# Resonance in the FE ELectrical Exam

Resonance, a fundamental concept in the world of electricity, has both intriguing possibilities and potential risks that engineers must navigate.  It manifests in various aspects of electrical engineering, such as power transmission lines, filters, and oscillators.

By understanding and harnessing resonance, engineers can optimize circuit performance, improve efficiency, and explore the intricacies of signal processing.

This detailed guide will explore resonance in the FE Electrical exam along with the causes & effects of resonance and insights into effective mitigation techniques.

Read till the end to uncover the principles of resonance and its practical applications, ensuring safe and reliable operation in the dynamic world of electrical engineering.

## Basic Concepts of Resonance

Resonance in the FE electrical exam is a very crucial topic that you must not overlook. Resonance is a phenomenon that occurs when a system is subjected to an external force or stimulus at its natural frequency, resulting in a significant increase in amplitude or energy transfer. To understand resonance, we first need to grasp the concept of natural frequency.

*Natural frequency refers to the frequency at which a system oscillates or vibrates when it is not subjected to any external forces. It is determined by the system’s inherent physical properties, such as its mass, stiffness, and geometry. When a system is excited or disturbed by a force that matches its natural frequency, it tends to respond with the greatest amplitude, leading to resonance.

For instance, you must have heard that soldiers break their march while marching across a hanging bridge. Why? Because a hanging bridge has a natural frequency associated with its structure. As the soldiers march in sync, their footsteps create a rhythmic force that can match the bridge’s natural frequency.

When this happens, the amplitude of the bridge’s vibrations can increase significantly due to resonance, potentially leading to destructive vibrations and even collapse.

In electrical circuits, resonance occurs when the frequency of an external source matches the natural frequency of the circuit. This natural frequency is determined by the circuit’s inductance, capacitance, and resistance (referred to as Series RLC Resonance which we will discuss below in detail).

When a circuit is in resonance, the voltage and current amplitudes can become very high, resulting in an efficient transfer of energy between the source and the circuit. This can have practical implications both in terms of desirable and undesirable effects.

For instance, in radio receivers, resonance is crucial for tuning in to a specific frequency. By adjusting the capacitance and inductance in the circuit, the resonant frequency can be matched to the desired radio frequency, allowing for efficient reception.

However, resonance can also cause issues if not properly managed. For instance, in power systems, resonance can occur between the capacitance of long transmission lines and the inductance of transformers or reactors. This can result in excessive voltages and currents, leading to equipment damage or system instability.

To mitigate these effects, power systems employ measures like shunt capacitors or reactors to adjust the resonance frequencies and prevent excessive energy transfer.

## Resonance in Electrical Systems

Resonant circuits are circuits that are designed to exhibit resonance at a specific frequency. They typically consist of an inductor (L), a capacitor (C), and sometimes a resistor (R). These components work together to create the resonant behavior of the circuit.

Let’s discuss the significance of these components along with the concepts of resonance frequency and quality factor.

### Series RLC Resonance

In this section, we will help you get a more practical and mathematical outlook of series RLC resonance in the FE Electrical exam. For the calculation of the resonance frequency with resonance equations, let’s consider a simple series RLC resonance circuit consisting of an inductor (L) and a capacitor (C).

We will calculate the resonant frequency (fr) and the quality factor (Q) of the circuit.

We have,

Inductance (L) = 10 mH (millihenries)

Capacitance (C) = 100 µF (microfarads)

Step 1 – Convert the given values to the appropriate SI units.

