($R$) and the reactive inductance ($X_L$):

$\tan \theta =\frac{{\mathrm{X}}_{\mathrm{L}}}{\mathrm{R}}$

From this, the cosine of the phase angle ($\cos \theta$) is calculated as:

$\cos \theta =\frac{\mathrm{R}}{\mathrm{Z}}$

This formula indicates that the power factor depends on the ratio of the resistance to the total impedance of the circuit. A higher resistance relative to the impedance results in a better power factor, whereas a higher reactive inductance results in a lower power factor.

Correct Option Analysis:

The correct option is:

Option 3: $\cos \theta =\frac{\mathrm{R}}{\mathrm{Z}}$

This option correctly defines the power factor of a series R-L circuit. The power factor is the ratio of the resistance ($R$) to the total impedance ($Z$) of the circuit. It is a measure of how efficiently the electrical power is being converted into useful work.

Additional Information

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

Option 1: $\cos \theta =\frac{{\mathrm{X}}_{\mathrm{L}}}{\mathrm{Z}}$

This option is incorrect. The ratio of the reactive inductance ($X_L$) to the impedance ($Z$) does not represent the power factor. Instead, it represents the sine of the phase angle ($\sin \theta$), which is associated with the reactive component of the circuit.

Option 2: $\cos \theta =\frac{\mathrm{Z}}{\mathrm{R}}$

This option is incorrect. The ratio of impedance ($Z$) to resistance ($R$) does not define the power factor. Instead, it is inversely related to the power factor and does not have a direct physical interpretation in this context.

Option 4: $\cos \theta =\frac{\mathrm{R}}{{\mathrm{X}}_{\mathrm{L}}}$

This option is incorrect. The ratio of resistance ($R$) to reactive inductance ($X_L$) is related to the tangent of the phase angle ($\tan \theta$), not the cosine of the phase angle ($\cos \theta$). It is not a measure of the power factor.

Conclusion:

In summary, the power factor of a series R-L circuit is given by $\cos \theta =\frac{\mathrm{R}}{\mathrm{Z}}$, which accurately represents the ratio of the resistance to the total impedance. This formula is essential for understanding the efficiency of power usage in AC circuits. Correctly identifying the power factor formula is crucial for analyzing and optimizing the performance of R-L circuits, especially in applications where power efficiency is critical.

Explanation:

Q Factor of a Coil

Definition: The Q factor, or quality factor, of a coil is a dimensionless parameter that describes the efficiency or quality of the coil in terms of its ability to store energy versus dissipating it. It is commonly used in the analysis of resonant circuits and is a measure of the sharpness of the resonance in the circuit.

Formula: The Q factor for a coil is defined as:

Q = X_L / R

Where:

• Q: Quality factor of the coil.
• X_L: Inductive reactance of the coil.
• R: Resistance of the coil.

Quality factor

In a resonant series circuit, the quality factor (Q) is a measure of how underdamped the system is and how sharp the resonance is.

It is given by:

$QF=\frac{1}{R}\sqrt{\frac{L}{C}}=\frac{{\omega }_{o}L}{R}$

From the above expression, it is observed that the quality factor is inversely proportional to capacitance.

If the capacitance in a series RLC circuit is increased, the Q-factor will decrease.

Concept

In a resonant series circuit, the quality factor (Q) is a measure of how underdamped the system is and how sharp the resonance is.

It is given by:

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

where, ω_o = Resonance frequency

BW = Bandwidth

Calculation

Given, QF = 100

ω_o = 1 MHz

$100=\frac{{10}^6}{BW}$

BW = 1kHz

Concept:

In series LCR circuits, Resonance is a condition in which the inductive reactance and capacitive reactance are equal and lie opposite in phase, so cancel out each other. Only, resistance is left as impedance.

The frequency at which the series LCR circuit goes into resonance is:

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

At resonance, the impedance of the series LCR circuit is minimum and hence the current is maximum.

