FE Electrical and Computer Exam Prep: Free Plan + 34 Questions

Start your FE Electrical and Computer exam prep with the question below, then work through the 34-question starter set and 10-week plan. These original, unofficial questions cover two examples per NCEES knowledge area—not a full-length exam or a passing prediction.

Round 1: one question per topic

Q1 · Mathematics: Complex numbers

Evaluate (3 + j4)(1 − j2), where j² = −1.

A. −5 − j2
B. 11 − j2
C. 3 − j8
D. 11 + j10

Show answer and explanation

Answer: B: 11 − j2. Multiply every term: 3 − j6 + j4 − j²8. Because j² = −1, the last term becomes +8. Collect terms: (3 + 8) + (−6 + 4)j = 11 − j2.

Why the others miss: A treats j² as +1, giving 3 − 8 = −5. C omits the cross terms and does not handle the j² product correctly. D flips the sign of the −j6 cross term.

Review target: Write all four products before you combine anything. Check the sign of the j² term.

Source: OpenStax, College Algebra 2e, §2.4 Complex Numbers, “Multiplying Complex Numbers”

Answer: B: 11 − j2. Multiply every term: 3 − j6 + j4 − j²8. Because j² = −1, the last term becomes +8. Collect terms: (3 + 8) + (−6 + 4)j = 11 − j2.

Why the others miss: A treats j² as +1, giving 3 − 8 = −5. C omits the cross terms and does not handle the j² product correctly. D flips the sign of the −j6 cross term.

Review target: Write all four products before you combine anything. Check the sign of the j² term.

Source: OpenStax, College Algebra 2e, §2.4 Complex Numbers, “Multiplying Complex Numbers”

Q2 · Probability and Statistics: Normal distribution

A resistor's value is normally distributed. About what fraction of resistors fall within ±1 standard deviation of the mean?

A. 50%
B. 68%
C. 95%
D. 99.7%

Show answer and explanation

Answer: B: 68%. For any normal distribution, about 68% of values lie within one standard deviation of the mean.

Why the others miss: 95% is the ±2σ band and 99.7% is the ±3σ band. 50% corresponds to roughly ±0.67σ.

Review target: Know the 68–95–99.7 pattern cold. For other z values, practice reading the unit normal table in the handbook instead of guessing.

Source: OpenStax, Introductory Statistics 2e, §6.1, “The Empirical Rule”

Answer: B: 68%. For any normal distribution, about 68% of values lie within one standard deviation of the mean.

Why the others miss: 95% is the ±2σ band and 99.7% is the ±3σ band. 50% corresponds to roughly ±0.67σ.

Review target: Know the 68–95–99.7 pattern cold. For other z values, practice reading the unit normal table in the handbook instead of guessing.

Source: OpenStax, Introductory Statistics 2e, §6.1, “The Empirical Rule”

Q3 · Ethics and Professional Practice: Codes of ethics

A licensed engineer warns that a utility's protective-relay settings leave a public building at risk of fire. Her manager overrules her and tells her to drop it. Under the NCEES Model Rules, what must she do?

A. Comply, because responsibility has passed to the manager
B. Notify her employer or client and any other authority that is appropriate
C. Post her concerns on social media before telling anyone else
D. Resign quietly without reporting the issue

Show answer and explanation

Answer: B: Notify her employer or client and any other authority that is appropriate. The Model Rules say licensees must notify their employer or client, and other authority as appropriate, when their professional judgment is overruled in a situation that endangers public health, safety, or welfare. The NSPE Code of Ethics contains a matching duty.

Why the others miss: A abandons the public-safety duty. C skips the notification route the rules describe. D removes her from the job but leaves the hazard unreported.

Review target: On ethics items, find the specific duty the facts trigger. Don't pick the answer that's merely convenient or dramatic.

Source: NCEES Model Rules (2026), §240.15 A.3, printed p. 11 (PDF p. 15); NSPE Code of Ethics for Engineers, Rules of Practice II.1.a

Why the others miss: A abandons the public-safety duty. C skips the notification route the rules describe. D removes her from the job but leaves the hazard unreported.

Review target: On ethics items, find the specific duty the facts trigger. Don't pick the answer that's merely convenient or dramatic.

Source: NCEES Model Rules (2026), §240.15 A.3, printed p. 11 (PDF p. 15); NSPE Code of Ethics for Engineers, Rules of Practice II.1.a

Q4 · Engineering Economics: Time value of money

What amount invested today at 6% compounded annually grows to $10,000 in 5 years?

A. $7,473
B. $7,692
C. $8,000
D. $13,382

Show answer and explanation

Answer: A: $7,473. Present value is P = F/(1 + i)ⁿ = 10,000/1.06⁵ = $7,472.58, which rounds to $7,473.

Why the others miss: B uses simple interest: 10,000/1.30. D is the future value of $10,000 after 5 years, so it moves money the wrong way in time. C is just a round number.

Review target: Sketch a cash-flow line with "now" and "year 5" marked before you choose a factor.

Source: OpenStax, Contemporary Mathematics, §6.4 Compound Interest, “Understand and Compute Present Value”

Answer: A: $7,473. Present value is P = F/(1 + i)ⁿ = 10,000/1.06⁵ = $7,472.58, which rounds to $7,473.

Why the others miss: B uses simple interest: 10,000/1.30. D is the future value of $10,000 after 5 years, so it moves money the wrong way in time. C is just a round number.

Review target: Sketch a cash-flow line with "now" and "year 5" marked before you choose a factor.

Source: OpenStax, Contemporary Mathematics, §6.4 Compound Interest, “Understand and Compute Present Value”

Q5 · Properties of Electrical Materials: Resistivity and geometry

A copper conductor is 100 m long with a 1 mm² cross-section. Take copper's resistivity as 1.68 × 10⁻⁸ Ω·m. Its resistance is closest to:

A. 1.68 mΩ
B. 0.168 Ω
C. 1.68 Ω
D. 16.8 Ω

Show answer and explanation

Answer: C: 1.68 Ω. R = ρL/A. Convert the area first: 1 mm² = 1 × 10⁻⁶ m². Then R = (1.68 × 10⁻⁸)(100)/(10⁻⁶) = 1.68 Ω.

Why the others miss: The other choices are powers-of-ten errors that come from skipping or botching the mm² → m² conversion.

Review target: Convert every length and area to SI units before substituting.

Source: OpenStax, University Physics Vol. 2, §9.3, Eq. 9.9 and Table 9.1

Answer: C: 1.68 Ω. R = ρL/A. Convert the area first: 1 mm² = 1 × 10⁻⁶ m². Then R = (1.68 × 10⁻⁸)(100)/(10⁻⁶) = 1.68 Ω.

Why the others miss: The other choices are powers-of-ten errors that come from skipping or botching the mm² → m² conversion.

Review target: Convert every length and area to SI units before substituting.

Source: OpenStax, University Physics Vol. 2, §9.3, Eq. 9.9 and Table 9.1

Q6 · Circuit Analysis (DC and AC Steady State): Thevenin equivalent

A 12 V ideal voltage source has its negative terminal at b and its positive terminal connected through a 4 Ω resistor to a. A 12 Ω resistor is connected directly across a–b and remains part of the network. With no external load attached, what are V_th = V_a − V_b and R_th at a–b?