L = 10 mH = 10 × 10^(-3) H (Hertz)

C = 100 µF = 100 × 10^(-6) F (Farads)

Step 2 – Calculation of the resonance frequency (fr) using the formula:

fr = 1 / (2π√(LC))

Substituting the given values into the formula:

fr = 1 / (2π√(10 × 10^(-3) × 100 × 10^(-6)))

fr = 1 / (2π√(10^(-1)))

fr = 1 / (2π × 0.316)

fr ≈ 0.159 Hz (Hertz)

Step 3 – Calculate the quality factor (Q) using the formula:

Q = 1 / R × √(L / C)

Since we are dealing with a series resonant circuit and have not been given a specific resistance value, we assume it to be negligible (R = 0). Therefore, the formula simplifies to:

Q = √(L / C)

Substituting the given values into the formula:

Q = √(10 × 10^(-3) / 100 × 10^(-6))

Q = √(10^(-1))

Q = 0.316

So, the resonant frequency (fr) of the circuit is approximately 0.159 Hz, and the quality factor (Q) is approximately 0.316.

### Application of Series RLC Resonance

The domain of Resonance in the FE Electrical Exam goes beyond just pen and paper. It has some vast practical applications in the electrical engineering field. Below are five applications of resonance in series RLC circuits:

Bandpass Filters – Series RLC circuits can be utilized as bandpass filters to selectively pass a specific range of frequencies while attenuating others. By adjusting the values of inductance (L), capacitance (C), and resistance (R), the circuit’s resonant frequency can be set to the desired frequency range, allowing efficient signal transmission within that band.

Oscillators – Series RLC circuits can function as oscillators, generating periodic waveforms at a specific frequency. By providing positive feedback to sustain oscillations, the circuit can produce stable and continuous signals. Oscillators find applications in various electronic systems, such as radio transmitters, clock generators, and signal generators.

Frequency Selectivity – Series RLC circuits exhibit sharp frequency selectivity around the resonant frequency. This characteristic allows them to separate or isolate signals of a particular frequency from a mixed signal input. These circuits are commonly used in communication systems to filter out unwanted frequencies and extract the desired signal for further processing.

Resonant Transformers – Series RLC circuits can be employed as resonant transformers, where the inductor and capacitor form a resonant circuit that facilitates efficient energy transfer between primary and secondary windings. Resonant transformers are utilized in applications like wireless power transfer systems, induction heating, and high-frequency power converters.

Impedance Matching – Series RLC circuits can be used for impedance matching purposes. By adjusting the values of L, C, and R, the circuit can match the impedance of a source to the load impedance. This technique ensures maximum power transfer between the source and load, minimizing reflections and optimizing signal quality in applications such as audio systems, antennas, and transmission lines.

To give you a detailed outlook of problems you can expect for the Resonance in the FE Electrical Exam must go through the examples below. We have compiled some of the most common applications of resonance in action including impedance matching and bandpass filters.

Let’s consider a scenario where we have a source with an impedance of 50 ohms and a load with an impedance of 100 ohms. We will use a series RLC circuit for impedance matching and calculate the values of inductance (L), capacitance (C), and resistance (R) required for maximum power transfer.

We have

The source impedance (Zs) = 50 ohms

Load impedance (Zl) = 100 ohms

Step 1 – Calculate the desired characteristic impedance (Z0) using the geometric mean of the source and load impedances:

Z0 = √(Zs * Zl)

Substituting the given values into the formula:

Z0 = √(50 * 100)

Z0 = √5000

Z0 ≈ 70.71 ohms

Step 2 – Determine the required values of inductance (L) and capacitance (C) for impedance matching using the desired characteristic impedance (Z0) and the resonant frequency (fr).

For impedance matching, we want the reactive components to cancel each other at the resonant frequency, resulting in a purely resistive impedance.

Assuming the resonant frequency is known or predetermined, we can calculate the values of L and C using the following formulas:

L = Z0 / (2πfr)

C = 1 / (2πfrZ0)

Step 3 – Substitute the values of fr and Z0 into the formulas to calculate the required values of L and C.

Suppose we have a resonant frequency (fr) of 10 MHz (10^7 Hz).

L = 70.71 / (2π * 10^7)

L ≈ 1.13 × 10^(-6) Henry (or 1.13 µH)

C = 1 / (2π * 10^7 * 70.71)

C ≈ 2.25 × 10^(-10) Farads (or 225 pF)

Step 4 – Calculate the value of resistance (R) required for impedance matching, which is the geometric mean of the source and load resistances.