A series LCR circuit is as shown:

The phasor diagram is as shown below:

At resonance, X_L = X_C, i.e. the voltage across the inductor and capacitor are equal and opposite. The resultant phasor diagram at resonant frequency will, therefore, be:

The impedance - frequency curve is shown below:

Concept:

RLC series circuit:

For a series RLC circuit, the net impedance is given by:

Z = R + j (XL - XC)

XL = Inductive Reactance given by:

XL = ωL

XC = Capacitive Reactance given by:

XL = 1/ωC

ω = 2 π f

ω = angular frequency

f = linear frequency

The magnitude of the impedance is given by:

$|Z|=\sqrt{R^2+(X_L-X_C{)}^2}$

$V=\sqrt{V_1^2+{\left(V_2-V_3\right)}^2}$

Where, V₁ = voltage across the resistor

V₂ = voltage across the inductor

V3 = voltage across the capacitor

V = resultant voltage

The current flowing across the series RLC circuit will be:

$I=\frac{V}{|Z|}$

At resonance, XL = XC, resulting in the net impedance to be minimum. This eventually results in the flow of the maximum current of:

∴$I=\frac{V}{R}$

Calculation:

V1 = 100 V, V2 = 50 V, V3 = 50 V 

V2 = V3 (As data provided in the question)

So, the circuit is in series resonance.

V = V1 = 100 V

The source voltage is 100 V

Concept:

In a series resonant circuit,

Bandwidth (BW) $=\frac{f_r}{Q}=f_H-f_L=\frac{R}{L}$

Where fr is the resonant frequency

Q is the quality factor

fH is the higher half-power frequency

fL is the lower half-power frequency

Resonant frequency (fr) is defined as:

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

Calculation:

Given that,

Higher half-power frequency (fH) = 90 kHz

Lower half-power frequency (fL) = 10 kHz

Bandwidth (BW) = fH – fL = 90 – 10 = 80 kHz

Resonant frequency (fr) will be:

$=\sqrt{f_H{f}_L}=\sqrt{90\times 10}=30kHz$

Concept: 

1. Consider the information Shown in the figure below

2. For the series RLC circuit ωr is geometrical mean of ωL and ωU or ${\omega }_{\mathrm{U}}=\frac{{\omega }_r^2}{{\omega }_{\mathrm{L}}}$

3. At upper and lower cutoff frequencies current through the circuit is 70.7% of I₀ because at these frequency Z = √ 2 R

4. At resonant frequency, the circuit behaves as purely resistive circuit and hence voltage and current are in same phase. Also power factor become unity.

Analysis:

1. At point A and B  current through the circuit is 70.7% of I0 because at these frequency Z = √ 2 R 

As we know power factor = cos ϕ = ${\displaystyle \frac{R}{Z}}$ = 0.707

ϕ = 45° hence option 1 is correct.

2.The curve is symmetrical about resonance frequency 

So ωU - ωr =  ωr - ωL

hence option 2 is incorrect.

3. At resonant frequency, the circuit behaves as purely resistive circuit and hence voltage and current are in same phase. Also power factor become unity.

4. for  ω > ω_r   X_L > X_C so circuit behave as inductive circuit.

 for  ω < ωr   XL < XC so circuit behave as Capacitive circuit.

The correct answer is option 4.

Concept: 

The RLC circuit means the series or parallel connection of the resistance(R), inductor(L), and capacitor(C).

A series circuit is given below.

Since it is mentioned that the circuit is a loss-free circuit the current will behave similarly to that of an LC series circuit i.e. an undamped sinusoidal waveform will be observed.

In a loss-free circuit, the energy keeps shifting between L and C.

 Important PointsIf the case would have been of loss-making RLC series circuit then the current would have a damped sinusoidal equation.

RLC series circuit:

 An RLC circuit is an electrical circuit consisting of Inductor (L), Capacitor (C), Resistor (R) it can be connected either parallel or series.

