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

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:

\(QF={1\over R}\sqrt{L\over C}={\omega_oL\over R}\)

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:

\(QF={ω_o\over BW}\)

where, ω_o = Resonance frequency

BW = Bandwidth

Calculation

Given, QF = 100

ω_o = 1 MHz

\(100={10^6\over BW}\)

BW = 1kHz

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:

\(QF={1\over R}\sqrt{L\over C}={\omega_oL\over R}\)

Voltage magnification in such a circuit is directly proportional to the quality factor Q Q" id="MathJax-Element-3-Frame" role="presentation" style="position: relative;" tabindex="0">Q$Q$ $Q$ . When Q increases, it indicates lower energy loss (primarily due to reduced resistance), and as a result, the voltage across the inductor or capacitor at resonance becomes significantly higher than the supply voltage, showing greater voltage magnification.

Calculation:
The formula for the quality factor (Q) of an LC circuit is given by:

Q = (1 / R) × √(L / C)

Where:

• R = resistance = 100 Ω
• L = inductance = 80 mH = 80 × 10^-3 H
• C = capacitance = 2 μF = 2 × 10^-6 F

Substituting the values into the formula:

Q = (1 / 100) × √((80 × 10^-3) / (2 × 10^-6))

Q = (1 / 100) × √(40 × 10³)

Q = (1 / 100) × 200

Q = 2

The quality factor of the circuit is 2.

Explanation:

Resonance in Electrical Circuits

Definition: Resonance in electrical circuits occurs when the inductive reactance and capacitive reactance are equal in magnitude but opposite in phase, causing them to cancel each other out. This results in the circuit impedance being purely resistive at a specific frequency, known as the resonant frequency. At resonance, the circuit can store and transfer energy between the inductor and capacitor with maximum efficiency, leading to a significant increase in current or voltage.

Working Principle: Resonance in an R-L-C circuit (where R stands for resistance, L for inductance, and C for capacitance) is achieved when the inductive reactance (X_L) and capacitive reactance (X_C) are equal. The condition for resonance can be mathematically expressed as:

X_L = X_C

Since inductive reactance (X_L) is given by:

X_L = 2πfL

And capacitive reactance (X_C) is given by:

X_C = 1 / (2πfC)

Where f is the frequency of the AC supply, L is the inductance, and C is the capacitance. At resonance:

2πfL = 1 / (2πfC)

Solving for the resonant frequency (f₀):

f₀ = 1 / (2π√LC)

At this frequency, the total impedance of the circuit is at its minimum if it is a series R-L-C circuit or at its maximum if it is a parallel R-L-C circuit, and the circuit exhibits resonant behavior.

Correct Option Analysis:

The correct option is:

Option 2: R-L-C parallel

This option correctly identifies a circuit where resonance can occur. In an R-L-C parallel circuit, the inductive and capacitive reactances can cancel each other out at a specific frequency, leading to a condition of resonance. This results in a high impedance at the resonant frequency, and the circuit can exhibit significant voltage magnification.

In a parallel R-L-C circuit, the total admittance (Y) is given by the sum of the individual admittances:

Y = Y_R + Y_L + Y_C

Where:

Y_R = 1 / R

Y_L = 1 / (jωL)

Y_C = jωC

At resonance, the inductive susceptance (B_L) and capacitive susceptance (B_C) cancel each other out:

B_L = -B_C

This results in the total susceptance being zero and the total admittance being purely resistive, leading to resonance.

Important Information

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

Option 1: R-L series

Resonance cannot occur in an R-L series circuit because it lacks a capacitive component. Resonance requires both inductance and capacitance to create a condition where inductive and capacitive reactances can cancel each other out. Without a capacitor, this balance cannot be achieved.

Option 3: R-C series

Similarly, resonance cannot occur in an R-C series circuit because it lacks an inductive component. Resonance requires both inductance and capacitance to achieve the condition where inductive and capacitive reactances cancel each other out. Without an inductor, this balance cannot be achieved.

Option 4: R-L parallel

Resonance cannot occur in an R-L parallel circuit either because it lacks a capacitive component. As with the series circuits, both inductance and capacitance are required to achieve the resonance condition. The absence of a capacitor means that the necessary balance between inductive and capacitive reactances cannot be established.

Conclusion:

Understanding resonance in electrical circuits is crucial for designing circuits that can efficiently store and transfer energy at specific frequencies. In the context of the options provided, the R-L-C parallel circuit is the correct choice where resonance can occur. This is because it includes both inductive and capacitive components, which are essential for achieving the condition of resonance. By analyzing the other options, it becomes evident that the absence of either the inductive or capacitive component makes resonance impossible in those circuits.

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:

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

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:

\(= \sqrt {{f_H}{f_L}} = \sqrt {90 \times 10} = 30\;kHz\)

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 \({{\rm{ω}}_{{\rm{U}}}} = \frac{{{\rm{ω}}_r^2}}{{{{\rm{ω}}_{\rm{L}}}}}\;\)

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 = \(\dfrac{R}{Z}\) = 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.

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 

 \(f = \dfrac{1}{{2π }}\sqrt {\dfrac{1}{{LC}}}\;Hz\)

\(\omega = \sqrt {\dfrac{1}{{LC}}}\;rad/sec\)

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

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.

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

\(R = \frac{{{V^2}}}{P} = 50\;{\Omega}\)

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