A. 3 V, 3 Ω
B. 9 V, 3 Ω
C. 9 V, 16 Ω
D. 12 V, 4 Ω

Show answer and explanation

Answer: B: 9 V, 3 Ω. V_th is the open-circuit voltage at a–b: a voltage divider gives 12 × 12/(4 + 12) = 9 V. R_th is the resistance seen from a–b with the source turned off, which replaces it with a short. That puts 4 Ω in parallel with 12 Ω: 4 × 12/16 = 3 Ω.

Why the others miss: A puts the wrong resistor in the divider numerator. C adds the resistors in series instead of in parallel. D ignores the 12 Ω path entirely.

Review target: Find V_th and R_th as two separate steps. For R_th, deactivate the independent sources first.

Source: MIT OCW 6.002, Lecture 3: Superposition, Thévenin and Norton, PDF pp. 11 and 16: zeroed independent sources and the port equivalent

Why the others miss: A puts the wrong resistor in the divider numerator. C adds the resistors in series instead of in parallel. D ignores the 12 Ω path entirely.

Review target: Find V_th and R_th as two separate steps. For R_th, deactivate the independent sources first.

Source: MIT OCW 6.002, Lecture 3: Superposition, Thévenin and Norton, PDF pp. 11 and 16: zeroed independent sources and the port equivalent

Q7 · Linear Systems: First-order transient response

A resistor and an initially uncharged capacitor are connected in series across an ideal 5 V step source. What is the capacitor voltage after exactly one time constant?

A. 1.84 V
B. 2.50 V
C. 3.16 V
D. 5.00 V

Show answer and explanation

Answer: C: 3.16 V. Charging follows v(t) = 5(1 − e^(−t/τ)). At t = τ the voltage is 5(1 − e⁻¹) = 3.16 V, or 63.2% of the final value.

Why the others miss: A is the value for a discharging capacitor (36.8% remaining). B assumes the capacitor charges in a straight line. D is the limiting final value, approached asymptotically. After five time constants the voltage is about 4.97 V, not exactly 5.00 V.

Review target: Name the initial value, the final value, and τ before you write any exponential.

Source: OpenStax, University Physics Vol. 2, §10.5, Eq. 10.8 and the time-constant discussion

Answer: C: 3.16 V. Charging follows v(t) = 5(1 − e^(−t/τ)). At t = τ the voltage is 5(1 − e⁻¹) = 3.16 V, or 63.2% of the final value.

Review target: Name the initial value, the final value, and τ before you write any exponential.

Source: OpenStax, University Physics Vol. 2, §10.5, Eq. 10.8 and the time-constant discussion

Q8 · Signal Processing: Sampling and aliasing

A 7 kHz sinusoid is sampled at 10 kHz with no anti-aliasing filter. What is its alias frequency in the baseband from 0 to 5 kHz?

A. 3 kHz
B. 5 kHz
C. 7 kHz
D. 17 kHz

Show answer and explanation

Answer: A: 3 kHz. The Nyquist frequency is f_s/2 = 5 kHz. A 7 kHz tone sits above it, so it folds back to an alias at |f_s − f| = |10 − 7| = 3 kHz.

Why the others miss: B is the Nyquist frequency itself, not an alias. C is outside the stated 0–5 kHz baseband. D is an image above the sampling rate, not what shows up in the baseband.

Review target: For baseband sampling, compare the signal frequencies with f_s/2 before you trust a sampled result.

Source: Analog Devices, MT-002: What the Nyquist Criterion Means, pp. 4–5, Figs. 4 and 5A

Answer: A: 3 kHz. The Nyquist frequency is f_s/2 = 5 kHz. A 7 kHz tone sits above it, so it folds back to an alias at |f_s − f| = |10 − 7| = 3 kHz.

Why the others miss: B is the Nyquist frequency itself, not an alias. C is outside the stated 0–5 kHz baseband. D is an image above the sampling rate, not what shows up in the baseband.

Review target: For baseband sampling, compare the signal frequencies with f_s/2 before you trust a sampled result.

Source: Analog Devices, MT-002: What the Nyquist Criterion Means, pp. 4–5, Figs. 4 and 5A

Q9 · Electronics: Operational amplifiers

An ideal inverting op-amp amplifier has R_f = 47 kΩ and R_in = 4.7 kΩ. Its non-inverting input is grounded, and its supply rails allow linear operation at the calculated output. With a 0.3 V DC input, the output is:

A. −3.0 V
B. +3.0 V
C. +3.3 V
D. −0.03 V

Show answer and explanation

Answer: A: −3.0 V. Inverting gain is −R_f/R_in = −47/4.7 = −10. The output is −10 × 0.3 V = −3.0 V. This assumes the supply rails leave room for a −3 V output.

Why the others miss: B drops the inversion. C applies the non-inverting gain, 1 + R_f/R_in = 11. D inverts the resistor ratio.

Review target: Identify the configuration (inverting or non-inverting) before you pick a gain formula.

Source: Analog Devices, MT-032: Ideal Voltage Feedback Op Amp, p. 3, Eq. 1 and Fig. 3

Answer: A: −3.0 V. Inverting gain is −R_f/R_in = −47/4.7 = −10. The output is −10 × 0.3 V = −3.0 V. This assumes the supply rails leave room for a −3 V output.

Why the others miss: B drops the inversion. C applies the non-inverting gain, 1 + R_f/R_in = 11. D inverts the resistor ratio.

Review target: Identify the configuration (inverting or non-inverting) before you pick a gain formula.

Source: Analog Devices, MT-032: Ideal Voltage Feedback Op Amp, p. 3, Eq. 1 and Fig. 3

Q10 · Power Systems: Three-phase power

A balanced three-phase load draws 50 A rms line current at 480 V rms line-to-line under sinusoidal steady-state conditions with a 0.85 lagging power factor. Enter its real power in kW, to the nearest 0.1.

Your answer: write the number in kW before opening the explanation.

Show answer and explanation

Answer: 35.3 kW. For balanced three-phase with line quantities, P = √3 · V_L · I_L · pf = √3 × 480 × 50 × 0.85 ≈ 35,334 W, or about 35.3 kW.

Common errors: If you entered 41.6, you found apparent power (kVA) and forgot the power factor. 61.2 uses 3 instead of √3 with line values. 20.4 is the single-phase formula.

Review target: Label each quantity line or phase, and real (W) or apparent (VA), before you substitute.

Source: Rose-Hulman, Electromechanical Devices lecture notes 7, p. 7-14

Answer: 35.3 kW. For balanced three-phase with line quantities, P = √3 · V_L · I_L · pf = √3 × 480 × 50 × 0.85 ≈ 35,334 W, or about 35.3 kW.

Common errors: If you entered 41.6, you found apparent power (kVA) and forgot the power factor. 61.2 uses 3 instead of √3 with line values. 20.4 is the single-phase formula.

Review target: Label each quantity line or phase, and real (W) or apparent (VA), before you substitute.

Source: Rose-Hulman, Electromechanical Devices lecture notes 7, p. 7-14

Q11 · Electromagnetics: Wave propagation

A 300 MHz wave travels through a lossless, nonmagnetic dielectric with εr = 4. Use c = 3 × 10⁸ m/s. Its wavelength is:

A. 0.25 m
B. 0.5 m
C. 1.0 m
D. 2.0 m

Show answer and explanation

Answer: B: 0.5 m. In a nonmagnetic dielectric, the wave speed is v = c/√εr = (3 × 10⁸)/2 = 1.5 × 10⁸ m/s. Wavelength is λ = v/f = 1.5 × 10⁸/(3 × 10⁸) = 0.5 m.

