FE Electrical and Computer · practice test · checked Oct 29, 2026
FE electrical practice exam, 60 worked problems
This practice exam has 60 problems, spread across the 17 areas in proportion to the middle of each range in the July 2020 specification. Answer one and its solution opens: the given data, the steps, and what is wrong with each of the other options. The real exam is longer and has question formats a web page cannot copy, so use the score to locate gaps, not to forecast a result.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Electrical and Computer
problems60
on the real exam110
areas covered17
real testing time5 h 20 min
Checked
60 problems across the specification
0 / 60 answered · 0 correct
Question 1 of 60 · Properties of Electrical Materials · 4–6 items
In a semiconductor with no applied electric field but with a gradient in carrier concentration, which mechanism is mainly responsible for current flow?
Solution
Jdiff = qDn(dn/dx) (no field needed), Jdrift = qnμnE (needs a field)
Recombination removes carrier pairs; it changes concentrations but does not carry current across the sample by itself.
Drift current needs an electric field (J = qnμE), and the stem says none is applied.
Tunneling needs a thin, high-field barrier such as a heavily doped junction, not a uniform sample.
With no field there is no drift; carriers spread from high to low concentration, J = qD dn/dx.
Question 2 of 60 · Properties of Electrical Materials · 4–6 items
In a semiconductor at thermal equilibrium, the product of electron concentration (n) and hole concentration (p) is equal to:
Solution
n · p = ni2 (thermal equilibrium)
Mass-action law: np = ni2 at equilibrium, whatever the doping.
The Fermi level moves with doping, while np stays fixed at a given temperature.
Doping sets the majority carrier concentration, not the product of both.
The bandgap sets ni through an exponential, but the product np is a concentration squared, not an energy.
Question 3 of 60 · Circuit Analysis (DC and AC Steady State) · 11–17 items
A series RL circuit with R = 6 Ω and XL = 8 Ω is connected to a 50 V rms source. What is the real power dissipated?
Given
Resistance R = 6 Ω
Inductive reactance XL = 8 Ω
Source voltage V = 50 V rms
Find
Real power P (W)
Solution
I = V/|Z| = 50 V/√(62 + 82) Ω = 5 A
P = I2R = (5 A)2 × 6 Ω = 150 W
|Z| = √(62 + 82) = 10 Ω, I = 50/10 = 5 A, P = I2R = 25 × 6 = 150 W.
This is I2XL = 25 × 8, the reactive power in var; the inductor dissipates nothing.
This is V·I = 50 × 5, the apparent power in VA.
This is V2/R = 2500/6, which assumes the whole source voltage appears across the resistor.
Question 4 of 60 · Circuit Analysis (DC and AC Steady State) · 11–17 items
Three branches connect a top node to a common bottom node: branch 1 is a 6 V source (positive terminal toward the top node) in series with 2 Ω; branch 2 is a 2 Ω resistor; branch 3 is a 24 V source (positive terminal toward the top node) in series with 4 Ω. Does the 6 V source absorb or deliver power, and how much?
Given
Branch 1: V1 = 6 V (+ at top) in series with R1 = 2 Ω
Branch 2: R3 = 2 Ω
Branch 3: V2 = 24 V (+ at top) in series with R2 = 4 Ω
Find
Power of the 6 V source (absorbed or delivered)
Solution
(V − 6)/2 + V/2 + (V − 24)/4 = 0 → 5V = 36 → V = 7.2 V
I1 = (7.2 − 6)/2 = 0.6 A into the + terminal of V1
PV1 = 6 V × 0.6 A = 3.60 W absorbed (passive sign convention)
This leaves out the 2 Ω middle branch: then V = 12 V and 3 A flows into V1, 6 × 3 = 18.00 W.
The 0.6 A flows into the + terminal, so by the passive sign convention the source absorbs power; this flips that sign.
This keeps only the 6 V source acting alone (1.8 A out of its + terminal, 6 × 1.8 = 10.80 W) and ignores the 24 V source.
This keeps only the 24 V source's contribution (2.4 A into V1, 6 × 2.4 = 14.40 W); the 6 V source drives 1.8 A the other way at the same time.
KCL at the top node: (V − 6)/2 + V/2 + (V − 24)/4 = 0 gives V = 7.2 V. Current (7.2 − 6)/2 = 0.6 A enters the + terminal of the 6 V source, so it absorbs 6 × 0.6 = 3.60 W.
Question 5 of 60 · Circuit Analysis (DC and AC Steady State) · 11–17 items
Compute the input impedance seen by a source connected to a 10 Ω resistor in series with a parallel combination of a 30 Ω resistor and an inductor of reactance j20 Ω. What are |Zin| and its phase angle?
Question 6 of 60 · Circuit Analysis (DC and AC Steady State) · 11–17 items
A DC voltage source is switched at t = 0 onto a series RL circuit whose inductor current was zero. What does the inductor current do right after switching?
Inductor current is continuous: i(0+) = i(0−) = 0, then i(t) = (V/R)(1 − e−tR/L).
An inductor resists any instant change of current (v = L di/dt would be infinite), so the current cannot jump.
That is the decay after the source is removed; when the source is applied the current grows.
