Circuit Analysis in FE Electrical Exam

Welcome to our blog on Circuit Analysis in the FE Electrical exam. If you’re an aspiring electrical engineer, understanding the importance of circuit analysis is crucial for a successful career in this dynamic field.

Circuit analysis is the backbone of electrical engineering, allowing us to scrutinize the inner workings of electronic systems with precision and finesse. It enables us to comprehend how currents flow, voltages interact, and components behave in intricate electrical networks.

Our future engineers can unlock a realm of possibilities by mastering the fundamentals of circuit analysis in the FE electrical exam. From creating innovative technologies to solving complex problems, this knowledge empowers you to push the boundaries of what’s possible in electrical engineering.

So, fasten your seatbelts as we embark on a journey through the intricacies of circuit analysis, unveiling its significance and the insights it holds.

Fundamentals of Circuit Analysis

To give you a brief outlook of circuit analysis accounts, the table below outlines the fundamental concepts and their key takeaways we will discuss in this guide.

DC Circuit Analysis

A DC (Direct Current) circuit is a closed loop through which an electric current can flow in only one direction. In a DC circuit, the voltage source (such as a battery or DC power supply) provides a constant voltage and the current flows in a steady state with a fixed polarity.

DC circuit analysis involves analyzing and solving circuits that consist of direct current (DC) sources, resistors, and other passive circuit elements. The analysis is based on various laws and theorems, which provide a systematic approach to understanding and solving these circuits. 

Performing DC circuit analysis involves the following steps and techniques:

• Identify the components – Identify the different components in the circuit, such as resistors, capacitors, inductors, and voltage sources. Note their values and properties.
• Draw the circuit diagram – Create a schematic representation of the DC circuit, illustrating the arrangement and connections of the components.
• Apply Kirchhoff’s laws – Apply Kirchhoff’s laws to analyze the circuit. Kirchhoff’s current law (KCL) states that the sum of currents entering a node equals the sum of currents leaving that node. Kirchhoff’s voltage law (KVL) says that the sum of the voltage rises and drops around any closed loop in a circuit is zero.
• Assign voltage polarities – Assign polarities to the voltage sources based on their orientation in the circuit. Usually, one voltage source terminal is designated as positive and the other negative.
• Calculate currents – Use Ohm’s law (V = IR) to calculate currents flowing through resistors, where V is the voltage across the resistor, I is the current, and R is the resistance. For other components like capacitors and inductors, you need to use appropriate equations specific to those components.
• Solve the equations – Set up and solve equations based on Kirchhoff’s laws and algebraic principles.
• Determine voltage drops and power – Determine the voltage drops across each component and calculate power dissipation (P = IV), where P is power, I is current, and V is voltage.
• Check the circuit behavior – Analyze the results to understand the circuit’s behavior. Determine current flow, voltage drops, power consumption, and other relevant characteristics.

DC circuit analysis in the FE electrical exam allows you to determine critical electrical quantities, such as currents, voltages, and power dissipation. It aids in understanding the behavior and performance of a DC circuit.

Don’t get confused with the laws and theorems used to perform DC Circuit Analysis. We will discuss these techniques and their respective circuit analysis problems in detail.

Ohm’s Law

Ohm’s Law is a fundamental principle in electrical engineering that relates the current flowing through a conductor to the voltage applied across it and the resistance of the conductor.

It states that the current (I) flowing through a conductor is directly proportional to the voltage (V) across the conductor and inversely proportional to the resistance (R) of the conductor. Mathematically, Ohm’s Law can be expressed as

V = IR

where:

• V is the voltage across the conductor,
• I is the current flowing through the conductor, and
• R is the resistance of the conductor.

Consider a resistor with a resistance of 10 ohms. If a voltage of 20 volts is applied across the resistor, How can you find the current flowing through it?

We have

Resistance (R) = 10 ohms

Voltage (V) = 20 volts

Using Ohm’s Law: V = IR

We rearrange the equation to solve for current (I): I = V/R

Substituting the given values:

I = 20 volts / 10 ohms

I = 2 amperes

Therefore, the current flowing through the resistor is 2 amperes.

