The FE Electrical and Computer Engineering exam tests your knowledge across a broad range of topics—from circuit analysis and power systems to signal processing and engineering economics. The best way to prepare is to work through realistic problems under exam-like conditions.

FE Electrical Exam at a Glance

Questions110 multiple-choice
Time5 hours 20 minutes (~2.9 min/question)
FormatComputer-based at Pearson VUE centers, year-round
Topic areas17 knowledge areas
ReferenceNCEES FE Reference Handbook (provided digitally)
Highest-weight topicsDigital Systems, Electronics, Circuit Analysis, Power Systems

Below you will find 10 practice problems that mirror the style, difficulty, and topic distribution you will encounter on the actual NCEES FE exam. Each problem includes four answer choices and a detailed, step-by-step solution. Use these to identify your strengths, target your weaknesses, and build the problem-solving speed you need on test day.

How to use these problems: Try solving each problem on your own before revealing the solution. Time yourself—you should average about 3 minutes per problem on the real exam. Use only the FE Reference Handbook and an NCEES-approved calculator.

Problem 1: Mathematics — Solving a Quadratic Equation

Find all real solutions to the equation 2x² − 5x − 3 = 0.

  • A) x = 3 and x = −0.5
  • B) x = 3 and x = 0.5
  • C) x = −3 and x = 0.5
  • D) x = −3 and x = −0.5
Show Solution

Answer: A

Apply the quadratic formula with a = 2, b = −5, c = −3:

x = (−b ± √(b² − 4ac)) / (2a)

Step 1: Compute the discriminant: b² − 4ac = (−5)² − 4(2)(−3) = 25 + 24 = 49.

Step 2: Take the square root: √49 = 7.

Step 3: Solve for both roots:

x = (5 + 7) / 4 = 12 / 4 = 3

x = (5 − 7) / 4 = −2 / 4 = −0.5

The solutions are x = 3 and x = −0.5.

Problem 2: Circuit Analysis — KVL and Ohm's Law

A series circuit contains a 12 V DC source, a 4 Ω resistor (R1), and a 6 Ω resistor (R2). What is the voltage drop across R2?

  • A) 4.8 V
  • B) 7.2 V
  • C) 6.0 V
  • D) 5.0 V
Show Solution

Answer: B

Step 1: Find the total resistance: R_total = R1 + R2 = 4 + 6 = 10 Ω.

Step 2: Find the current using Ohm's law: I = V / R_total = 12 / 10 = 1.2 A.

Step 3: Find the voltage drop across R2: V_R2 = I × R2 = 1.2 × 6 = 7.2 V.

You can verify with KVL: V_R1 + V_R2 = (1.2)(4) + (1.2)(6) = 4.8 + 7.2 = 12 V. ✓

Problem 3: Power Systems — Three-Phase Power Calculation

A balanced three-phase, wye-connected load draws a line current of 10 A from a 480 V (line-to-line) source at a power factor of 0.85 lagging. What is the total real power consumed by the load?

  • A) 4,080 W
  • B) 7,067 W
  • C) 8,314 W
  • D) 4,800 W
Show Solution

Answer: B

For a balanced three-phase system, total real power is given by:

P = √3 × V_LL × I_L × pf

Step 1: Substitute the given values:

P = √3 × 480 × 10 × 0.85

Step 2: Compute √3 × 480 = 1.732 × 480 = 831.4.

Step 3: Multiply: 831.4 × 10 × 0.85 = 7,067 W (approximately 7.07 kW).

Problem 4: Electronics — Inverting Op-Amp Gain

An inverting op-amp circuit has a feedback resistor R_f = 100 kΩ and an input resistor R_in = 10 kΩ. If the input voltage is V_in = 0.5 V, what is the output voltage? Assume the op-amp is ideal.

  • A) +5.0 V
  • B) −5.0 V
  • C) +0.05 V
  • D) −50.0 V
Show Solution

Answer: B

The closed-loop gain of an inverting op-amp is:

A_v = −R_f / R_in

Step 1: Calculate the gain: A_v = −100k / 10k = −10.

Step 2: Calculate the output: V_out = A_v × V_in = (−10)(0.5) = −5.0 V.

The negative sign confirms the inverting configuration—the output is 180° out of phase with the input.