Oscillation needs a second energy-storage element (a capacitor); a first-order RL circuit rises without overshoot.
Question 7 of 60 · Circuit Analysis (DC and AC Steady State) · 11–17 items
Three resistors of 12 Ω each are connected in delta. What is the equivalent resistance of one leg when converted to a wye?
Given
Delta resistance RΔ = 12 Ω
Find
Equivalent wye resistance RY (Ω)
Solution
RY = RΔ/3 = 12 Ω/3 = 4 Ω
This is 3R, the wye-to-delta direction (RΔ = 3RY) applied backwards.
This keeps the delta value; each wye leg must be smaller because two legs in series replace one delta branch.
This divides by 2 instead of 3.
For a balanced network, RY = RΔ/3 = 12/3 = 4 Ω.
Question 8 of 60 · Circuit Analysis (DC and AC Steady State) · 11–17 items
A 24 V DC source with 2 Ω internal resistance feeds a 4 Ω resistor R1 in series with a parallel pair of a 12 Ω resistor R2 and a 6 Ω resistor R3. What power does the 12 Ω resistor dissipate?
Given
Source V = 24 V DC, internal resistance Rs = 2 Ω
R1 = 4 Ω in series with (R2 = 12 Ω ∥ R3 = 6 Ω)
Find
Power dissipated in R2 (W)
Solution
Rp = (12 × 6)/(12 + 6) = 4 Ω, Rtotal = 2 + 4 + 4 = 10 Ω, I = 24 V/10 Ω = 2.4 A
Vp = I · Rp = 9.6 V, P12 = Vp2/R2 = 9.62/12 = 7.68 W
This puts the full 24 V across the 12 Ω resistor: 242/12 = 48.00 W.
This is the power of the whole parallel pair, 9.6 V × 2.4 A = 23.04 W.
This is the power in the 6 Ω resistor, 9.62/6 = 15.36 W, not in the 12 Ω one.
This ignores the 2 Ω internal resistance: I = 24/8 = 3 A, Vp = 12 V, 122/12 = 12.00 W.
Rtotal = 2 + 4 + (12 ∥ 6) = 2 + 4 + 4 = 10 Ω, I = 24/10 = 2.4 A, Vp = 2.4 × 4 = 9.6 V, so P12 = 9.62/12 = 7.68 W.
Question 9 of 60 · Circuit Analysis (DC and AC Steady State) · 11–17 items
Two AC sources feed a node: a 100∠0° V source connects to the node through 10 Ω, a 50∠−90° V source connects through 20 Ω, and the node connects to ground through 30 Ω. What is the magnitude of the current through the 30 Ω resistor?
Given
Source V1 = 100∠0° V through R1 = 10 Ω
Source V2 = 50∠−90° V through R2 = 20 Ω
Node-to-ground resistance R3 = 30 Ω
Find
Current magnitude through the 30 Ω resistor (A)
Solution
(V − 100)/10 + (V − (−j50))/20 + V/30 = 0
V = (10 − j2.5)/(0.1 + 0.05 + 0.0333) = 56.22∠−14.04° V
I30 = |V|/30 Ω = 56.22/30 = 1.87 A
This keeps only the 100 V source; the 50∠−90° V source also drives current into the node.
KCL at the node: (V − 100)/10 + (V + j50)/20 + V/30 = 0 gives V = 56.22∠−14.04° V, so I = 56.22/30 = 1.87 A.
This treats the second source as 50∠0° V; its −90° angle makes its contribution add in quadrature, not in phase.
This divides the node voltage by 10 Ω instead of the 30 Ω branch the question asks about.
Question 10 of 60 · Linear Systems · 5–8 items
A system has the transfer function T(s) = (s2 + 4s + 8)/(s3 + 6s2 + 12s + 8). What is its impulse response y(t) for t ≥ 0?
This inverts 4/(s + 2)3 as if it were the double pole 4/(s + 2)2, giving 4te−2t; a triple pole gives 4(t2/2)e−2t.
With p = s + 2, the numerator becomes p2 + 4, so T = 1/p + 4/p3; 4/p3 inverts to 4·(t2/2)e−2t = 2t2e−2t.
This drops the 1/2! from L−1{1/(s + a)3} = (t2/2)e−at, so 4/(s + 2)3 becomes 4t2e−2t instead of 2t2e−2t.
The pole is at s = −2 (s3 + 6s2 + 12s + 8 = (s + 2)3), so the exponent is −2t.
Question 11 of 60 · Linear Systems · 5–8 items
A series RLC circuit has R = 10 Ω, L = 50 mH, and C = 80 nF and is driven at resonance by a 120 V rms source. What are the resonant frequency f0, the quality factor Q, and the rms voltage across the inductor?
Given
Resistance R = 10 Ω
Inductance L = 50 mH
Capacitance C = 80 nF
Source voltage V = 120 V rms
Find
Resonant frequency f0 (Hz), quality factor Q, inductor voltage VL (V rms)
This uses R√(C/L), the reciprocal of Q, so the inductor voltage comes out tiny.
At resonance the source voltage appears across R, while the L and C voltages are Q times larger and cancel each other.
f0 = 1/(2π√(0.05 × 80 × 10−9)) = 2516 Hz; Q = (1/10)√(L/C) = 79.1; I = 120/10 = 12 A, VL = IXL = QV = 9487 V rms.