Now let’s discuss Ohm’s Law in the context of series and parallel circuits:

Series Circuit

In a series circuit, the components (resistors, capacitors, or inductors) are connected end-to-end, forming a single path for current flow. In a series circuit:

• Current – The current remains the same at all points in a series circuit. The total current flowing through the circuit equals the current through each component.
• Voltage – The total voltage across the series circuit is the sum of the voltage drops across each component. The voltage is divided among the components based on their resistance or impedance.
• Resistance – The total resistance of a series circuit is the sum of the individual resistances of the components. The total resistance increases as more components are added.

Consider a series circuit consisting of three resistors connected: R1 = 10 ohms, R2 = 20 ohms, and R3 = 30 ohms. If a current of 2 amperes flows through the circuit, you can calculate the voltage drops across each resistor.

We have

R1 = 10 ohms

R2 = 20 ohms

R3 = 30 ohms

I = 2 amperes

Using Ohm’s Law: V = IR

For resistor R1:

V1 = I * R1 = 2 amperes * 10 ohms = 20 volts

For resistor R2:

V2 = I * R2 = 2 amperes * 20 ohms = 40 volts

For resistor R3:

V3 = I * R3 = 2 amperes * 30 ohms = 60 volts

Therefore, the voltage drop across resistor R1 is 20 volts, across R2 is 40 volts, and across R3 is 60 volts.

Parallel Circuit

In a parallel circuit, the components are connected, providing multiple paths for current to flow. In a parallel circuit:

• Current – The total current flowing into a parallel circuit is divided among the different branches, depending on the resistance or impedance of each branch. The total current equals the sum of the currents in each branch.
• Voltage – The voltage across each component in a parallel circuit is the same. The total voltage across the parallel circuit equals across any component.
• Resistance –  The reciprocal of the total resistance of a parallel circuit is equal to the sum of the reciprocals of the individual resistances. The total resistance decreases as more branches are added.

Consider a parallel circuit consisting of three resistors connected in parallel: R1 = 10 ohms, R2 = 20 ohms, and R3 = 30 ohms. If a voltage of 50 volts is applied across the circuit, how can you calculate the current flowing through each resistor?

We have

R1 = 10 ohms

R2 = 20 ohms

R3 = 30 ohms

V = 50 volts

Using Ohm’s Law: I = V/R

For resistor R1:

I1 = V / R1 = 50 volts / 10 ohms = 5 amperes

For resistor R2:

I2 = V / R2 = 50 volts / 20 ohms = 2.5 amperes

For resistor R3:

I3 = V / R3 = 50 volts / 30 ohms ≈ 1.67 amperes

Therefore, the current flowing through resistor R1 is 5 amperes, through R2 is 2.5 amperes, and through R3 is approximately 1.67 amperes.

Equivalent Resistance, Current, and Voltage

You can also calculate the equivalent resistance, total current, and voltage for the series and parallel circuits in the previous circuit analysis problems.

For Series Circuit

R1 = 10 ohms

R2 = 20 ohms

R3 = 30 ohms

I = 2 amperes

Equivalent resistance (Req):

In a series circuit, the equivalent resistance is the sum of the individual resistances.

R = R1 + R2 + R3

R = 10 ohms + 20 ohms + 30 ohms

R = 60 ohms

Total voltage (V):

In a series circuit, the total voltage across the circuit equals the sum of the voltage drops across each resistor.

V = V1 + V2 + V3

V = 20 volts + 40 volts + 60 volts

V = 120 volts

Total current (I):

In a series circuit, the total current is the same as the current through each component.

I = 2 amperes

Therefore, in the series circuit:

The equivalent resistance (Req) is 60 ohms.

The total current (I) is 2 amperes.

The total voltage (V) is 120 volts.

For Parallel Circuit

R1 = 10 ohms

R2 = 20 ohms

R3 = 30 ohms

V = 50 volts

Equivalent resistance (Req):

In a parallel circuit, the reciprocal of the equivalent resistance is equal to the sum of the reciprocals of the individual resistances.

1/R = 1/R1 + 1/R2 + 1/R3

1/R = 1/10 ohms + 1/20 ohms + 1/30 ohms

1/R = (3/30 + 1/30 + 2/60) ohms

1/R = 6/60 ohms

R = 60/6 ohms

R = 10 ohms

Total voltage (V):

In a parallel circuit, the total voltage across the circuit is equal to the voltage across any of the components.