R = √(Zs * Zl)

R = √(50 * 100)

R = √5000

R ≈ 70.71 ohms

Therefore, to achieve impedance matching between a source impedance of 50 ohms and a load impedance of 100 ohms, we would require an RLC circuit with the inductance of approximately 1.13 µH, a capacitance of approximately 225 pF, and a resistance of approximately 70.71 ohms.

By matching the impedance, maximum power transfer can be achieved between the source and load, minimizing reflections and optimizing the performance of the circuit.

Similarly, let’s consider a sample problem involving the design of a bandpass filter using a series RLC circuit. We will determine the values of inductance (L), capacitance (C), and resistance (R) required to create a bandpass filter with a center frequency of 1 kHz and a bandwidth of 100 Hz.

We have

Center frequency (fc) = 1 kHz (1000 Hz)

Bandwidth (B) = 100 Hz

Step 1 – Calculate the lower and upper cutoff frequencies of the bandpass filter.

Lower cutoff frequency (f1) = fc – (B/2)

Upper cutoff frequency (f2) = fc + (B/2)

Substituting the given values into the formulas:

f1 = 1000 – (100/2)

f1 = 950 Hz

f2 = 1000 + (100/2)

f2 = 1050 Hz

Step 2 – Calculate the values of inductance (L) and capacitance (C) required for the bandpass filter using the lower and upper cutoff frequencies.

Inductance:

L = 1 / (4π^2f1f2C)

Substituting the given values of f1 and f2 into the formula:

L = 1 / (4π^2 * 950 * 1050 * C)

Capacitance:

C = 1 / (4π^2f1f2L)

Substituting the given values of f1, f2, and L into the formula:

C = 1 / (4π^2 * 950 * 1050 * L)

Step 3 –  Substitute known values and solve for either L or C.

Let’s assume we want to solve for the inductance (L). We’ll arbitrarily assign a value of 1 µF for the capacitance (C).

L = 1 / (4π^2 * 950 * 1050 * 1 × 10^(-6))

L ≈ 0.376 H (Henry)

So, for the bandpass filter with a center frequency of 1 kHz and a bandwidth of 100 Hz, the required values are approximately 0.376 H (Henry) for the inductance (L) and 1 µF (microfarad) for the capacitance (C).

By designing the series RLC circuit with these values, the bandpass filter will selectively pass frequencies within the specified bandwidth, effectively attenuating frequencies outside that range.

*Read the following section for a detailed account of resonance filters

## Resonance and Filter Design

This is another vital topic you must know to prepare for the topic of resonance in the FE Electrical exam. Resonant filters are a type of electronic filter that utilizes the phenomenon of resonance to selectively pass or reject specific frequencies.

These filters consist of a combination of passive components, such as resistors, capacitors, and inductors, arranged in a specific configuration to achieve the desired frequency response

### Low-pass Resonant Filters

A low-pass resonant filter allows frequencies below a certain cutoff frequency to pass through while attenuating higher frequencies. It is designed to pass signals with lower frequencies, such as bass tones, while filtering out higher-frequency components. The cutoff frequency, denoted as fc, is the point where the filter’s response begins to roll off.

For instance, consider designing a second-order low-pass resonant filter with a cutoff frequency of 1 kHz. The desired gain at the cutoff frequency is -3 dB. Using a standard configuration such as a Sallen-Key topology, determine the values of the components required for the filter.

We have,

Cutoff frequency (fc) = 1 kHz = 1000 Hz

Gain at fc = -3 dB

Step 1 – Determine the filter order and calculate the damping factor (ζ).

For a second-order filter, the damping factor can be calculated as:

ζ = √(10^(Gain/10) – 1)

Substituting the given gain into the formula:

ζ = √(10^(-3/10) – 1)

ζ ≈ 0.45

Step 2 – Calculate the values of the components required for the filter.