This reports ω0 = 1/√(LC) in rad/s as the frequency in hertz.
Question 12 of 60 · Signal Processing · 5–8 items
A digital IIR filter has H(z) = 1/(1 − 1.6 cos(0.4π) z−1 + 0.64 z−2) and a sampling frequency of 8000 Hz. Is the filter stable, and what is its resonant frequency?
Given
H(z) = 1/(1 − 1.6·cos(0.4π)·z−1 + 0.64·z−2)
Sampling frequency fs = 8000 Hz
Find
Stability and resonant frequency (Hz)
Solution
poles at z = 0.8e±j0.4π (r2 = 0.64, 2r cos θ = 1.6 cos 0.4π)
The pole radius is √0.64 = 0.8, inside the unit circle, so the filter is stable.
The denominator is 1 − 2r cos θ z−1 + r2z−2 with r = 0.8 < 1 and θ = 0.4π: f = (θ/2π)fs = 0.2 × 8000 = 1600 Hz.
This uses fs/2 in place of fs: f = (θ/2π) × 4000.
This uses θ/π instead of θ/(2π) to convert the pole angle to hertz.
Question 13 of 60 · Signal Processing · 5–8 items
A 2000 Hz tone is sampled at 8000 Hz and then downsampled by M = 3 (one of every three samples kept). Does aliasing occur, and at what frequency does the tone appear in the decimated sequence?
Given
Tone frequency f0 = 2000 Hz
Sampling rate fs = 8000 Hz
Decimation factor M = 3
Find
Whether aliasing occurs and the aliased frequency (Hz)
This adds the new rate to the tone; an alias lies in the 0 to fs′/2 band.
After decimation the rate is 2666.7 Hz, whose Nyquist limit 1333.3 Hz lies below 2000 Hz, so the tone aliases.
fs′ = 8000/3 = 2666.7 Hz, Nyquist 1333.3 Hz < 2000 Hz; alias = |2000 − 2666.7| = 666.7 Hz.
This is the new Nyquist frequency, the folding point, not the alias.
Question 14 of 60 · Signal Processing · 5–8 items
Using the bilinear transform with prewarping Ωp = 2fs tan(πfd/fs), what analog angular frequency corresponds to a digital cutoff of 1 kHz at a sampling rate of 8000 Hz?
At MOSFET turn-off, 5.0 A flows in a stray loop inductance Ls = 200 nH. An RC snubber across the MOSFET must absorb that energy so the voltage rises no more than 50 V above the 300 V DC bus. Using an energy balance, what minimum snubber capacitance is needed, and what series resistance critically damps the LsC loop?
Given
Current I = 5.0 A
Stray inductance Ls = 200 nH
DC bus VDC = 300 V
Allowed overshoot ΔV = 50 V
Find
Minimum snubber capacitance C (nF) and resistance R (Ω)
Solution
½ Ls I2 = ½ C ΔV2 → C = LsI2/ΔV2 = (200 × 10−9 × 5.02)/502 = 2.0 nF
This squares VGS instead of the overdrive VGS − Vt.
Question 18 of 60 · Electronics · 7–11 items
A BJT differential amplifier has a tail current of 2 mA split equally between two matched transistors. What is the collector current in each transistor under balanced conditions?
One transistor carries no current only when a large differential input steers the whole tail to the other side; balanced inputs split it evenly.
With matched transistors and equal inputs the 2 mA tail splits in half: 1 mA each (IC ≈ IE for large β).
This gives each transistor the whole tail current; the two collector currents must sum to it.
This doubles the tail current instead of halving it.
Question 19 of 60 · Electronics · 7–11 items
A power MOSFET switches Iload = 20.0 A at VDS = 48.0 V and fs = 50.0 kHz. Its gate charge is Qg = 30.0 nC, driven at 12.0 V with a peak gate current of 1.00 A. Using Pgate = QgVgatefs, tsw = Qg/Idrv, Esw ≈ 0.5VDSIloadtsw, and Psw = Eswfs, what are Pgate, Psw, and their total?
Given
Load current I = 20.0 A
Drain-source voltage VDS = 48.0 V
Gate charge Qg = 30.0 nC
Gate drive current Idrv = 1.00 A
Gate-drive voltage = 12.0 V
Switching frequency fs = 50.0 kHz
Find
Pgate, Psw and their total (W)
Solution
Pgate = QgVgatefs = 30 nC × 12 V × 50 kHz = 0.0180 W
tsw = Qg/Idrv = 30 nC/1.00 A = 30 ns
Esw = 0.5 × 48 V × 20 A × 30 ns = 14.4 µJ, Psw = 14.4 µJ × 50 kHz = 0.720 W
This uses the voltage ratio 13.8/13.2 without squaring it: 0.16 × 1.045 = 0.167; impedance scales with V2.
This inverts the voltage ratio, (13.2/13.8)2: 0.16 × 0.915 = 0.146; a lower base voltage raises the per-unit value.
This inverts the MVA ratio (50/100); a larger base power makes the per-unit impedance larger: 0.08 × 0.5 × 1.093 = 0.044.