V = 50 volts

Total current (I):

In a parallel circuit, the total current is the sum of the currents flowing through each branch.

I = I1 + I2 + I3

I = V/R1 + V/R2 + V/R3

I = 50 volts / 10 ohms + 50 volts / 20 ohms + 50 volts / 30 ohms

I ≈ 5 amperes

Therefore, in the parallel circuit:

The equivalent resistance (Req) is 10 ohms.

The total current (I) is 5 amperes

The total voltage (V) is 50 volts.

These calculations provide information about the behavior of the circuits and allow us to understand the relationship between voltage, current, and resistance in series and parallel configurations.

Power Law or Watt’s Law

Power Law is another critical circuit analysis account to address circuit analysis in the FE electrical exam. The power law states that the power (P or W) dissipated or consumed in an electrical circuit is equal to the product of the voltage (V) applied across the circuit and the current (I) flowing through it.

Mathematically, it can be expressed as P = V ^* I (or P=(V^{^}2)/R or P=I^{^}2 ^* R, based on variations from Ohm’s Law)

Where:

• P is the power in watts (W).
• V is the voltage in volts (V).
• I is the current in amperes (A).
• R is the Resistance in Ohm.

This equation relates the power consumption or dissipation to the voltage and current in a circuit. Calculating power in various electrical devices and systems is a fundamental principle.

Now consider a circuit with a voltage source of 12 volts and a current of 2 amperes. You can calculate the power consumed in the circuit.

We have

V = 12 volts

I = 2 amperes

Using the power equation: P = V * I

Substituting the given values:

P = 12 volts * 2 amperes

P = 24 watts

Therefore, the power consumed in the circuit is 24 watts.

This calculation demonstrates how the power equation can determine the power dissipated or consumed in an electrical circuit when the voltage and current values are known.

Kirchhoff’s Law

Kirchhoff’s laws are fundamental principles in circuit analysis in the FE electrical exam, widely used in the electrical engineering domain to analyze the behavior of DC circuits. These laws are used to analyze and understand the behavior of electric circuits.

Kirchhoff’s Current Law (KCL)

KCL, also known as the junction rule, states that the sum of currents entering a node or junction in a circuit equals the sum of currents leaving that node. In other words, the total current flowing into a node equals the total current flowing out of that node, as no charge is accumulated at a node.

Mathematically, KCL can be expressed as:

∑I_in = ∑I_out,

Where

• ∑I_in is the sum of currents entering the node, and
• ∑I_out is the sum of currents leaving the node.

KCL is based on the principle of charge conservation and is essential for analyzing the current distribution in complex circuit configurations.

Consider a circuit with a junction or node where three currents enter: I1 = 2 A, I2 = 3 A, and I3 = 4 A. You can calculate the total current leaving the junction.

We have

I1 = 2 A (entering current)

I2 = 3 A (entering current)

I3 = 4 A (entering current)

Using KCL, the sum of currents entering the node is equal to the sum of currents leaving the node:

Total entering current = Total leaving current

I1 + I2 + I3 = ∑I_out

Substituting the given values:

2 A + 3 A + 4 A = ∑I_out

9 A = ∑I_out

Therefore, the total current leaving the junction is 9 amperes.

Kirchhoff’s Voltage Law (KVL)

KVL, also known as the loop rule, states that the algebraic sum of the voltage rises and voltage drops around any closed loop in a circuit equals zero. In other words, the sum of the potential differences (voltages) across the components in a closed loop equals the sum of the potential differences applied to the loop.

Mathematically, KVL can be expressed as:

∑V_rises – ∑V_drops = 0,

Where:

• ∑V_rises is the sum of the voltage rises (positive contributions),
• ∑V_drops is the sum of the voltage drops (negative contributions).

KVL is based on the principle of energy conservation and is crucial for analyzing voltage distributions and solving circuit loop equations.

Consider a circuit with a closed loop consisting of three resistors: R1 = 5 Ω, R2 = 10 Ω, and R3 = 8 Ω. Now you have to calculate the voltage drops across each resistor if a voltage source of 30 volts is connected across the loop.