Using the Sallen-Key topology, the component values can be calculated as follows:

R1 = R2 = R (equal resistors)

C1 = C2 = C (equal capacitors)

Choose a resistor value, e.g., R = 10 kΩ, and calculate the values of capacitors using the following formulas:

C = 1 / (2πfcRζ)

Therefore, for the second-order low-pass resonant filter with a cutoff frequency of 1 kHz and a gain of -3 dB, the required component values are approximately R = 10 kΩ and C = 10 nF.

### High-pass Resonant Filters

A high-pass resonant filter allows frequencies above a certain cutoff frequency to pass through while attenuating lower frequencies. It is designed to pass signals with higher frequencies, such as treble tones while filtering out lower-frequency components. The cutoff frequency, denoted as fc, is the point where the filter’s response begins to roll off.

Consider, as an electrical engineer, you have to design a first-order high-pass resonant filter with a cutoff frequency of 500 Hz. Determine the values of the resistor (R) and capacitor (C) required for the filter.

We have

Cutoff frequency (fc) = 500 Hz

Step 1 – Choose a capacitor value.

Select a capacitor value, e.g., C = 1 µF (microfarad).

Step 2 – Calculate the resistor value.

Using the formula for the resistor:

R = 1 / (2πfcC)

Substituting the given values into the formula:

R = 1 / (2π * 500 * 10^(-6))

R ≈ 318.31 ohms

Therefore, for the first-order high-pass resonant filter with a cutoff frequency of 500 Hz, the required component values are approximately R = 318.31 ohms and C = 1 µF.

### Band-pass Resonant Filters

A band-pass resonant filter allows a specific range of frequencies, called the passband, to pass through while attenuating frequencies outside the passband. It is designed to selectively filter and pass signals within a desired frequency range while rejecting others.

Consider a scenario where you have to design a second-order band-pass resonant filter with a passband from 1 kHz to 10 kHz. Determine the values of the inductor (L), capacitor (C), and resistor (R) required for the filter.

We have,

Lower cutoff frequency (f1) = 1 kHz = 1000 Hz

Upper cutoff frequency (f2) = 10 kHz = 10,000 Hz

Step 1 – Determine the center frequency (fc) and the bandwidth (B).

Center frequency (fc) = √(f1 * f2)

Bandwidth (B) = f2 – f1

Substituting the given values into the formulas:

fc = √(1000 * 10000) ≈ 3162.28 Hz

B = 10000 – 1000 = 9000 Hz

Step 2 – Calculate the values of the components required for the filter.

Using the formulas for the inductor and capacitor:

L = 1 / (4π²fc²C)

C = 1 / (4π²fc²L)

Choose a resistor value, e.g., R = 1 kΩ, and calculate the values of the inductor and capacitor:

L = 1 / (4π² * (3162.28 Hz)² * C)

C = 1 / (4π² * (3162.28 Hz)² * L)

For simplicity, assume C = 1 nF (nano farad).

Calculating the values of L and C using the given resistor value:

L ≈ 51.97 µH (microhenries)

Therefore, for the second-order band-pass resonant filter with a passband from 1 kHz to 10 kHz, the required component values are approximately L = 51.97 µH, C = 1 nF, and R = 1 kΩ.

## Resonance and Power Systems

Resonance in power transmission lines occur when the line’s inherent inductance (L) and capacitance (C) interact in such a way that the line’s natural frequency matches the frequency of the applied voltage or current. This can lead to significant issues and disturbances in the power system.

### Types of Resonance in Power Transmission Lines

The two most common types of resonance in power transmission includes:

#### 1. Series Resonance

In series resonance, the line inductance and capacitance combine in series. The resonant frequency (fr) can be calculated using the following formula:

fr = 1 / (2π√(LC))

#### 2. Parallel Resonance

In parallel resonance, the line inductance and capacitance combine in parallel. The resonant frequency (fr) can be calculated using the following formula:

fr = 1 / (2π√(LC))

The table below gives the resonance causes and their impact on power transmission along with techniques to mitigate these risks.