Why the others miss: C is the free-space wavelength. A divides by εr instead of √εr. D multiplies where it should divide.

Review target: Ask "what medium is this in?" before you use c.

Source: Ellingson, Electromagnetics I, §9.2: Lossless wave equations, β = ω√(με); v = ω/β and λ = 2π/β are derived from the phase constant

Answer: B: 0.5 m. In a nonmagnetic dielectric, the wave speed is v = c/√εr = (3 × 10⁸)/2 = 1.5 × 10⁸ m/s. Wavelength is λ = v/f = 1.5 × 10⁸/(3 × 10⁸) = 0.5 m.

Why the others miss: C is the free-space wavelength. A divides by εr instead of √εr. D multiplies where it should divide.

Review target: Ask "what medium is this in?" before you use c.

Source: Ellingson, Electromagnetics I, §9.2: Lossless wave equations, β = ω√(με); v = ω/β and λ = 2π/β are derived from the phase constant

Q12 · Control Systems: Steady-state error

A negative-unity-feedback system has open-loop transfer function G(s) = 10/(s + 2). What is its steady-state error to a unit step input?

A. 0
B. 0.091
C. 0.167
D. 0.2

Show answer and explanation

Answer: C: 0.167. This is a type 0 system (no free integrator). The position constant is K_p = G(0) = 10/2 = 5, so e_ss = 1/(1 + K_p) = 1/6 ≈ 0.167. The closed-loop pole is at s = −12, so the final value exists.

Why the others miss: A is the zero step error expected of a stable type 1 unity-feedback system, not this type 0 system. B uses K_p = 10, which skips evaluating G(s) at s = 0. D is 1/K_p, which drops the 1 in the denominator.

Review target: Find the system type first, then evaluate the matching error constant.

Source: MIT OCW 2.004, Lecture 16: Steady-state error, PDF pp. 7–8: negative unity feedback and static error constants

Review target: Find the system type first, then evaluate the matching error constant.

Source: MIT OCW 2.004, Lecture 16: Steady-state error, PDF pp. 7–8: negative unity feedback and static error constants

Q13 · Communications: Modulation bandwidth

A low-pass message occupying frequencies up to 5 kHz modulates a 1 MHz carrier using double-sideband suppressed-carrier (DSB-SC) AM. What is the transmitted bandwidth?

A. 5 kHz
B. 10 kHz
C. 20 kHz
D. 1 MHz

Show answer and explanation

Answer: B: 10 kHz. Double-sideband modulation places a copy of the message on each side of the carrier, so the bandwidth is 2 × 5 kHz = 10 kHz.

Why the others miss: A is single-sideband bandwidth. C double-counts the sidebands. D is the carrier frequency, which isn't a bandwidth at all.

Review target: Separate where the signal sits (the carrier) from how wide it is (the bandwidth).

Source: MIT OCW Signals and Systems, Lecture 13: Continuous-Time Modulation, p. 13-7 (PDF p. 7), Transparency 13.9: translated sidebands

Answer: B: 10 kHz. Double-sideband modulation places a copy of the message on each side of the carrier, so the bandwidth is 2 × 5 kHz = 10 kHz.

Why the others miss: A is single-sideband bandwidth. C double-counts the sidebands. D is the carrier frequency, which isn't a bandwidth at all.

Review target: Separate where the signal sits (the carrier) from how wide it is (the bandwidth).

Source: MIT OCW Signals and Systems, Lecture 13: Continuous-Time Modulation, p. 13-7 (PDF p. 7), Transparency 13.9: translated sidebands

Q14 · Computer Networks: Network models

In the OSI model, IP addressing and routing between networks happen at which layer?

A. Layer 2, data link
B. Layer 3, network
C. Layer 4, transport
D. Layer 7, application

Show answer and explanation

Answer: B: Layer 3, network. The network layer chooses paths between networks using IP addresses. That job is routing.

Why the others miss: Layer 2 handles framing and link-layer addressing, such as MAC addresses on Ethernet. Layer 4 handles end-to-end delivery with ports (TCP and UDP). Layer 7 is where application protocols live.

Review target: Tie each function to its layer: link-layer frames and addresses, network-layer packets and IP addresses, and transport-layer ports for TCP and UDP.

Source: ITU-T X.200 / ISO/IEC 7498-1: OSI Basic Reference Model, §6.1.2, printed p. 28; §§7.5.3.2 and 7.5.4.2, printed pp. 42 and 44; RFC 791: Internet Protocol, §1.4 Operation: IP addressing and routing

Answer: B: Layer 3, network. The network layer chooses paths between networks using IP addresses. That job is routing.

Review target: Tie each function to its layer: link-layer frames and addresses, network-layer packets and IP addresses, and transport-layer ports for TCP and UDP.

Q15 · Digital Systems: Boolean simplification

Simplify F = A + A′B.

A. A
B. B
C. A + B
D. AB

Show answer and explanation

Answer: C: A + B. Use the identity A + A′B = A + B. Algebraically, A + A′B = (A + A′)(A + B) = 1 · (A + B) = A + B.

Why the others miss: Test each wrong choice against a truth-table row. With A = 0 and B = 1, F = 1, so the answer can't be A alone. With A = 1 and B = 0, F = 1, so it can't be B or AB.

Review target: Check any simplification against at least the rows where the candidate answers disagree.

Source: University of Toronto CSC 258: Proofs of absorption rules, Proof of x + x′y = x + y

Answer: C: A + B. Use the identity A + A′B = A + B. Algebraically, A + A′B = (A + A′)(A + B) = 1 · (A + B) = A + B.

Why the others miss: Test each wrong choice against a truth-table row. With A = 0 and B = 1, F = 1, so the answer can't be A alone. With A = 1 and B = 0, F = 1, so it can't be B or AB.

Review target: Check any simplification against at least the rows where the candidate answers disagree.

Source: University of Toronto CSC 258: Proofs of absorption rules, Proof of x + x′y = x + y

Q16 · Computer Systems: Memory addressing

A byte-addressable processor has a 16-bit address bus. Each address selects one byte. What is the maximum directly addressable memory?

A. 16 bytes
B. 32,768 bytes
C. 65,536 bytes
D. 1,048,576 bytes

Show answer and explanation

Answer: C: 65,536 bytes. Sixteen address lines select 2¹⁶ = 65,536 locations. At one byte per location, that's 65,536 bytes, or 64 KiB.

Why the others miss: B is 2¹⁵, one address line short. D is 2²⁰, which needs a 20-bit bus. A confuses the bus width with capacity.

Review target: Write the chain: address lines → locations → bytes per location → total bytes.

Source: NPTEL, Computer Organization, Module 1 Lecture 1, slide 13

Answer: C: 65,536 bytes. Sixteen address lines select 2¹⁶ = 65,536 locations. At one byte per location, that's 65,536 bytes, or 64 KiB.

Why the others miss: B is 2¹⁵, one address line short. D is 2²⁰, which needs a 20-bit bus. A confuses the bus width with capacity.

Review target: Write the chain: address lines → locations → bytes per location → total bytes.