Question 23 of 60 · Power Systems · 8–12 items
A 138 kV transmission line has a surge (characteristic) impedance of 400 Ω. What is its three-phase surge impedance loading (SIL)?
Given
Line voltage VLL = 138 kV
Surge impedance Z0 = 400 Ω
Find
Three-phase surge impedance loading (MW)
Solution
SIL = VLL2/Z0
SIL = (138 kV)2/400 Ω = 47.61 MW
This divides the per-phase 15.87 MW by 3 instead of multiplying: 15.87/3 = 5.29 MW.
This is the per-phase value, (138/√3 kV)2/400 Ω; the three-phase SIL is three times larger.
This divides the three-phase result by √3 (47.61/√3 = 27.49 MW), as if a phase-voltage correction were still needed; VLL2/Z0 already is the three-phase total.
SIL = VLL2/Z0 = (138 × 103)2/400 = 47.61 × 106 W.
Question 24 of 60 · Power Systems · 8–12 items
A balanced three-phase transmission line has a resistance of 0.5 Ω per phase and carries 100 A. What is the total power loss in the three phases?
This is the loss in one phase, I2R = 1002 × 0.5; all three conductors carry 100 A.
This multiplies one phase's loss by √3; the losses of three phases add, so the factor is 3.
P = 3I2R = 3 × 1002 × 0.5 = 15,000 W = 15.0 kW.
This counts a return conductor for every phase (2I2R per phase); a balanced three-phase line has no return current.
Question 25 of 60 · Electromagnetics · 4–6 items
A quarter-wave transformer matches a 100 Ω load to a 50 Ω transmission line at one frequency. What characteristic impedance must the quarter-wave section have?
Given
Line impedance Z0 = 50 Ω
Load impedance ZL = 100 Ω
Find
Quarter-wave section impedance Zq (Ω)
Solution
Zin = Zq2/ZL = Z0 → Zq = √(Z0ZL)
Zq = √(50 Ω × 100 Ω) = 70.7 Ω
A λ/4 section with Zq = √(Z0ZL) = √(50 × 100) = 70.7 Ω turns the 100 Ω load into 50 Ω.
This is the arithmetic mean (50 + 100)/2; the match needs the geometric mean.
This is the load impedance; a quarter-wave section equal to ZL presents ZL2/ZL = 100 Ω, still mismatched.
This is ZL2/Z0 = 1002/50 = 200.0 Ω, the quarter-wave relation solved the wrong way round.
Question 26 of 60 · Electromagnetics · 4–6 items
A transmission line with characteristic impedance 75 Ω has a voltage standing wave ratio (VSWR) of 3. What is the magnitude of the voltage reflection coefficient?
|Γ| = (VSWR − 1)/(VSWR + 1) = (3 − 1)/(3 + 1) = 0.50; Z0 is not needed.
This is 1/VSWR, which has no physical meaning here.
This is 1/(VSWR + 1); the formula is (VSWR − 1)/(VSWR + 1).
Question 27 of 60 · Control Systems · 6–9 items
A Type 0 unity feedback control system has Kp = 9. What is the steady-state error for a unit step input?
Given
System type = 0 (unity feedback)
Position error constant Kp = 9
Find
Steady-state error for a unit step
Solution
ess = 1/(1 + Kp)
ess = 1/(1 + 9) = 0.10
Zero step error requires at least one integrator (Type 1); a Type 0 loop always leaves some error.
ess = 1/(1 + Kp) = 1/(1 + 9) = 0.10.
This is 1/Kp = 1/9, the ramp-type formula pattern; the step error of a Type 0 loop is 1/(1 + Kp).
This is Kp/(1 + Kp), the fraction of the input that the output reaches, not the error.
Question 28 of 60 · Control Systems · 6–9 items
A loop with forward gain G1 = 4 and feedback gain H = 0.1 (negative feedback) is bypassed by a parallel feedforward path G2 = 6 from R straight to the output summing point. What is the overall gain C/R?
Given
Forward gain G1 = 4
Feedback gain H = 0.1
Parallel path G2 = 6
Find
Overall gain C/R
Solution
C/R = G1/(1 + G1H) + G2
C/R = 4/(1 + 0.4) + 6 = 8.86
This is the loop alone, G1/(1 + G1H); the parallel path G2 adds to it.
This puts G2 inside the loop with G1, (G1 + G2)/(1 + (G1 + G2)H); the feedforward bypasses the loop.
This keeps only the feedforward path G2 = 6 and drops the loop's 2.86 contribution.
A control system must have a step-response overshoot below 10 %. What minimum damping ratio is required?
Given
Maximum overshoot PO = 10%
Find
Minimum damping ratio ζ
Solution
ζ = −ln(PO)/√(π2 + ln2(PO))
ζ = 2.303/√(9.870 + 5.302) = 0.59
Critical damping gives no overshoot at all; it meets the limit but is not the minimum ζ.
This is −ln(0.1)/π, which drops the ln2 term under the square root.
ζ = −ln(PO)/√(π2 + ln2(PO)) = 2.303/√(9.870 + 5.302) = 0.59; any smaller ζ overshoots more than 10 %.
This is the overshoot fraction itself, not a damping ratio.