## Resonance and Electrical Safety

Resonance can result in overvoltage and overcurrent conditions, posing risks to the system and connected equipment. When harmonic frequencies match the system’s natural frequency, voltage, and current magnification can occur, exceeding safe operating limits.

This can cause insulation breakdown, equipment damage, and even electrical shocks to personnel. Resonance or harmonic current in electrical systems can lead to overvoltage and overcurrent issues, posing significant challenges and risks, such as:

• Increased RMS Current – Harmonic currents increase the root mean square (RMS) current flowing through the electrical system. This elevated current level can exceed the design limits of conductors, transformers, and other equipment, potentially leading to overheating and premature failure.
• Deterioration of Supply Voltage Quality – Harmonic currents distort the waveform of the supply voltage, causing voltage quality issues. The presence of harmonics can result in voltage distortion, voltage unbalance, and deviation from the ideal sinusoidal waveform, affecting the performance and reliability of connected equipment.
• Stress on Electrical Network –  The presence of harmonics places stress on the electrical network by introducing additional current and voltage components. This increased stress can lead to higher losses in conductors, transformers, and other components, reducing system efficiency and potentially compromising its overall stability.
• Equipment Damage and Disruption – Harmonic currents can cause overheating of transformers, motors, and cables due to increased losses and thermal effects. This can result in equipment damage, insulation degradation, and reduced lifespan of electrical devices. Furthermore, harmonic-induced voltage fluctuations and disruptions can cause logic faults in digital devices, leading to malfunction or erratic behavior.
• Increased Operating Costs – Resonance-induced harmonics can lead to higher operating costs. The increased losses in equipment and additional energy consumption caused by harmonic currents result in elevated energy bills and reduced efficiency of the electrical system.
• Sensitivity of Capacitors – Capacitors are particularly vulnerable to harmonic components in the supply voltage due to their capacitive reactance decreasing with increasing frequency. Even a small percentage of harmonic voltage can cause a significant current to flow in the capacitor circuit. This can lead to excessive heating, reduced capacitor lifespan, and potential failure.

### Mitigating Resonance Hazards in Electrical Systems

To mitigate the hazards associated with resonance in electrical systems and ensure electrical safety, several measures can be taken. These include:

#### Harmonic Filters and Active Power Filters

Installing harmonic filters and active power filters can effectively mitigate harmonic currents and voltages. These devices are designed to attenuate specific harmonic frequencies and maintain the supply voltage within acceptable limits, reducing the risks of overvoltage and overcurrent conditions.

#### Power Factor Correction

Improving the power factor of the system helps minimize harmonic distortions. By introducing power factor correction techniques such as capacitor banks, the reactive power demand is reduced, leading to a smoother current waveform and reducing the likelihood of resonance.

#### Harmonic Current Limiting Devices

Employing harmonic current limiting devices such as line reactors or harmonic traps can help mitigate the flow of harmonic currents. These devices introduce impedance in the system to suppress harmonics and prevent them from reaching critical levels, thus safeguarding equipment and reducing the risks of overheating and failures.

#### Proper Equipment Design and Selection

Careful consideration should be given to the design and selection of equipment to minimize harmonic generation and susceptibility. Choosing power electronic devices, transformers, and other equipment with low harmonic distortion characteristics can reduce the impact of harmonics and minimize resonance hazards.

#### Regular Monitoring and Maintenance

Implementing a comprehensive monitoring and maintenance program is crucial to identify and rectifying resonance issues promptly. Periodic inspections, thermal imaging, and power quality measurements can help detect early signs of resonance, allowing for corrective actions to be taken before significant damage or failures occur.

### Conclusion

Now it is evident that resonance in the FE electrical exam holds immense value not only in academic studies but also in shaping your successful career as a professional electrical engineer. By mastering the principles of resonance, you will gain a solid foundation in electrical engineering, enabling you to solve complex problems, design efficient systems, and contribute to technological advancements.

The knowledge and skills you acquire in this area will open doors to exciting opportunities in industries ranging from power systems and telecommunications to electronics and control systems.

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