Source: NPTEL, Computer Organization, Module 1 Lecture 1, slide 13

Q17 · Software Engineering: Algorithm complexity

Which choice is the tightest asymptotic bound on the worst-case time complexity of binary search on a sorted array of n elements?

A. O(1)
B. O(log n)
C. O(n)
D. O(n log n)

Show answer and explanation

Answer: B: O(log n). Each comparison halves the remaining search range, so the worst case takes about log₂ n + 1 comparisons. That's O(log n).

Why the others miss: O(n) and O(n log n) are looser upper bounds, not the tightest choice. O(1) would mean a lookup that doesn't depend on n. If the array isn't sorted, binary search doesn't apply at all.

Review target: Note the precondition (here, sorted data) along with the growth rate.

Source: Python documentation: bisect — Array bisection algorithm, insort_left/insort_right and Performance Notes distinguish O(log n) search from O(n) insertion

Answer: B: O(log n). Each comparison halves the remaining search range, so the worst case takes about log₂ n + 1 comparisons. That's O(log n).

Review target: Note the precondition (here, sorted data) along with the growth rate.

Source: Python documentation: bisect — Array bisection algorithm, insort_left/insort_right and Performance Notes distinguish O(log n) search from O(n) insertion

Round 2: another pass through the topics

Q18 · Mathematics: Ordinary differential equations

Solve dy/dt = −4y with y(0) = 5. What is y(0.25)?

A. 0.37
B. 1.84
C. 3.16
D. 3.89

Show answer and explanation

Answer: B: 1.84. The solution is y = 5e^(−4t). At t = 0.25, y = 5e⁻¹ = 1.84.

Why the others miss: A forgets the initial value (e⁻¹ alone). C is 5(1 − e⁻¹), a charging curve. D uses e^(−0.25) and drops the 4.

Review target: Write the general solution, apply the initial condition, and only then plug in t.

Source: OpenStax, Calculus Vol. 2, §2.8, Rule: Exponential Decay Model

Q19 · Probability and Statistics: Binomial distribution

Each chip in a lot is defective with probability 0.1, independently of the others. In a sample of 4 chips, what is the probability that exactly one is defective?

A. 0.0729
B. 0.1000
C. 0.2916
D. 0.3439

Show answer and explanation

Answer: C: 0.2916. Use the binomial formula: C(4,1)(0.1)¹(0.9)³ = 4 × 0.1 × 0.729 = 0.2916.

Why the others miss: A leaves out C(4,1), the four positions the defective chip could occupy. D is P(at least one) = 1 − 0.9⁴. B is just p.

Review target: Read whether the stem asks for "exactly," "at least," or "at most" before you calculate.

Source: OpenStax, Introductory Statistics 2e, §4.3, Binomial distribution, near Example 4.11

Answer: C: 0.2916. Use the binomial formula: C(4,1)(0.1)¹(0.9)³ = 4 × 0.1 × 0.729 = 0.2916.

Why the others miss: A leaves out C(4,1), the four positions the defective chip could occupy. D is P(at least one) = 1 − 0.9⁴. B is just p.

Review target: Read whether the stem asks for "exactly," "at least," or "at most" before you calculate.

Source: OpenStax, Introductory Statistics 2e, §4.3, Binomial distribution, near Example 4.11

Q20 · Ethics and Professional Practice: Intellectual property

A U.S. firm wants to protect a proprietary wafer-cleaning process indefinitely without publicly disclosing how it works. The process has economic value because it is not generally known or readily ascertainable through proper means, and the firm takes reasonable measures to keep it secret. Which protection fits best?

A. Patent
B. Trade secret
C. Copyright
D. Trademark

Show answer and explanation

Answer: B: Trade secret. Trade-secret protection has no set term. It lasts while the information continues to meet the secrecy, economic-value, and reasonable-protection requirements, and it doesn't require public disclosure. It does not prevent lawful independent discovery or reverse engineering.

Why the others miss: A patent is granted in exchange for publicly disclosing the invention, and it expires (for a utility patent, generally 20 years from filing). Copyright protects creative expression, not a process. A trademark protects brand identifiers.

Review target: Match the protection to what the owner actually wants: secrecy, duration, or exclusivity in exchange for disclosure.

Source: USPTO, Trade secret policy, Overview: economic value, nonpublic/not readily ascertainable information, reasonable secrecy efforts, and duration; USPTO, IP Basics (Midwest Regional Office), PDF pp. 9–10, 15, 19, 24–26: patent disclosure/terms, trademarks, copyright, trade secrets; USPTO: Trade Secrets Intellectual Property Toolkit, PDF pp. 2 and 6: qualifying information and lawful acquisition

Review target: Match the protection to what the owner actually wants: secrecy, duration, or exclusivity in exchange for disclosure.

Q21 · Engineering Economics: Present worth analysis

A power-factor-correction bank costs $12,000 now, produces net savings of $3,000 at the end of each year for 5 years, and has no salvage value. At an 8% annual discount rate, which statement is correct?

A. NPW ≈ +$3,000; accept
B. NPW ≈ +$11,978; accept
C. NPW ≈ −$22; essentially breakeven and slightly unfavorable at 8%
D. The rate of return is 25%

Show answer and explanation

Answer: C: NPW ≈ −$22; essentially breakeven and slightly unfavorable at 8%. The uniform-series present-worth factor is (P/A, 8%, 5) = [(1.08)⁵ − 1]/[0.08(1.08)⁵] = 3.9927. Present worth of the savings is 3,000 × 3.9927 = $11,978. Subtracting the $12,000 cost gives NPW ≈ −$22.

Why the others miss: A ignores the time value of money (15,000 − 12,000). B forgets to subtract the first cost. D divides annual savings by cost, which isn't a rate of return. The true rate of return here is just under 8%.

Review target: Bring every cash flow to the same point in time before you add or compare them.

Source: Penn State EME 460, Uniform Series Present-Worth Factor, Uniform Series Present-Worth Factor, Equation 1-5; first payment at end of year 1

Review target: Bring every cash flow to the same point in time before you add or compare them.

Source: Penn State EME 460, Uniform Series Present-Worth Factor, Uniform Series Present-Worth Factor, Equation 1-5; first payment at end of year 1

Q22 · Properties of Electrical Materials: Semiconductor doping

Doping pure silicon with arsenic, which has five valence electrons, produces what kind of material?

A. n-type
B. p-type
C. Intrinsic
D. An insulator

Show answer and explanation

Answer: A: n-type. Silicon has four valence electrons. Each arsenic atom brings a fifth one that's only loosely bound and can become a conduction electron. Electrons become the majority carriers, so the material is n-type.

Why the others miss: In silicon, p-type comes from a dopant with three valence electrons, such as boron, which creates holes. Intrinsic means undoped. Doping increases conductivity, so the material doesn't become an insulator.

Review target: For silicon, count valence electrons: five means donor, n-type. Three means acceptor, p-type.

Source: OpenStax, University Physics Vol. 3, §9.6, n-type doping

Review target: For silicon, count valence electrons: five means donor, n-type. Three means acceptor, p-type.

Source: OpenStax, University Physics Vol. 3, §9.6, n-type doping

Q23 · Circuit Analysis (DC and AC Steady State): Phasors and impedance

A 120 V rms source drives a load with impedance Z = 30 + j40 Ω. What are the current magnitude and the power factor?