Question 30 of 60 · Control Systems · 6–9 items
In a negative-feedback block diagram, G(s) is in the forward path and H(s) is in the feedback path. What is the closed-loop transfer function T(s) = C(s)/R(s)?
Solution
C/R = G/(1 + GH) (negative feedback)
The minus sign belongs to positive feedback; the standard loop subtracts the feedback signal.
For negative feedback, C/R = G/(1 + GH): forward path over one plus the loop gain.
This is the ratio of the fed-back signal B to R, not of the output C to R.
This is the sensitivity (error) function E/R for unity feedback, not the output-to-input ratio.
Question 31 of 60 · Communications · 5–8 items
A voice-grade channel with 3 kHz bandwidth is sampled at the Nyquist rate and quantized with 12 bits per sample. Twenty-four such channels are combined by synchronous TDM, and 5 % overhead is added for framing and control. What is the total bit rate of the multiplexed signal?
This is 24 × 6 kHz × 12 bits without the 5 % framing overhead.
This subtracts the 5 % overhead instead of adding it.
This samples at 3 kHz, the bandwidth itself; Nyquist requires twice the bandwidth, 6 kHz.
Question 32 of 60 · Communications · 5–8 items
Five channels with bandwidths 2.4 kHz, 3.2 kHz, 4.8 kHz, 1.6 kHz, and 2.0 kHz share an FDM band. Adjacent channels are separated by 0.5 kHz guard bands, and a 0.2 kHz pilot tone sits at one end of the band. What is the total occupied bandwidth?
Given
Channel bandwidths = 2.4, 3.2, 4.8, 1.6, 2.0 kHz
Guard band between adjacent channels = 0.5 kHz
Pilot tone = 0.2 kHz
Find
Total occupied bandwidth (kHz)
Solution
B = ∑Bi + (N − 1)Bg + Bpilot
B = 14.0 + 4 × 0.5 + 0.2 = 16.2 kHz
This adds only the five channel bandwidths and omits both the guard bands and the pilot.
This adds the pilot tone but omits the four 0.5 kHz guard bands.
This includes the four guard bands but leaves out the 0.2 kHz pilot tone.
Channels: 2.4 + 3.2 + 4.8 + 1.6 + 2.0 = 14.0 kHz; four gaps × 0.5 = 2.0 kHz; pilot 0.2 kHz; total 16.2 kHz.
Question 33 of 60 · Communications · 5–8 items
A communication receiver has an input signal power of −80 dBm and an input noise power of −100 dBm. What is the input signal-to-noise ratio?
Given
Signal power Ps = −80 dBm
Noise power Pn = −100 dBm
Find
Input SNR (dB)
Solution
SNR (dB) = Ps (dBm) − Pn (dBm)
SNR = −80 − (−100) = 20 dB
Both powers are already in dBm, so the ratio is a difference: −80 − (−100) = 20 dB.
This takes 10 log10(20), treating the 20 dB difference as a ratio and converting it to decibels a second time.
This is log10(100) = 2 without the factor 10 of the decibel, after converting the 20 dB difference to a ratio of 100.
This subtracts in the wrong order, noise minus signal; a signal 100 times stronger than the noise gives a positive SNR.
Question 34 of 60 · Computer Networks · 4–6 items
In the CIA triad of information security, what does the 'I' represent?
Solution
The CIA triad: confidentiality, integrity, availability.
Identification belongs to access control (identification, authentication, authorization), not the CIA triad.
Intrusion is a threat event that the triad protects against, not one of its properties.
CIA = confidentiality, integrity, availability; integrity means data is not altered without authorization.
Implementation is not a security property at all.
Question 35 of 60 · Computer Networks · 4–6 items
Which network topology provides the highest fault tolerance but requires the most cabling?
Solution
Full mesh: redundant paths between all nodes; links = n(n − 1)/2.
A star survives a broken edge link, but the central switch is a single point of failure.
A bus uses the least cabling, and one break in the backbone takes the whole segment down.
A single ring fails when one link or node breaks, unless a second ring is added.
In a full mesh every pair of nodes has its own link, so traffic reroutes around any failure, at the cost of n(n − 1)/2 cables.
Question 36 of 60 · Digital Systems · 8–12 items
A 3-bit shift register with outputs Q2 Q1 Q0 shifts right on each clock (Q2 → Q1 → Q0), and the bit entering Q2 is Q2 ⊕ Q0. If the present state is Q2 Q1 Q0 = 101, what is the next state?
Given
Present state Q2 Q1 Q0 = 101
Shift direction Q2 → Q1 → Q0
Feedback into Q2 = Q2 ⊕ Q0
Find
Next state Q2 Q1 Q0
Solution
f = Q2 ⊕ Q0 = 1 ⊕ 1 = 0
Q2′ = f = 0, Q1′ = Q2 = 1, Q0′ = Q1 = 0 → 010
This writes the feedback bit into Q2 but does not shift: Q1 and Q0 keep their old values.
Feedback = Q2 ⊕ Q0 = 1 ⊕ 1 = 0 enters Q2; Q1 takes the old Q2 (1) and Q0 the old Q1 (0): 010.