When the LCR circuit is set to resonate (XL = XC), the resonant frequency is expressed as 

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

$\omega =\sqrt{{\displaystyle \frac{1}{LC}}}rad/sec$

Quality factor:

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

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

Mathematically, for a coil, the quality factor is given by:

 $Q=\frac{{\omega }_{0}L}{R}=\frac{1}{R}\sqrt{\frac{L}{C}}$

Where,

XL & XC = Impedance of inductor and capacitor respectively

L, R & C = Inductance, resistance, and capacitance respectively

fr = frequency

ω0 = angular resonance frequency

Calculation:

Given, L = 20 mH = 20 × 10^-3 H

C = 200 μF = 200 × 10^-6 F

For a series RLC circuit, the net impedance is given by:

Z = R + j (XL - XC)

XL = Inductive Reactance given by:

XL = ωL

XC = Capacitive Reactance given by:

XL = 1/ωC

The magnitude of the impedance is given by:

$|Z|=\sqrt{R^2+(X_L-X_C{)}^2}$

At resonance, XL = XC,

|Z| = R

Hence, the impedance at resonance is |Z| = R = 0.5 kΩ

Note:

The current flowing across the series RLC circuit will be:

$I=\frac{V}{|Z|}$

∴$I=\frac{V}{R}$

Hence impedance is minimum and current will be maximum at resonance.

In the series RLC circuit, the current (I) Vs frequency (f) graph for the series resonance circuit is shown below. :-

From above we can conclude:-

• For resonance to occur in any circuit it must have at least one inductor and one capacitor.
• Resonance is the result of oscillations in a circuit as stored energy is passed from the inductor to the capacitor.
• Resonance occurs when XL = XC and the imaginary part of the transfer function is zero.
• At resonance, the impedance of the circuit is equal to the resistance value as Z = R.
• At low frequencies the series circuit is capacitive as XC > XL, this gives the circuit a leading power factor.
• At high frequencies the series circuit is inductive as XL > XC, this gives the circuit a lagging power factor.
• The high value of current at resonance produces very high values of voltage across the inductor and capacitor.
• Because impedance is minimum and current is maximum, series resonance circuits are also called Acceptor Circuits.

Concept:

An LC circuit is represented as:

There is a characteristic frequency at which the circuit will oscillate, called the resonant frequency given by:

$f_0=\frac{1}{2\pi \sqrt{LC}}$

Application:

For a fixed value of L, the resonant frequency is proportional to:

$f_0\propto \frac{1}{\sqrt{C}}$

To double the resonant frequency, C must be reduced to 1/4 (quarter), i.e. for C' = C/4, f₀ becomes:

f'₀ = 2 f₀

Concept:

At resonance, only the resistance will remain (as both the inductive and capacitive reactance will be equal), because of which all the power will be delivered to the resistance only.

So, power will be:

$P=\frac{V^2}{R}W$

The bandwidth of a resonant circuit is the difference between the upper cut-off frequency and lower cut-off frequency (with the cut-off frequency being the frequency at which the power is half of the maximum power).

For a series RLC circuit, the bandwidth is calculated as:

$B=\frac{{\omega }_o}{Q}$

Where Q = quality factor and is defined as:

$Q=\frac{{\omega }_{o}L}{R}$.

From both these equations, the Bandwidth can be calculated as:

$B=\frac{R}{L}$.

Calculation:

Given resonant frequency (f₀) = 1.5 kHz

Bandwidth (BW) = 0.75 kHz

Voltage (V) = 50 V

Power (P) = 50 W

$R=\frac{V^2}{P}=50\mathrm{\Omega }$

B.W. in radians = R/L

B.W. in Hertz = $\frac{R}{2\pi L}$

Key Points

 At half-power points of a resonance curve, the current is 0.707 times the maximum current.

Important points regarding half-power points are:

• The current is I_max/√2.
• Impedance is √2 R or √2 Z_min.
• The circuit phase angle → ±45° or π/4 rad