A. 2.4 A, 0.6 lagging
B. 2.4 A, 0.6 leading
C. 1.71 A, 0.6 lagging
D. 4.0 A, 0.8 lagging

Show answer and explanation

Answer: A: 2.4 A, 0.6 lagging. |Z| = √(30² + 40²) = 50 Ω, so I = 120/50 = 2.4 A rms. The power factor is R/|Z| = 30/50 = 0.6. Positive reactance means the load is inductive, so the current lags.

Why the others miss: B gets the lead/lag direction backward. C adds R and X as if they were in phase (70 Ω). D uses only the resistance for current and incorrectly uses X/|Z| = 0.8 for the power factor.

Review target: Use phasor magnitude, never R + X. Then use the sign of X to decide leading or lagging.

Source: OpenStax, University Physics Vol. 2, §15.3, Eqs. 15.10–15.11; OpenStax, University Physics Vol. 2, §15.4, power factor cos φ = R/Z

Review target: Use phasor magnitude, never R + X. Then use the sign of X to decide leading or lagging.

Source: OpenStax, University Physics Vol. 2, §15.3, Eqs. 15.10–15.11; OpenStax, University Physics Vol. 2, §15.4, power factor cos φ = R/Z

Q24 · Linear Systems: Resonance

A series RLC circuit has L = 10 mH and C = 1 µF. Enter its resonant frequency in hertz, to the nearest whole number.

Your answer: write the number in Hz before opening the explanation.

Show answer and explanation

Answer: 1,592 Hz. f₀ = 1/(2π√(LC)) = 1/(2π√(10 × 10⁻³ × 1 × 10⁻⁶)) = 1/(2π × 10⁻⁴) ≈ 1,592 Hz.

Common errors: If you entered 10,000, you found ω₀ in rad/s and didn't divide by 2π.

Review target: Check whether the question wants ω (rad/s) or f (Hz).

Source: OpenStax, University Physics Vol. 2, §15.5, Eq. 15.17

Q25 · Signal Processing: Digital filters

A causal digital filter obeys y[n] = 0.5y[n−1] + x[n] and starts at rest. What is its impulse response at n = 3?

A. 0.125
B. 0.25
C. 0.5
D. 1.5

Show answer and explanation

Answer: A: 0.125. Feed in a unit impulse. h[0] = 1, h[1] = 0.5, h[2] = 0.25, and h[3] = 0.125. In general h[n] = 0.5ⁿ for n ≥ 0.

Why the others miss: B is h[2], one step early. C is h[1]. D adds h[0] and h[1], rather than evaluating h[3].

Review target: For a recursive filter, build a small table of n and h[n] instead of guessing a pattern.

Source: Steven W. Smith, The Scientist and Engineer’s Guide to DSP, Chapter 19, Recursive Filters: the recursive method and impulse response

Answer: A: 0.125. Feed in a unit impulse. h[0] = 1, h[1] = 0.5, h[2] = 0.25, and h[3] = 0.125. In general h[n] = 0.5ⁿ for n ≥ 0.

Why the others miss: B is h[2], one step early. C is h[1]. D adds h[0] and h[1], rather than evaluating h[3].

Review target: For a recursive filter, build a small table of n and h[n] instead of guessing a pattern.

Source: Steven W. Smith, The Scientist and Engineer’s Guide to DSP, Chapter 19, Recursive Filters: the recursive method and impulse response

Q26 · Electronics: Rectifiers

An ideal full-wave rectifier with a resistive load and no smoothing capacitor is driven by a sine wave with a 170 V peak. Its average (DC) output voltage is closest to:

A. 54 V
B. 108 V
C. 120 V
D. 170 V

Show answer and explanation

Answer: B: 108 V. Averaging |V_p sin θ| over one half-cycle gives 2V_p/π. Here that's 2 × 170/π = 108 V.

Why the others miss: A is the half-wave average, V_p/π. C is the rms value of the input sine, not the average. D is the peak.

Review target: Separate peak, rms, and average before you answer any rectifier or waveform question.

Source: Benha University ECE 121, Lecture 3: Rectifiers, PDF p. 16, §2–5 Full-Wave Rectifiers: V_AVG = 2V_P/π

Answer: B: 108 V. Averaging |V_p sin θ| over one half-cycle gives 2V_p/π. Here that's 2 × 170/π = 108 V.

Why the others miss: A is the half-wave average, V_p/π. C is the rms value of the input sine, not the average. D is the peak.

Review target: Separate peak, rms, and average before you answer any rectifier or waveform question.

Source: Benha University ECE 121, Lecture 3: Rectifiers, PDF p. 16, §2–5 Full-Wave Rectifiers: V_AVG = 2V_P/π

Q27 · Power Systems: Transformers

An ideal transformer has a 10:1 primary-to-secondary turns ratio. An 8 Ω load on the secondary appears on the primary as:

A. 0.08 Ω
B. 8 Ω
C. 80 Ω
D. 800 Ω

Show answer and explanation

Answer: D: 800 Ω. Impedance referred to the primary is Z₁ = (N₁/N₂)² Z₂ = 10² × 8 = 800 Ω.

Why the others miss: C uses the turns ratio instead of its square. A divides instead of multiplying, which refers impedance the wrong direction. B ignores the transformer.

Review target: Remember that impedance scales with the square of the turns ratio.

Source: Ellingson, Electromagnetics I (LibreTexts), §8.6, input impedance relation

Answer: D: 800 Ω. Impedance referred to the primary is Z₁ = (N₁/N₂)² Z₂ = 10² × 8 = 800 Ω.

Why the others miss: C uses the turns ratio instead of its square. A divides instead of multiplying, which refers impedance the wrong direction. B ignores the transformer.

Review target: Remember that impedance scales with the square of the turns ratio.

Source: Ellingson, Electromagnetics I (LibreTexts), §8.6, input impedance relation

Q28 · Electromagnetics: Transmission lines

A 50 Ω lossless line is terminated in a 150 Ω resistive load. What are the voltage reflection coefficient at the load and the VSWR?

A. 0.5 and 3
B. 0.5 and 2
C. −0.5 and 3
D. 0.33 and 2

Show answer and explanation

Answer: A: 0.5 and 3. Γ = (Z_L − Z₀)/(Z_L + Z₀) = (150 − 50)/(150 + 50) = 0.5. VSWR = (1 + |Γ|)/(1 − |Γ|) = 1.5/0.5 = 3.

Why the others miss: C reverses the order of subtraction in Γ. B gets the VSWR step wrong. D uses Z₀/Z_L ≈ 0.33 as Γ; its VSWR of 2 follows from that incorrect coefficient.

Review target: Remember the order: load minus line, divided by load plus line.

Source: Ellingson, Electromagnetics I (LibreTexts), §3.12, reflection coefficient; Ellingson, Electromagnetics I (LibreTexts), §3.14, standing wave ratio

Answer: A: 0.5 and 3. Γ = (Z_L − Z₀)/(Z_L + Z₀) = (150 − 50)/(150 + 50) = 0.5. VSWR = (1 + |Γ|)/(1 − |Γ|) = 1.5/0.5 = 3.

Why the others miss: C reverses the order of subtraction in Γ. B gets the VSWR step wrong. D uses Z₀/Z_L ≈ 0.33 as Γ; its VSWR of 2 follows from that incorrect coefficient.

Review target: Remember the order: load minus line, divided by load plus line.