This shifts toward Q2 (Q2 ← Q1, Q1 ← Q0) and recirculates the old Q2 into Q0, a plain rotation with no XOR.
This feeds Q0 straight back into Q2 (a right rotation) or uses XNOR: either way the new Q2 is 1, but 1 ⊕ 1 = 0.
Question 37 of 60 · Digital Systems · 8–12 items
A synchronous state machine cycles S0 → S1 → S2 → S3 → S0 using two flip-flops. Which state assignment changes only one bit on every transition (a Gray-code assignment)?
Solution
Gray code: successive states differ in one bit, including the wrap S3 → S0.
00 → 01 → 11 → 10 → 00
Gray sequence: each step, including S3 → S0 (10 → 00), changes exactly one bit.
This is plain binary counting: 01 → 10 changes both bits, and so does 11 → 00 on the wrap.
S0 → S1 (00 → 11) and S2 → S3 (01 → 10) each change two bits.
S1 → S2 (10 → 01) changes both bits.
Question 38 of 60 · Digital Systems · 8–12 items
A flip-flop requires a 4 ns setup time and a 1 ns hold time. Its data input changes 3 ns before the active clock edge and changes again 0.5 ns after the edge. Which timing requirements are violated?
Given
Data change before clock edge = 3 ns
Data stable after clock edge = 0.5 ns
Setup time = 4 ns
Hold time = 1 ns
Find
Whether timing is violated
Solution
tsetup,avail = 3 ns < 4 ns → setup violated
thold,avail = 0.5 ns < 1 ns → hold violated
Both windows are broken: 3 ns < 4 ns of setup and 0.5 ns < 1 ns of hold.
The data changed 3 ns before the edge (less than the 4 ns setup time) and again 0.5 ns after it (less than the 1 ns hold time).
Setup is broken (3 ns < 4 ns), and the change 0.5 ns after the edge also falls inside the 1 ns hold window.
Hold is broken (0.5 ns < 1 ns), but setup is broken too: the data arrived only 3 ns before the edge, not 4 ns.
Question 39 of 60 · Digital Systems · 8–12 items
Convert the binary number 11010110 to hexadecimal.
Given
Binary number = 11010110
Find
Hexadecimal equivalent
Solution
1101 0110 → D 6 → D616 (= 21410)
This maps 1101 (13) to E; in hexadecimal, 13 is D and 14 is E.
This is the decimal value of the number, not its hexadecimal form.
Group into nibbles from the right: 1101 = D, 0110 = 6, so 11010110 = D6.
This reverses the bit order before grouping (01101011); grouping starts from the right of the original word.
Question 40 of 60 · Digital Systems · 8–12 items
What is the minimum number of 2-input NAND gates needed to implement a 2-input XNOR function using only 2-input NAND gates?
Solution
XOR = 4 NAND gates (N1 = (XY)', N2 = (XN1)', N3 = (YN1)', N4 = (N2N3)'); a fifth NAND as an inverter gives XNOR.
total = 4 + 1 = 5
Three 2-input NAND gates cannot produce XOR or XNOR; the minimum for XOR alone is four.
Four NAND gates make XOR; XNOR needs one more gate to invert it.
XOR takes four NAND gates; a fifth NAND with tied inputs inverts it to XNOR, so 5 gates in total.
Six gates work but are not the minimum.
Question 41 of 60 · Computer Systems · 5–8 items
A 4-way set-associative cache has 64 sets and a block size of 32 bytes. What is the total cache capacity?
Given
Associativity = 4-way
Number of sets = 64
Block size = 32 bytes
Find
Total cache capacity (KB)
Solution
C = ways × sets × block size
C = 4 × 64 × 32 B = 8192 B = 8 KB
This reads the 32-byte block as 32 bits, giving 8192 bits = 1 KB.
This is sets × block size (64 × 32 B) and leaves out the 4 ways.
Capacity = ways × sets × block size = 4 × 64 × 32 B = 8192 B = 8 KB.
This multiplies by 8 for bits but keeps the unit as bytes.
Question 42 of 60 · Computer Systems · 5–8 items
A 32-bit wide data bus operates at 100 MHz with one transfer per clock cycle. What is the maximum theoretical data transfer rate?
Given
Bus width = 32 bits
Bus clock f = 100 MHz, one transfer per cycle
Find
Maximum data transfer rate (MB/s)
Solution
rate = width × f = (32 bits/8) × 100 × 106 /s
rate = 4 B × 100 × 106 /s = 400 MB/s
This moves one byte per clock; a 32-bit bus moves 4 bytes per clock.
This assumes two transfers per clock (double data rate); the stem specifies one transfer per cycle.
This is 3200 Mbit/s labeled as megabytes; divide by 8 to get bytes.
Question 43 of 60 · Computer Systems · 5–8 items
An embedded real-time system must service an interrupt within 50 µs. The interrupt service routine (ISR) takes 200 clock cycles, and the processor runs at 8 MHz. Does the system meet the timing requirement?
Given
Deadline = 50 µs
ISR length = 200 clock cycles
Clock frequency f = 8 MHz
Find
ISR execution time (µs) and whether the deadline is met
Solution
t = N/f = 200/(8 × 106 Hz) = 25 µs
25 µs < 50 µs → the deadline is met
This uses a 12.5 ns clock period, ten times too short; the deadline is met, but not with this time.
