The FE Electrical and Computer exam spreads 110 questions across seventeen topics, which is more breadth than any other FE discipline. That shape rewards a particular kind of practice: enough problems in each area to recognise what is being asked, rather than deep drilling in the two or three topics you already enjoy.
The ten problems below are worked the way you would work them on exam day. Each shows the setup, the arithmetic, the calculator keystrokes, and the handbook section the relation lives on. Try each one before opening the solution.
How Should You Use These Practice Problems?
Treat a missed problem as information rather than a score. The useful question is which step failed: choosing the relation, reading the handbook, or the arithmetic. Those three failures need different fixes, and only the last one is helped by working more problems quickly.
Every solution below names the wrong answers and what produces them, because on a multiple-choice exam the distractors are built from real mistakes. Recognising your own error in that list is worth more than the correct answer on its own.
Problem 1: Circuit Analysis — Norton to Thevenin Conversion
A Norton equivalent circuit has \(I_N = 4\) A and \(R_N = 10\,\Omega\). What is the Thevenin voltage \(V_{TH}\) of the same circuit?
Show Solution
Answer: C
A Norton source and a Thevenin source describe the same circuit seen from its two terminals, so the resistance carries straight across and only the source term changes.
$$R_{TH} = R_N = 10\,\Omega$$
$$V_{TH} = I_N R_N = 4 \times 10 = \mathbf{40\ \text{V}}$$
The sanity check is that shorting the Thevenin source must return the Norton current: \(40/10 = 4\) A.
On the TI-36X Pro: 4 × 10 = 40
Why the other choices are wrong: Option A (14 V) adds the two values instead of multiplying. Option B (2.5 V) computes \(R_N/I_N\), inverting Ohm’s law. Option D (0.4 V) computes \(I_N/R_N\).
Handbook section: Circuit Analysis > Source Transformations
Problem 2: Mathematics — Determinant of a 2×2 Matrix
Given \(A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\), what is \(\det A\)?
Show Solution
Answer: D
For a 2×2 matrix the determinant is the product of the main diagonal minus the product of the off-diagonal.
$$\det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - bc$$
$$\det A = (2)(4) - (1)(3) = 8 - 3 = \mathbf{5}$$
A non-zero determinant also tells you the matrix is invertible, which is often the real point of the question.
On the TI-36X Pro: 2 × 4 = 8, then 1 × 3 = 3, then 8 − 3 = 5
Why the other choices are wrong: Option A (−1) reverses the subtraction and computes \(bc - ad\). Option B (11) adds all four entries. Option C (8) stops at the diagonal product and never subtracts.
Handbook section: Mathematics > Linear Algebra > Determinants
Problem 3: Power Systems — Transformer Efficiency at Full Load
A single-phase transformer rated 50 kVA, 2400/240 V has a full-load copper loss of 750 W and a core loss of 250 W. What is its efficiency at full load and 0.80 power factor lagging?
Show Solution
Answer: B
Efficiency compares real output power against that output plus every loss. Start by turning the kVA rating into real power.
$$P_{out} = S \times \text{pf} = 50{,}000 \times 0.80 = 40{,}000\ \text{W}$$
$$P_{loss} = 750 + 250 = 1{,}000\ \text{W}$$
$$\eta = \frac{P_{out}}{P_{out} + P_{loss}} = \frac{40{,}000}{41{,}000} = \mathbf{97.6\%}$$
Note that the power factor changes the output but not the losses, so a lower power factor always lowers efficiency at the same kVA.
On the TI-36X Pro: 50000 × 0.8 = 40000 STO 1, then RCL 1 ÷ (RCL 1 + 1000) = 0.9756
Why the other choices are wrong: Against the same 40 kW output, option A (95.2%) implies about 2,020 W of loss, option C (98.4%) about 650 W, and option D (93.8%) about 2,640 W. Only 1,000 W matches the stated copper and core losses.
Handbook section: Power Systems > Transformers > Efficiency
Problem 4: Digital Systems — Two’s Complement Representation
What is the 8-bit two’s complement representation of −53?
Show Solution
Answer: D
Convert the magnitude first, then negate it in two steps.
$$53 = 32 + 16 + 4 + 1 = 00110101_2$$
Invert every bit: \(00110101 \rightarrow 11001010\).
Add one: \(11001010 + 1 = \mathbf{11001011}\).
Check it by negating again. Inverting 11001011 gives 00110100, and adding one returns 00110101, which is +53.
On the TI-36X Pro: 32 + 16 + 4 + 1 = 53 to confirm the place values, then invert and add one by hand
Why the other choices are wrong: Option A is +53 unchanged. Option B stops at the one’s complement and never adds the one. Option C is a different negative magnitude altogether.