We have

R1 = 5 Ω

R2 = 10 Ω

R3 = 8 Ω

Voltage across the loop (V) = 30 volts

We’ll use Ohm’s Law (V = IR) to calculate the voltage drops across each resistor, where V represents voltage, I means current, and R represents resistance.

Let’s assume the current flowing through the loop is I.

For resistor R1:

V1 = I * R1

For resistor R2:

V2 = I * R2

For resistor R3:

V3 = I * R3

Substituting the given values into Ohm’s Law:

V1 = I * 5

V2 = I * 10

V3 = I * 8

Since we don’t have specific information about the current flowing through the loop, we cannot directly calculate the voltage drops across each resistor. However, we can determine the relationship between the voltage drops.

Applying Kirchhoff’s Voltage Law (KVL), the sum of the voltage rises equals the sum of the voltage drops around a closed loop. In this case, the voltage rise is the applied voltage V.

Let’s assume the voltage drop across resistor R1 is V1, across R2 is V2, and across R3 is V3.

Using KVL

V – V1 – V2 – V3 = 0

Substituting the expressions for V1, V2, and V3:

V – (I * 5) – (I * 10) – (I * 8) = 0

Simplifying the equation:

V – 23I = 0

Since we know the applied voltage V is 30 volts, we can substitute it into the equation:

30 – 23I = 0

Solving for I:

23I = 30

I = 30 / 23

Now, we can calculate the voltage drops across each resistor using the found value of I.

For resistor R1:

V1 = I * R1

V1 = (30 / 23) * 5

For resistor R2:

V2 = I * R2

V2 = (30 / 23) * 10

For resistor R3:

V3 = I * R3

V3 = (30 / 23) * 8

Calculating the voltage drops:

V1 ≈ 6.52 V

V2 ≈ 13.04 V

V3 ≈ 10.43 V

Therefore, the voltage drop across resistor R1 is approximately 6.52 volts, across R2 is approximately 13.04 volts, and across R3 is approximately 10.43 volts.

Thevenian and Norton Theorems 

Thevenin’s Theorem

Thevenin’s theorem states that any linear electrical network consisting of resistors, voltage sources, and current sources can be replaced by an equivalent circuit containing a single voltage source and a single resistor connected in series. This equivalent circuit is called the Thevenin equivalent circuit.

The Thevenin equivalent circuit is helpful because it simplifies complex circuits into a single voltage source and a resistor, making analysis and calculations much more straightforward.

Consider the circuit with 2 resistances across terminals A and B:

You can calculate the Thevenin voltage (Vth):

• Disconnect all the resistors from the circuit.
• Measure the voltage between terminals A and B.
• This is the Thevenin voltage (Vth).

Calculate the Thevenin resistance (Rth):

• Remove the voltage source and all current sources from the circuit.
• Replace them with a short circuit (wire).
• Calculate the equivalent resistance between terminals A and B.
• This is the Thevenin resistance (Rth).

Draw the Thevenin equivalent circuit:

• Draw a voltage source with the Thevenin voltage (Vth) connected in series with the Thevenin resistance (Rth).

Norton’s Theorem

Norton’s theorem is similar to Thevenin’s theorem. Still, instead of a voltage source and a resistor, it states that any linear electrical network can be replaced by an equivalent circuit containing a current source and a resistor connected in parallel. This equivalent circuit is called the Norton equivalent circuit.

The Norton equivalent circuit is also helpful for simplifying complex circuits, mainly when current values are interesting.

Consider the circuit with two resistances across terminals A and B and the current source.

To calculate the Norton current (In):

Disconnect all the resistors from the circuit.

• Measure the current between terminals A and B.
• This is the Norton current (In).

Calculate the Norton resistance (Rn):

• Remove the current source and all voltage sources from the circuit.
• Replace them with an open circuit.
• Calculate the equivalent resistance between terminals A and B.
• This is the Norton resistance (Rn).

Draw the Norton equivalent circuit:

• Draw a current source with the Norton current (In) connected (in parallel) with the Norton resistance (Rn).

AC Circuit Analysis

AC voltage stands for Alternating Current voltage. It is a type of electrical voltage that continuously changes its magnitude and direction over time. Unlike DC (Direct Current) voltage, which remains constant, AC voltage fluctuates back and forth in a cyclical pattern.