This uses a 1.25 µs clock period, ten times too long; 1/8 MHz = 0.125 µs.
This multiplies the cycle count by the clock frequency (200 × 8) instead of dividing by it.
t = 200 cycles/(8 × 106 cycles/s) = 25 µs, which is inside the 50 µs deadline.
Question 44 of 60 · Software Engineering · 4–6 items
In a singly linked list, what is the time complexity to insert a new node at the beginning of the list?
Solution
Insert at head: new.next = head; head = new → O(1).
The new node points to the old head and becomes the head: a fixed number of steps.
O(n) is the cost of inserting at the tail without a tail pointer, or of inserting at the front of an array.
No step here depends on n at all, let alone n2.
Logarithmic cost comes from halving a search space, which insertion at the head never does.
Question 45 of 60 · Software Engineering · 4–6 items
A recursive function calls itself n times before reaching the base case, with each call performing a constant amount of work. What is the time complexity?
Solution
T(n) = T(n − 1) + c → T(n) = cn = O(n)
Logarithmic depth happens when each call halves the problem, not when it reduces it by one.
Quadratic time needs O(n) work inside each of the n calls.
Constant time would need a fixed number of calls; here the number of calls grows with n.
n calls × O(1) work each = O(n).
Question 46 of 60 · Software Engineering · 4–6 items
In Big O notation, which function grows the fastest as n approaches infinity?
Solution
n < n log n < n2 < 2n for large n
Linear is the slowest of the listed growth rates.
Polynomial growth is eventually overtaken by any exponential.
Exponential growth eventually exceeds any polynomial: 2n outgrows n2, n log n, and n.
n log n grows only slightly faster than n.
Question 47 of 60 · Mathematics · 11–17 items
What are the Fourier sine coefficients bn of f(x) = x on (−π, π), extended as an odd 2π-periodic function?
Given
Function f(x) = x on (−π, π)
Odd 2π-periodic extension
Find
Fourier coefficients bn
Solution
bn = (1/π)∫−ππ x sin(nx) dx = (2/π)∫0π x sin(nx) dx
This is the Laplacian ∇2f = 2 + 2 + 2, a scalar; the gradient is a vector.
This uses the function value f(1, 1, 1) = 3 as each component instead of the partial derivatives.
∇f = 2xi + 2yj + 2zk; at (1, 1, 1) this is 2i + 2j + 2k.
This drops the factor 2 from d(x2)/dx = 2x.
Question 53 of 60 · Probability and Statistics · 4–6 items
An engineer claims that the mean breaking strength of a cable is at least 5000 N. A sample of 40 cables has mean 4920 N and standard deviation 200 N. Using the z-approximation at the 1 % significance level, what is the conclusion?
This divides s by n instead of √n, so z = −16 is far too extreme.
This doubles the p-value for a two-tailed test; the claim "at least 5000 N" makes the test one-tailed.
This uses z = −80/200 with s instead of the standard error s/√n.
z = (4920 − 5000)/(200/√40) = −2.53; the lower-tail p = 0.0057 < 0.01, so the claim is rejected.
Question 54 of 60 · Probability and Statistics · 4–6 items
A quality engineer compares two production lines. Line A has 14 defectives in 200 inspected; line B has 5 in 180. What is the 95 % confidence interval for pA − pB (z = 1.96)?
This uses z = 1.645, the 90 % value, for a 95 % interval.
This subtracts the two variances under the square root; variances of a difference add.
This centers the margin on zero; the interval is centered on the observed difference of 4.22 %.
Question 55 of 60 · Ethics and Professional Practice · 4–6 items
When working on electrical equipment, the primary purpose of Personal Protective Equipment (PPE) such as arc-rated clothing and face shields is to:
Solution
Arc-rated PPE is chosen by the incident energy of an arc flash (cal/cm2); shock protection is a separate measure.
Arc-rated PPE is not designed to block electromagnetic fields or RF radiation.
Shock protection comes from de-energizing, insulation, and rubber insulating gloves, not from arc-rated clothing.
Insulating mats and rubber gloves do that; arc-rated fabric and face shields are not rated as shock insulation.
Arc-rated clothing and face shields are rated by the incident energy they can block, to prevent burns from an arc flash.
Question 56 of 60 · Ethics and Professional Practice · 4–6 items
According to OSHA regulations and electrical safety standards, proper grounding of electrical equipment is primarily intended to:
Solution
Equipment grounding: a low-impedance fault path so the overcurrent device opens, and exposed metal stays at a safe touch voltage.
A low-impedance path lets fault current trip the breaker quickly and keeps exposed metal near ground potential.
Harmonics are handled with filters and equipment design; the equipment ground is not a harmonic remedy.
Shield grounding can help with noise, but that is a secondary use, not the safety purpose of equipment grounding.
The grounding conductor carries no current in normal operation, so it does not change efficiency.
Question 57 of 60 · Engineering Economics · 5–8 items
An investment of $20,000 returns $25,000 after 3 years. What is the approximate annual rate of return?
