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.

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.

Tip: 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.

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.

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.

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.

Tips for Using 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:

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:

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.

Next Steps

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.

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.