The most widely used and valuable technique is Waveform Analysis to perform AC circuit analysis in the FE electrical exam.

Waveform Analysis

Waveform analysis involves studying the shape and characteristics of an AC voltage waveform. AC voltage waveforms are typically sinusoidal, following a smooth, repetitive, and symmetrical pattern resembling a sine wave.

Waveform analysis helps in understanding various parameters of the AC voltage, such as amplitude, frequency, phase, and harmonic content.

Mathematically, the relation of voltage over time in waveform analysis is given as.

A(t) = A sin(ωt)

Where, 

ω is angular displacement given by:

ω = Δθ/ Δt

For one complete phase shift,

Δθ = 2π radian (that is 360 Degrees)

Δt = T (time period) = 1/f

So

1/Δt = f

Hence

ω = 2πf

So we can write

A(t) = A sin(2πft)

Where A shows the amplitude of the voltage at time instant “t”, also written as

V(t) = V sin(2πft)

Now, let’s move on to a sample problem that involves AC voltage and waveform analysis.

A sinusoidal AC voltage with an amplitude of 10 volts and a frequency of 50 Hz is given by the equation V(t) = 10 sin(2πft), where t represents time. Analyze the waveform and determine the peak voltage, period, and frequency.

To analyze the waveform, let’s break down the given equation and understand its components.

V(t) = 10 sin(2πft)

The waveform’s amplitude is 10 volts. It represents the maximum value the voltage reaches in either the positive or negative direction. In this case, it is +10 volts and -10 volts.

The term sin(2πft) represents the variation of the waveform over time. The sinusoidal function oscillates smoothly between the maximum positive and negative voltages.

The value of f is the frequency of the waveform, which is given as 50 Hz. Frequency measures the number of cycles (periods) the waveform completes in one second.

We can calculate the peak voltage, period, and frequency based on the above AC circuit analysis accounts at designated time instances.

Peak Voltage:

The peak voltage equals the waveform’s amplitude (maximum displacement from the mean position). In this case, it is 10 volts.

Period:

The period (T) is the time the waveform takes to complete one cycle. It is the inverse of the frequency (f).

T = 1 / f = 1 / 50 = 0.02 seconds

Frequency:

The frequency (f) is the number of complete cycles per second. In this case, it is 50 Hz.

You can plot the waveform and understand its shape. (by taking the x-axis representing time and the y-axis representing voltage).

Start from time t = 0, and plot the corresponding voltage value with each interval of T based on the equation V(t) = 10 sin(2πft).

Continue this process for multiple cycles to see the repetitive nature of the waveform.

The waveform graph resembles a sinusoidal curve that smoothly oscillates between the maximum positive and negative voltages of +10 and -10 volts.

For the given AC voltage waveform with an amplitude of 10 volts and a frequency of 50 Hz, the peak voltage is 10 volts, the period is 0.02 seconds, and the frequency is 50 Hz. The waveform follows a sinusoidal pattern, oscillating smoothly between the maximum positive and negative voltages.

Conclusion

Mastering circuit analysis in FE electrical exam is a fundamental step toward becoming a proficient electrical engineer. It equips you with the tools and knowledge needed to tackle complex electrical systems and unleash your potential in the field.

Did you find the techniques and mathematical approach used in the above guide complex? Don’t worry; check our detailed guide on Algebra and Trigonometry in the FE Electrical exam to crack all the complex algebraic and trigonometric concepts. It will help you focus on core circuit analysis problems without getting distracted by the why and how of complex calculations.

If you want to solidify your understanding of circuit analysis and advance your engineering career, look no further than Study for FE – your go-to place for specialized FE electrical exam preparation courses led by PE-licensed professionals. Our comprehensive curriculum and expert guidance will help you excel in the FE exam, enhance your problem-solving skills, and boost your career prospects. Don’t miss this opportunity to gain the confidence and competence you need to succeed.

Take the leap and visit Study for FE today to explore their courses tailored to your needs. Your future as a knowledgeable and skilled electrical engineer awaits.

Wasim Asghar – P.E, P.ENG, M.ENG

Licensed Professional Engineer in Texas (PE), Florida (PE) and Ontario (P. Eng) with consulting experience in design, commissioning and plant engineering for clients in Energy, Mining and Infrastructure.

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