Given
Investment P = $20,000
Return F = $25,000
Time n = 3 years
Find
Annual rate of return i (%)
Solution
F = P(1 + i)n → i = (F/P)1/n − 1
i = (25,000/20,000)1/3 − 1 = 1.0772 − 1 = 7.7 %
This divides the $5,000 gain by the $25,000 return and then by 3 years; the base is the $20,000 invested.
This is ln(1.25)/3, the continuously compounded rate, not the effective annual rate.
F = P(1 + i)n: (1 + i)3 = 25,000/20,000 = 1.25, so i = 1.251/3 − 1 = 7.7 %.
This is simple interest, 25 %/3; compounding makes the annual rate lower.
Question 58 of 60 · Engineering Economics · 5–8 items
What is the future value of $1,000 deposited today if the nominal annual interest rate is 12% compounded quarterly for 2 years?
Given
Deposit P = $1,000
Nominal annual rate r = 12%, compounded quarterly
Period = 2 years
Find
Future value ($)
Solution
i = r/m = 0.12/4 = 0.03, n = m × years = 8
F = P(1 + i)n = $1,000 × 1.038 = $1,267
This compounds 3 % for only 4 quarters (one year): 1000 × 1.034 = $1,126.
This is simple interest, 1000 × (1 + 0.12 × 2), with no compounding.
This compounds annually at 12 % for 2 years; quarterly compounding uses 3 % for 8 periods.
i = 12 %/4 = 3 % per quarter, n = 8: F = 1000 × 1.038 = $1,267.
Question 59 of 60 · Engineering Economics · 5–8 items
What is the present worth of a perpetual annual payment of $5,000 (first payment at the end of year 1) at 10 % interest per year?
Given
Perpetual annual payment A = $5,000
Interest rate i = 10 % per year
Find
Present worth ($)
Solution
P = A/i
P = $5,000/0.10 = $50,000
This adds a payment at time zero (A/i + A); the end-of-year perpetuity has no payment today.
For a perpetuity, P = A/i = 5,000/0.10 = $50,000.
This is one payment plus one year of interest, A(1 + i), not the value of an endless series.
This multiplies the payment by the rate; a perpetuity divides by it.
Question 60 of 60 · Power Systems · 8–12 items
A balanced 480 V line-to-line, 60 Hz, four-wire supply feeds an unbalanced wye load with the neutral connected. The phase impedances are ZA = 5 + j10 Ω, ZB = 10 − j5 Ω, and ZC = 8 + j6 Ω. What total real power does the load absorb?
Given
Line voltage VLL = 480 V, 60 Hz, four-wire (neutral connected)
Phase impedance ZA = 5 + j10 Ω
Phase impedance ZB = 10 − j5 Ω
Phase impedance ZC = 8 + j6 Ω
Find
Total real power P (W)
Solution
Vph = 480/√3 = 277.1 V, Pk = |Vph|2Rk/|Zk|2
P = 76,800 × (5/125 + 10/125 + 8/100) = 3072 + 6144 + 6144 = 15,360 W
This adds |V|2/|Z| for each phase, which is apparent power in VA, not real power.
This uses |V|2/R per phase; the current is set by |Z|, so P = I2R = |V|2R/|Z|2.
This uses the 480 V line-to-line voltage across each wye phase, three times the power.
Answer every question to see your score
How this set is built
Each area gets questions in proportion to the midpoint of its range. Circuit Analysis and Mathematics, the two widest areas at 11–17 each, get the most; most 4–6 areas get two each (software gets three).
47 discipline problems: circuits 7, power 6, electronics 5, digital 5, control 4, then 2–3 each for linear systems, signal processing, communications, computer systems, software, networks, electromagnetics and electrical materials.
13 problems on the shared FE areas: mathematics 6, engineering economics 3, probability and statistics 2, ethics 2, written for an electrical reader. More of those are on the FE practice test and on engineering economics.
Use the section chips to drill one area, then go to its page for more problems of the same kind.
How it differs from exam day
Length and pace: the real exam gives 5 h 20 min for 110 questions, a little under 3 minutes each, with the first half locked once you submit it before the break.
Formats:multiple choice plus alternative item types (multiple correct, point and click, drag and drop, fill in the blank). No partial credit on alternative items. Every problem here is single-answer multiple choice.
Reference: on test day you look formulas up in the FE Reference Handbook PDF with a search box, not Ctrl+F. Practice with the same PDF open (how to get and search it).
NCEES sells two practice exams for this discipline (2 × 50 questions, $50 each). They are written by the exam's owner; this page is independent of it.
Calculator mode costs points
Phasors here come with angles in degrees, and digital problems with binary or hex numbers. Set the calculator for complex numbers and degrees before you start, and practice on the exact approved model you will bring (rule:NCEES-approved models only; model list).
No. It is an independent set. NCEES sells two interactive practice exams for this discipline, Vol. 1 and Vol. 2 (2 × 50 questions, $50 each as of Oct 2026).
How close is this to the real FE electrical exam?
It follows the July 2020 specification areas and ranges, but it is about half the length and has no alternative item types, so a score here is a diagnosis by area, not a prediction.
Is there a PDF of these questions?
No. The problems and full solutions are on this page, and each section page adds more of the same area.
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FE Electrical and Computer questions, answers and worked solutions in a study app.