Source: Ellingson, Electromagnetics I (LibreTexts), §3.12, reflection coefficient; Ellingson, Electromagnetics I (LibreTexts), §3.14, standing wave ratio

Q29 · Control Systems: Second-order response

A system starts at rest and has transfer function T(s) = ω_n²/(s² + 2ζω_n s + ω_n²), where ω_n > 0 and ζ = 0.5. Its percent overshoot to a unit step input is about:

A. 4.3%
B. 16.3%
C. 25%
D. 50%

Show answer and explanation

Answer: B: 16.3%. For an underdamped standard second-order system, %OS = 100 · e^(−ζπ/√(1 − ζ²)) = 100 · e^(−1.814) ≈ 16.3%.

Why the others miss: A is the overshoot for ζ ≈ 0.707. D assumes overshoot equals ζ. C has no basis in the formula.

Review target: Memorize two landmarks: ζ = 0.5 gives about 16% overshoot, and ζ = 0.707 gives about 4%.

Source: MIT OCW 2.004, Lecture 7: Second-order transients, PDF p. 26: percent overshoot of the standard underdamped second-order response

Answer: B: 16.3%. For an underdamped standard second-order system, %OS = 100 · e^(−ζπ/√(1 − ζ²)) = 100 · e^(−1.814) ≈ 16.3%.

Why the others miss: A is the overshoot for ζ ≈ 0.707. D assumes overshoot equals ζ. C has no basis in the formula.

Review target: Memorize two landmarks: ζ = 0.5 gives about 16% overshoot, and ζ = 0.707 gives about 4%.

Source: MIT OCW 2.004, Lecture 7: Second-order transients, PDF p. 26: percent overshoot of the standard underdamped second-order response

Q30 · Communications: FM bandwidth

An FM signal has a peak frequency deviation of 75 kHz and a maximum modulating frequency of 15 kHz. Using Carson's rule, what is its approximate bandwidth?

A. 30 kHz
B. 90 kHz
C. 150 kHz
D. 180 kHz

Show answer and explanation

Answer: D: 180 kHz. Carson's rule is B ≈ 2(Δf + f_m) = 2(75 + 15) = 180 kHz.

Why the others miss: C counts only the deviation (2Δf). A counts only the message (2f_m). B drops the factor of 2.

Review target: Carson's rule is an approximation. Use it when the question asks for approximate FM bandwidth.

Source: Montana State EE 446, FM notes, PDF p. 7 (printed p. 25): Carson’s Rule

Q31 · Computer Networks: Network security

Which events primarily compromise availability? Select all that apply.

A. A volumetric DDoS attack that floods a web server until it stops responding
B. Eavesdropping on unencrypted Wi-Fi traffic
C. Altering a firmware image while it's in transit
D. Cutting the only fiber link into a substation's control network

Show answer and explanation

Answer: A and D. Availability means timely, reliable access to information and systems. A loss of availability is a disruption of that access. Both A and D deny access.

Why the others miss: B is a confidentiality breach because it exposes information. C is an integrity breach because it modifies information.

Review target: Ask which part of the security triad the attacker damaged: secrecy (confidentiality), correctness (integrity), or access (availability).

Source: NIST FIPS 199, p. 2, security objectives

Answer: A and D. Availability means timely, reliable access to information and systems. A loss of availability is a disruption of that access. Both A and D deny access.

Why the others miss: B is a confidentiality breach because it exposes information. C is an integrity breach because it modifies information.

Review target: Ask which part of the security triad the attacker damaged: secrecy (confidentiality), correctness (integrity), or access (availability).

Source: NIST FIPS 199, p. 2, security objectives

Q32 · Digital Systems: Number systems

Interpret 11110110 as an 8-bit two's-complement integer. Enter its decimal value.

Your answer: write the number before opening the explanation.

Show answer and explanation

Answer: −10. The leading 1 means the number is negative. Invert the bits (00001001) and add 1 to get 00001010, which is 10. So the value is −10.

Common errors: If you entered 246, you read the bits as an unsigned number.

Review target: Check the most significant bit first. It tells you whether to take the two's complement.

Source: Cornell (T. Finley), Two's Complement, main text

Q33 · Computer Systems: Memory hierarchy

A cache has a 2 ns hit time, a 5% miss rate, and an additional 50 ns miss penalty after the hit-time lookup. What is the average memory access time?

A. 2.5 ns
B. 4.5 ns
C. 7.0 ns
D. 52 ns

Show answer and explanation

Answer: B: 4.5 ns. AMAT = hit time + miss rate × miss penalty = 2 + 0.05 × 50 = 4.5 ns.

Why the others miss: A includes the 2.5 ns weighted miss penalty but leaves out the 2 ns hit time. D treats every access as a miss. C has no basis in the formula.

Review target: Keep the hit time in the total. Every access pays it.

Source: MIT OCW Computation Structures, §14.1: Caches and the Memory Hierarchy, Average memory access time discussion: hit time + miss ratio × additional miss penalty

Answer: B: 4.5 ns. AMAT = hit time + miss rate × miss penalty = 2 + 0.05 × 50 = 4.5 ns.

Why the others miss: A includes the 2.5 ns weighted miss penalty but leaves out the 2 ns hit time. D treats every access as a miss. C has no basis in the formula.

Review target: Keep the hit time in the total. Every access pays it.

Source: MIT OCW Computation Structures, §14.1: Caches and the Memory Hierarchy, Average memory access time discussion: hit time + miss ratio × additional miss penalty

Q34 · Software Engineering: Loops and complexity

Consider this Python code:

count = 0
for i in range(10):
    for j in range(i, 10):
        count += 1

What is the final value of count, and how does the count grow if 10 is replaced by a nonnegative integer n? In Python, range(a, b) runs from a up to but not including b.

A. 45; O(n)
B. 55; O(n log n)
C. 55; O(n²)
D. 100; O(n²)

Show answer and explanation

Answer: C: 55; O(n²). The inner loop runs 10, 9, 8, … , 1 times, which adds up to 10 × 11/2 = 55. In general the count is n(n + 1)/2, which grows as O(n²).

Why the others miss: A is the count you'd get with range(i + 1, 10), and that count is still quadratic, not linear. D assumes a full 10 × 10 grid. B has the right count but the wrong growth rate.

Review target: Trace the loop bounds for the first two or three outer passes before you count.

Source: Python 3 Tutorial, §4.3 The range() Function, §§4.2–4.3

Answer: C: 55; O(n²). The inner loop runs 10, 9, 8, … , 1 times, which adds up to 10 × 11/2 = 55. In general the count is n(n + 1)/2, which grows as O(n²).

Review target: Trace the loop bounds for the first two or three outer passes before you count.

Source: Python 3 Tutorial, §4.3 The range() Function, §§4.2–4.3

What your score on these questions means

Your result here is feedback on these 34 questions only. It isn't a scaled score, a passing prediction, or a measure of how much of a whole topic you know. Two questions per area is far too few for that. NCEES doesn't publish a passing score, and "70% to pass" isn't an NCEES rule (NCEES exam scoring).

Your number of correct scored answers on the actual exam is converted to a scaled score and compared with a standard that subject-matter experts set. Exam forms vary slightly in difficulty, and the scaling adjusts for that. There's no fixed percentage of examinees who pass (NCEES exam scoring).

For this set, count an item correct only when your answer matches the key; the select-all item requires the complete correct set. Keep the handbook open while you work. Each answer comes with the working, why the other choices miss, and what to review.