Problem 5: Digital Systems — Boolean Algebra Simplification

Simplify the Boolean expression: F = A·B + A·B' + A'·B

  • A) F = A + B
  • B) F = A · B
  • C) F = A ⊕ B
  • D) F = A' + B'
Show Solution

Answer: A

Step 1: Group the first two terms, factoring out A:

A·B + A·B' = A·(B + B') = A·1 = A

Step 2: Substitute back into the original expression:

F = A + A'·B

Step 3: Apply the absorption theorem (X + X'Y = X + Y):

F = A + B

You can verify by checking all four input combinations in a truth table—F is 0 only when both A and B are 0.

Tip for digital systems problems: Many Boolean algebra and logic gate questions can be solved faster by building a quick truth table than by algebraic manipulation. If the expression has 2–3 variables, a truth table takes under a minute and eliminates algebraic mistakes.

Problem 6: Signal Processing — Nyquist Sampling Theorem

An analog signal contains frequency components up to 4 kHz. According to the Nyquist-Shannon sampling theorem, what is the minimum sampling rate required to perfectly reconstruct the signal?

  • A) 2 kHz
  • B) 4 kHz
  • C) 8 kHz
  • D) 16 kHz
Show Solution

Answer: C

The Nyquist-Shannon sampling theorem states that to avoid aliasing, the sampling frequency must be at least twice the maximum frequency component of the signal:

f_s ≥ 2 × f_max

Step 1: Identify f_max = 4 kHz.

Step 2: Calculate the minimum sampling rate: f_s = 2 × 4 kHz = 8 kHz.

Any sampling rate below 8 kHz would cause aliasing, where higher-frequency components fold back into the lower-frequency range and corrupt the reconstructed signal.

Problem 7: Electromagnetics — Parallel-Plate Capacitance

A parallel-plate capacitor has plates of area A = 0.02 m², a separation of d = 1 mm, and is filled with a dielectric of relative permittivity ε_r = 4. What is the capacitance? (Use ε_0 = 8.854 × 10&supmin;¹² F/m.)

  • A) 177 pF
  • B) 708 pF
  • C) 1.77 nF
  • D) 354 pF
Show Solution

Answer: B

The capacitance of a parallel-plate capacitor is:

C = ε_0 × ε_r × A / d

Step 1: Substitute the values:

C = (8.854 × 10&supmin;¹²)(4)(0.02) / (0.001)

Step 2: Compute the numerator: 8.854 × 10&supmin;¹² × 4 × 0.02 = 7.083 × 10&supmin;¹³.

Step 3: Divide by d: 7.083 × 10&supmin;¹³ / 10&supmin;³ = 7.083 × 10&supmin;¹° F.

Step 4: Convert: 7.083 × 10&supmin;¹° F ≈ 708 pF.

Problem 8: Control Systems — Transfer Function Stability

A system has the open-loop transfer function G(s) = 10 / (s² + 5s + 6). What are the poles of this system, and is it stable?

  • A) Poles at s = −2, −3; stable
  • B) Poles at s = 2, 3; unstable
  • C) Poles at s = −1, −6; stable
  • D) Poles at s = −2, 3; unstable
Show Solution

Answer: A

The poles are the roots of the denominator polynomial s² + 5s + 6 = 0.

Step 1: Factor the denominator: s² + 5s + 6 = (s + 2)(s + 3).

Step 2: Set each factor to zero: s = −2 and s = −3.

Step 3: Check stability: A linear time-invariant (LTI) system is stable if and only if all poles have negative real parts. Both −2 and −3 are negative, so the system is stable.

On the s-plane, both poles lie in the left-half plane (LHP), confirming BIBO stability.

Tip for control systems problems: When asked about stability, check the poles first. Factor the denominator polynomial (or use the quadratic formula), then check if all real parts are negative. For the FE exam, most stability problems can be solved by factoring alone—Routh-Hurwitz is only needed for higher-order systems.

Problem 9: Probability — Independent Events

A circuit board undergoes two independent quality tests. The probability of passing Test 1 is 0.95, and the probability of passing Test 2 is 0.90. What is the probability that a randomly selected board passes both tests?

  • A) 0.855
  • B) 0.925
  • C) 0.950
  • D) 0.810
Show Solution

Answer: A

For two independent events, the probability that both occur is the product of their individual probabilities:

P(A ∩ B) = P(A) × P(B)

Step 1: Identify the probabilities: P(Test 1) = 0.95, P(Test 2) = 0.90.

Step 2: Multiply: P(both) = 0.95 × 0.90 = 0.855.

There is an 85.5% chance that a board passes both tests. Note that this is lower than either individual probability—adding more independent tests always reduces the overall pass rate.