Handbook section: Digital Systems > Binary Arithmetic > Two’s Complement
Problem 5: Control Systems — Bode Magnitude at a Given Frequency
A system has \(G(s) = \dfrac{100}{s(s + 10)}\). What is the gain in decibels at \(\omega = 1\) rad/s?
Show Solution
Answer: A
Substitute \(s = j\omega\) and take the magnitude of each factor separately.
$$|G(j1)| = \frac{100}{|j1| \cdot |j1 + 10|} = \frac{100}{1 \times \sqrt{1^2 + 10^2}}$$
$$= \frac{100}{10.05} = 9.95$$
$$20\log_{10}(9.95) = \mathbf{20\ \text{dB}}$$
Because \(\omega = 1\) is well below the pole at 10, the straight-line approximation \(|G| \approx 100/(10\omega) = 10\) gives the same 20 dB, which is a useful check.
On the TI-36X Pro: √(1 x² + 10 x²) = 10.05, then 100 ÷ 10.05 = 9.95, then 20 × log 9.95 = 20.0
Why the other choices are wrong: Option B (40 dB) reads the low-frequency asymptote \(|G| = 100\) and ignores the integrator. Option C (0 dB) is the gain crossover, not \(\omega = 1\). Option D (14 dB) corresponds to a magnitude of 5.
Handbook section: Control Systems > Frequency Response > Bode Plots
Problem 6: Communications — FM Bandwidth by Carson’s Rule
An FM signal has a maximum frequency deviation of 75 kHz and a modulating frequency of 15 kHz. Using Carson’s rule, what is the approximate bandwidth?
Show Solution
Answer: C
Carson’s rule doubles the sum of the deviation and the highest modulating frequency.
$$BW = 2(\Delta f + f_m) = 2(75 + 15) = \mathbf{180\ \text{kHz}}$$
The modulation index here is \(\beta = \Delta f / f_m = 75/15 = 5\), comfortably wideband, which is exactly the regime Carson’s rule is meant for.
On the TI-36X Pro: 75 + 15 = 90, then 2 × 90 = 180
Why the other choices are wrong: Option A (90 kHz) is the sum inside the brackets with the factor of 2 never applied. Option B (150 kHz) doubles the deviation alone and drops \(f_m\). Option D (210 kHz) adds an extra \(2f_m\) that the rule does not contain.
Handbook section: Communications > Frequency Modulation > Carson’s Rule
Problem 7: Computer Systems — Cache Tag Field Width
A direct-mapped cache has 64 lines of 16 bytes each and the main memory address is 16 bits wide. How many bits are in the tag field?
Show Solution
Answer: C
A direct-mapped address splits into three fields: tag, index and byte offset. Size the two you can read off the cache geometry, then take what is left.
$$\text{offset} = \log_2(16) = 4\ \text{bits}$$
$$\text{index} = \log_2(64) = 6\ \text{bits}$$
$$\text{tag} = 16 - 6 - 4 = \mathbf{6\ \text{bits}}$$
The index selects the line and the offset selects the byte within it, so only the remaining high-order bits need storing to tell two addresses apart.
On the TI-36X Pro: log 16 ÷ log 2 = 4, then log 64 ÷ log 2 = 6, then 16 − 6 − 4 = 6
Why the other choices are wrong: Option A (4) reports the offset alone. Option B (5) uses \(\log_2(32)\) for the index. Option D (7) subtracts only a 5-bit index from the total.
Handbook section: Computer Systems > Cache Memory > Address Mapping
Problem 8: Engineering Economics — Net Present Value of a Uniform Series
A project costs $50,000 up front and returns $15,000 per year for 5 years. At a 10% discount rate, and using \((P/A, 10\%, 5) = 3.7908\), what is the NPV?
Show Solution
Answer: C
Discount the uniform series back to the present with the supplied factor, then subtract the cost that is already in present terms.
$$PW_{\text{benefits}} = 15{,}000 \times 3.7908 = 56{,}862$$
$$NPV = -50{,}000 + 56{,}862 = \mathbf{\$6{,}862}$$
A positive NPV at the stated discount rate means the project clears the 10% hurdle, which is the decision the number exists to support.
On the TI-36X Pro: 15000 × 3.7908 = 56862, then − 50000 = 6862
Why the other choices are wrong: Option B ($25,000) never discounts at all, taking \(15{,}000 \times 5 - 50{,}000\). Option A ($12,500) applies a partial or incorrect factor. Option D (−$2,500) turns a viable project into a rejected one and would reverse the decision.