Use your misses to decide which week of the plan to start with.

The 17 topics, ranked by question count

The FE Electrical and Computer exam has 110 questions. NCEES publishes a question range for each of its 17 knowledge areas. Here they are, largest first. Ties follow NCEES's order.

The 17 topics, ranked by question count
Knowledge area	Questions on the exam
Mathematics	11–17
Circuit Analysis (DC and AC Steady State)	11–17
Power Systems	8–12
Digital Systems	8–12
Electronics	7–11
Control Systems	6–9
Engineering Economics	5–8
Linear Systems	5–8
Signal Processing	5–8
Communications	5–8
Computer Systems	5–8
Probability and Statistics	4–6
Ethics and Professional Practice	4–6
Properties of Electrical Materials	4–6
Electromagnetics	4–6
Computer Networks	4–6
Software Engineering	4–6

Source: NCEES FE Electrical and Computer CBT Exam Specifications, pp. 1–3.

What the ranges tell you, and what they don't:

Math and circuits come first. Together they can supply 22 to 34 questions, and almost every other topic leans on them. If either one feels shaky, start there.
Both halves of the degree get tested. Power Systems and Digital Systems each carry 8–12 questions. A power-track graduate still faces logic and memory questions, and a computer-track graduate still faces transformers and three-phase power.
Small doesn't mean skippable. Six areas carry 4–6 questions each, and together they add up to 24–36 questions. The ranges are NCEES's published question counts, not a promise that every area will appear at its minimum or maximum on the same exam. Do not add all the minima or all the maxima and treat that sum as the exam length.

Your first session

Work Questions 1–17 above with the FE Reference Handbook open. Don't time yourself yet.
For every miss, write down whether it came from the setup, a concept, units, arithmetic, or finding the right formula.
Pick two areas where you missed a question or struggled to explain the setup, and put your next two study sessions on the calendar for them. One question per area is a starting clue, not a ranking of your readiness.

Your 10-week FE Electrical and Computer study plan

Time assumption: about 8–10 hours a week, for roughly 80–100 hours total. One way to fit that in: four 1-hour weeknight sessions plus two weekend blocks of 2–3 hours. This is our planning assumption, not an NCEES requirement. If the material is fresh from class, you may need less. If you're years out of school, see the 16-week version below.

Each week follows the same loop:

Learn: review the relevant explanations on this page, using the handbook's equations and symbols as you go. Use your course notes for any subtopic that needs a fuller review.
Practice: re-solve the matching questions from this page with the explanation hidden, then change one given value and recompute using the same method. For less familiar practice, add new problems from your own course materials or another source you're authorized to use.
Check it: finish the week's "done when" task with the handbook available and the worked answers hidden. Explain your setup before substituting values.

Your 10-week FE Electrical and Computer study plan
Week	Topics	Re-solve from this page	Done when you can…
Setup within Week 1	Setup and baseline	All 34, untimed, handbook open	Pick three areas for further review from misses and uncertain setups, with the current handbook downloaded
1	Mathematics; Probability and Statistics	Q1, Q2, Q18, Q19	Multiply complex numbers in rectangular form, solve a first-order differential equation with an initial condition, and set up a binomial probability
2	Circuit Analysis	Q6, Q23	Find a Thevenin equivalent and a phasor current for a new circuit, and explain lead vs. lag from the sign of X
3	Linear Systems; Signal Processing	Q7, Q24, Q8, Q25	Write an RC transient from its initial value, final value, and τ; find a resonant frequency in Hz; predict an alias frequency; tabulate an impulse response
4	Electronics; Properties of Electrical Materials	Q9, Q26, Q5, Q22	Pick the right op-amp gain formula by configuration, keep peak, rms, and average straight, convert mm² to m² without thinking, and explain n-type vs. p-type doping
5	Power Systems; Electromagnetics	Q10, Q27, Q11, Q28	Compute three-phase real power from line values, refer an impedance through a transformer, find a wavelength in a dielectric, and get Γ and VSWR for a mismatched line
6	Control Systems; Communications	Q12, Q29, Q13, Q30	Find steady-state error from the loop gain and system type, estimate overshoot for the stated second-order model, and size AM and FM bandwidths
7	Digital Systems; Computer Systems	Q15, Q32, Q16, Q33	Simplify and verify a Boolean expression with a truth table, read two's complement, size a memory from its address lines, and compute AMAT
8	Software Engineering; Computer Networks; Engineering Economics; Ethics	Q17, Q34, Q14, Q31, Q4, Q21, Q3, Q20	Trace a nested loop and state its growth rate, place a protocol in the right OSI layer, classify an attack by the security triad, discount cash flows, and name the specific ethics or IP rule a scenario triggers
9	Timed on-page rehearsal	Q1–Q34	Finish the 34-item rehearsal described below using only the handbook and an approved calculator. Then review every miss with the error log below.
10	Weak areas, a second rehearsal, and logistics	Two areas needing further review after Week 9	Re-solve those items without the explanations, repeat the timed rehearsal with a written setup for each item, run the handbook search drill, and confirm your ID, appointment time, and calculator

The timed rehearsal stays on this page. In Week 9, allow about 99 minutes for all 34 items: roughly 49½ minutes for Q1–Q17, followed by a break, then 49½ minutes for Q18–Q34. An optional 25-minute break is outside that working time. This practice budget comes from 320 minutes ÷ 110 × 34 ≈ 98.9 minutes; it is not an official time limit for this set. The real exam uses a shared time budget rather than separately timed halves (NCEES Examinee Guide, May 2026, pp. 11 and 16).

You have already seen these items, so faster answers may reflect memory. Show the setup and units rather than treating a repeated score as readiness evidence. This is a short timing exercise, not a full-length endurance test. A separate full-length rehearsal is optional when you already have enough authorized material; it is not required to use this plan.

Short on time: the 6-week version

Use this sequence: Week 1 = the main plan's Weeks 1–2; Week 2 = Weeks 3–4; Week 3 = Weeks 5–6; Week 4 = Week 7; Week 5 = Week 8; Week 6 = the on-page timed rehearsal, error review, and logistics from Weeks 9–10. A compressed plan covers the same ground with less repetition, so lean hardest on the areas your first sitting flagged. At 8–10 hours weekly, that is 48–60 hours, not the main plan's 80–100.

Years out of school: the 16-week version

Give two weeks each to Mathematics, Circuit Analysis, and whichever track you're rustier on: Power Systems or Digital Systems. Make the sequence concrete: Weeks 1–2 = main-plan Week 1; Weeks 3–4 = main-plan Week 2; Week 5 = main-plan Week 3; Week 6 = main-plan Week 4; Weeks 7–8 = the rustier track's main-plan week (5 or 7); Week 9 = the other track's week; Week 10 = main-plan Week 6; Week 11 = main-plan Week 8. Run the on-page timed rehearsals in Weeks 12 and 15, review the error log and missed concepts in Weeks 13–14, and use Week 16 for weak areas and logistics.

Retaking after a fail

If you don't pass, NCEES sends a diagnostic report. It reports each knowledge area on a scaled 0–15 measure and compares your performance with the average of examinees who passed (NCEES Examinee Guide, May 2026, pp. 20–22). Use both the reported performance gaps and each area's question range to prioritize review. A moderate gap in an 11–17-question area may offer more opportunities to improve than a large gap in a 4–6-question area. The 0–15 values are not percentages or numbers correct, and the comparison graphic is not drawn to scale; do not turn its gaps into an estimated passing score. Then restart this plan at your top-ranked area. Our NCEES exam results guide explains the report in more detail.