Problem 10: Engineering Economics — Present Worth Analysis

A piece of equipment costs $10,000 today and generates annual savings of $3,000 for 5 years. If the interest rate is 8% per year, what is the net present worth (NPW) of this investment?

  • A) $1,978
  • B) $5,000
  • C) $2,485
  • D) −$1,978
Show Solution

Answer: A

The net present worth is calculated as:

NPW = −Initial Cost + Annual Savings × (P/A, i, n)

where (P/A, i, n) is the present worth of an annuity factor.

Step 1: Compute the P/A factor:

(P/A, 8%, 5) = [(1 + 0.08)&sup5; − 1] / [0.08 × (1 + 0.08)&sup5;]

Step 2: Calculate (1.08)&sup5; = 1.4693.

Step 3: Substitute: (P/A) = (1.4693 − 1) / (0.08 × 1.4693) = 0.4693 / 0.11755 = 3.9927.

Step 4: Compute NPW: NPW = −$10,000 + $3,000 × 3.9927 = −$10,000 + $11,978 = $1,978.

Since NPW > 0, the investment is economically justified at an 8% interest rate.


How Should You Use These Practice Problems?

Working through practice problems is one of the highest-value activities in your FE exam preparation, but how you practice matters just as much as how much you practice. Here are some tips to get the most out of these problems:

  • Simulate exam conditions. Set a timer for 3 minutes per problem. The real FE exam gives you about 5 hours and 20 minutes for 110 questions, so building speed is critical.
  • Use the NCEES FE Reference Handbook. During the actual exam, you will have access to a searchable PDF of the handbook. Practice finding formulas in it rather than memorizing everything.
  • Review your mistakes carefully. When you get a problem wrong, identify whether the error was conceptual, computational, or due to a misread. Each type requires a different fix.
  • Track your performance by topic. If you consistently miss power systems or control systems questions, allocate more study time there.
  • Don't just read solutions—rework them. After reading a step-by-step solution, close it, and solve the problem again from scratch to reinforce the method.

What Topics Does the FE Electrical and Computer Exam Cover?

The FE Electrical and Computer Engineering exam, administered by NCEES, covers the following major topic areas:

  • Mathematics (11%)
  • Probability and Statistics (4%)
  • Ethics and Professional Practice (4%)
  • Engineering Economics (4%)
  • Properties of Electrical Materials (4%)
  • Engineering Sciences (4%)
  • Circuit Analysis (DC and AC) (12%)
  • Linear Systems (5%)
  • Signal Processing (6%)
  • Electronics (8%)
  • Power Systems (8%)
  • Electromagnetics (5%)
  • Control Systems (6%)
  • Communications (5%)
  • Computer Networks (5%)
  • Digital Systems (9%)

The 10 problems above touch on many of these high-weight categories. To fully prepare, you need to practice across all of them—especially circuit analysis, electronics, digital systems, and mathematics, which together account for roughly 40% of the exam.

Don't skip engineering economics: Present worth, annual cost, and rate-of-return problems are among the most formulaic on the entire FE exam. The P/A, P/F, and A/P factor equations are in the reference handbook—learn to find them quickly and these become nearly free points.

Essential Study Tools

Having the right tools makes a significant difference in your practice sessions:

What Should You Do Next?

These 10 problems give you a solid starting point, but the FE exam covers far more ground. To build true exam readiness, you need hundreds of problems across every topic, with instant feedback and detailed explanations for every answer choice.

Continue your FE Electrical preparation:

FE Electrical Study GuideHow to Pass the FE Electrical ExamBest FE Exam Prep Books🔢 Calculator Guide✅ Exam Day ChecklistGuide for Returning Engineers

Frequently Asked Questions

How many practice problems should I do for the FE Electrical exam?

Aim for 300–500 practice problems across all 17 topics. Prioritize Circuit Analysis, Electronics, Signal Processing, and Digital Systems, which carry the most exam weight. Practice under timed conditions (about 3 minutes per problem) to build exam pacing.

What types of questions are on the FE Electrical exam?

The FE Electrical exam includes multiple-choice, fill-in-the-blank, drag-and-drop, and select-all-that-apply question types. Most questions are calculation-based, requiring you to apply circuit analysis, signal processing, or power system formulas from the reference handbook.

Disclaimer: This content is for educational purposes only and is not affiliated with, endorsed by, or sponsored by NCEES. "FE" and "Fundamentals of Engineering" are trademarks of NCEES. Always refer to the official NCEES website for the most current exam specifications and policies.