Handbook section: Engineering Economics > Present Worth > Uniform Series
Problem 9: Electronics — NMOS Drain Current in Saturation
An enhancement-mode NMOS transistor has \(V_{tn} = 1.5\) V and a conduction parameter \(k_n = 2\) mA/V\(^2\), biased at \(V_{GS} = 3.5\) V and \(V_{DS} = 4\) V. What is the drain current, using \(I_D = k_n(V_{GS} - V_{tn})^2\) in saturation?
Show Solution
Answer: B
Establish the operating region before reaching for a formula. Saturation requires \(V_{DS}\) to exceed the overdrive voltage.
$$V_{ov} = V_{GS} - V_{tn} = 3.5 - 1.5 = 2.0\ \text{V}$$
Since \(V_{DS} = 4\ \text{V} > 2.0\ \text{V}\), the device is saturated and the square-law expression applies.
$$I_D = k_n V_{ov}^2 = 2 \times (2.0)^2 = \mathbf{8\ \text{mA}}$$
Had \(V_{DS}\) fallen below the overdrive, the triode expression \(I_D = k_n[2V_{ov}V_{DS} - V_{DS}^2]\) would apply instead, so the region check is not a formality.
On the TI-36X Pro: 3.5 − 1.5 = 2, then 2 × 2 x² = 8
Why the other choices are wrong: Option A (4 mA) multiplies by the overdrive instead of its square. Option C (2 mA) uses a 1 V overdrive. Option D (16 mA) carries an extra factor of 2 that the stated square-law form does not contain.
Handbook section: Electronics > MOSFET > Operating Regions
Problem 10: Signal Processing — Cutoff Frequency of a First-Order IIR Filter
A first-order low-pass digital filter has the difference equation \(y[n] = 0.2x[n] + 0.8y[n-1]\) and runs at a 10 kHz sampling rate. What is its 3 dB cutoff frequency?
Show Solution
Answer: B
Take the transfer function first, then find the digital frequency at which the squared magnitude falls to one half.
$$H(z) = \frac{0.2}{1 - 0.8z^{-1}}$$
At DC the gain is \(0.2/0.2 = 1\), so the half-power condition is \(|H(e^{j\omega})|^2 = 0.5\). Solving gives \(\cos\omega = 0.975\), so \(\omega_c = 0.224\) rad/sample.
$$f_c = \frac{\omega_c f_s}{2\pi} = \frac{0.224 \times 10{,}000}{2\pi} \approx \mathbf{356\ \text{Hz}}$$
The digital frequency is in radians per sample, so the sampling rate is what converts it to hertz. Forgetting that conversion is the most common way this problem goes wrong.
On the TI-36X Pro: 2nd cos⁻¹ 0.975 = 0.224 in radian mode, then × 10000 ÷ (2 × π) = 356
Why the other choices are wrong: Option A (224 Hz) reports the digital frequency scaled by the sampling rate but without dividing by \(2\pi\). Option C (500 Hz) assumes the cutoff is a round fraction of \(f_s\). Option D (1,592 Hz) applies the analogue \(1/(2\pi RC)\) result with no z-domain step at all.
Handbook section: Signal Processing > Digital Filters > Frequency Response
What Topics Does the FE Electrical Exam Cover?
Seventeen topics, and the weighting is flatter than most FE disciplines. Circuit Analysis and Mathematics carry the most questions at roughly 11 to 17 each, with Power Systems and Digital Systems close behind at 8 to 12. Nothing else exceeds 11, and six topics sit at 4 to 6 questions apiece.
That flatness has a practical consequence. Ignoring one small topic costs only a few questions, but ignoring four of them costs about a fifth of the exam. Breadth beats depth here in a way it does not on, say, FE Mechanical.
The FE Electrical study guide lists the official topic ranges alongside how many practice questions our bank carries for each, so you can see where your own coverage is thin.
Essential Study Materials
- An NCEES-approved calculator you have already practised on. The TI-36X Pro handles complex magnitudes, logarithms and stored intermediates without mode-switching, which matters on circuit and Bode questions.
- The NCEES FE Reference Handbook edition assigned to your exam date, so section names and search terms are familiar before you sit down.
- A problem set broad enough to touch all seventeen topics rather than a thick book on three of them.
Our FE prep book comparison covers the review manuals in more detail.
What Should You Do Next?
Ten problems tell you where you stand on ten narrow subtopics. What they cannot tell you is how you perform across a full mixed set under time pressure, which is the thing the exam actually measures.
The next step is a timed set drawn from every topic, not another block of circuit problems. Our free 20-question FE Electrical practice exam is drawn from the same bank as the problems above and reports which topics came out weakest.
Ready for More Practice?
The full FE Electrical and Computer question bank covers all seventeen topics with worked solutions, calculator keystrokes and handbook references on every question, plus timed exams that mirror the real 110-question format.
Start with the study guide for a topic-by-topic plan, or go straight to the free practice exam.