Keep an error log you can act on

A wrong answer is only useful if you can name what went wrong. Log every miss, from this page and from any other practice set:

Keep an error log you can act on
Question	What went wrong	Correct step or model	Search word that finds it in the handbook	Next task
Q6 (example)	Used the wrong resistor in the divider	Open-circuit voltage first, then R_th with the source shorted	Thevenin	Re-solve Q6 tomorrow with the answer hidden
	Setup / concept / units / arithmetic / lookup

If the same kind of error shows up three times, such as a unit conversion, add a check for it to every problem until it stops.

Use the handbook like a tool

The FE Reference Handbook is the only reference you get on exam day. NCEES supplies the current version on screen as a searchable PDF, and you can't bring your own copy or notes. You search it with the search box on the left side of the viewer. Ctrl+F doesn't work in the exam (NCEES Examinee Guide, May 2026, pp. 8 and 10).

Download the current handbook free from your MyNCEES account and practice with it from day one. NCEES lets you print it for personal use but not repost it (NCEES Help: exam reference handbooks).

A lookup routine that works:

Name the quantity the question asks for.
Write the relationship you think governs it.
Start with a concept word rather than a symbol, then adjust the wording if needed.
Read the conditions and symbol definitions next to the equation. Is it peak or rms? Line or phase? Rad/s or Hz?
Substitute with consistent units.
Check the sign and size of the answer before you move on.

Search drill. Time yourself finding each of these in your downloaded copy. Practice opening and using your PDF viewer's search panel; the exam viewer has its own left-side search box rather than Ctrl+F. For each one, write down the word that got you there fastest:

Thevenin equivalent
Three-phase power
Transformer impedance referral
RC time constant
Resonant frequency
Sampling and aliasing
Z-transform pairs
Percent overshoot
FM bandwidth (Carson's rule)
Two's complement
Karnaugh maps
Interest factor tables

Pacing

You get 5 hours 20 minutes for 110 questions, about 2 minutes 55 seconds per question on average. That's an average, not a per-question limit. The exam comes in two sections with approximately half the questions in each. Both sections share the 5-hour-20-minute working-time budget; the halves are not separately timed. Once you review and submit the first section, you can't go back to it, so review before you submit. A short scheduled break follows the first section, and coming back early doesn't add time to the second section (NCEES Examinee Guide, May 2026, pp. 11–12).

Some questions are unscored pretest items, and you can't tell which ones. Wrong answers aren't penalized, so answer every question (NCEES Examinee Guide, May 2026, pp. 11 and 14).

Exam-day rules that change how you prepare

Exam-day rules that change how you prepare
Rule	What NCEES says
Fee	$225, paid to NCEES when you register. Your board may charge its own application fee.
Questions	110, including unscored pretest items whose identities are not disclosed
Scheduled appointment	About 6 hours. The Guide table totals 5 hours 55 minutes: a 2-minute nondisclosure agreement, an 8-minute tutorial, 5 hours 20 minutes of exam time, and an optional 25-minute scheduled break. NCEES also instructs you to arrive at least 30 minutes before the appointment.
Where	Year-round at NCEES-approved Pearson test centers
Question types	Mostly single-answer multiple choice, plus multiple correct, point and click, drag and drop, and fill in the blank
Scoring	Based on the number of correct scored answers, with no partial credit and no deduction for wrong answers. Converted to a scaled score; the passing score isn't published.
Reference	The current FE Reference Handbook, on screen. Search with the left-side box; Ctrl+F isn't available.
Calculator — 2026 list	One approved calculator in the room: Casio models beginning with fx-115 or fx-991, the HP 33s or HP 35s, or Texas Instruments models beginning with TI-30X or TI-36X. A TI-30XS is also available on screen.
Scratch work	Pearson supplies two reusable booklets and three markers
Results	Usually 7–10 days after the exam. NCEES emails you when results are ready in MyNCEES. Pass/fail only, with a diagnostic report if you don't pass.
Retakes	One attempt per testing window (Jan–Mar, Apr–Jun, Jul–Sep, Oct–Dec), and no more than three in any 12 months. Some boards are stricter.
Rescheduling	At least 48 hours before your appointment, with a $50 fee paid to Pearson

Sources: NCEES FE exam page; NCEES Examinee Guide, May 2026, pp. 3, 5–6, 8–12, 14, 16; NCEES 2026 calculator-policy memo, October 20, 2025, p. 1; NCEES exam scoring.

Accommodations: the timing above describes the standard appointment. Request accommodations during initial registration and complete the approval process before scheduling (NCEES Examinee Guide, May 2026, p. 3).

For the full registration steps, including board approval, scheduling, and cancellations, see our NCEES exam registration guide.

FAQ

Is this the right FE exam for computer engineering graduates?

Electrical and Computer is one FE discipline covering both electrical and computer engineering. Four of its areas are mostly computing: Computer Networks, Digital Systems, Computer Systems, and Software Engineering. By NCEES's ranges, those four can supply 21–32 questions (FE Electrical and Computer specifications, p. 3). The rest still apply to you, especially math, circuits, and power. Your licensing board decides eligibility, so check its requirements before you register (NCEES Examinee Guide, May 2026, p. 2).

Which FE Electrical and Computer specifications should I use?

The specifications currently linked by NCEES are effective July 2020. That is the topic map used here, not a newly issued “2026 blueprint.” Check the NCEES FE exam page when you book; if NCEES posts a future specification, use the version effective on your test date rather than assuming this one applies indefinitely.

Does passing the FE make me a licensed engineer?

No. NCEES describes the FE as generally your first step toward becoming a licensed professional engineer (NCEES FE exam page). Your state or territory board decides what passing it means for you, such as engineer intern (EI or EIT) certification, and what experience and exams come next (NCEES Examinee Guide, May 2026, p. 24).

Sources and independence

Official exam sources

NCEES FE exam page: fee, format, delivery, and results
NCEES FE Electrical and Computer CBT Exam Specifications, effective July 2020: topics and question ranges
NCEES Examinee Guide, May 2026: registration, test-center rules, reference handbook, item types, breaks, scoring, results, and the diagnostic report. The current edition is always at the Examinee Guide page.
NCEES 2026 calculator-policy memo, October 20, 2025, p. 1: approved calculator families for 2026
NCEES exam scoring
NCEES Help: exam reference handbooks

Teaching sources. Each practice question links the source for the principle it tests, right beside its answer. Those sources support the engineering principle. They don't say how often a topic appears on the FE.

By Castleport Test Prep Editorial Team.

How we checked this page. Exam facts were last verified on October 6, 2026, against the NCEES sources listed above. We recomputed every numeric answer, and we checked each question's underlying principle against the source linked to it. AI tools assisted with drafting and source checks; source checking is not professional review. Our editorial methodology explains that distinction. It wasn't reviewed by NCEES or by a licensed professional engineer.

Castleport Test Prep is an independent exam prep publisher. We are not affiliated with, endorsed by, or approved by NCEES. The practice questions on this page are original and are not NCEES exam questions. FE, Fundamentals of Engineering, and NCEES are used only to identify the exam; trademarks belong to their respective owners.