FRIED ENGINEERS Β· FE PRACTICE

FE Exam Practice Questions

Practice realistic FE-style engineering questions with step-by-step reasoning and detailed solutions. Choose your discipline or start solving now.

FE-style problemsβ€’Reasoning explainedβ€’Solutions included
FE QUESTION BANK

Available FE practice questions

Choose a question and solve it before reading the final engineering verdict.

QUESTION01
E-115Mechanics Of Materials

The Plate That Peaked at 92.4 MPa, Not 110

Adding the shear stress directly to the larger normal stress gave 110 MPa. Principal stress comes from Mohr-circle geometry and was only 92.4 MPa.

Principal Stresses5–10 min
Solve β†’
QUESTION02
E-114Mechanics Of Materials

The Shaft That Reached 50.9 MPa, Not 25.5

The polar moment was correct, but shear stress was evaluated at half the outer radius. Maximum torsional stress occurs at the outside surface and was 50.9 MPa.

Torsional Shear Stress5–10 min
Solve β†’
QUESTION03
E-113Mechanics Of Materials

The Bar That Stretched 0.80 mm, Not 0.40

The stress was correct, but only half the member length was used in the deformation equation. The full bar stretched 0.80 mm.

Axial Deformation5–10 min
Solve β†’
QUESTION04
E-112Water Resources Environmental Engineering

The Aquifer That Moved 1,500 ftΒ³/day, Not 72,000

The hydraulic gradient was correct, but the plan-view area was used instead of the area perpendicular to groundwater flow. The aquifer moved 1,500 ftΒ³/day.

Darcy Groundwater Flow5–10 min
Solve β†’
QUESTION05
E-111Water Resources Environmental Engineering

The Basin That Needed 12,600 ftΒ³, Not 21,600

The basin received 21,600 ftΒ³ during the design interval, but 9,000 ftΒ³ left through the outlet at the same time. Required storage was only the net 12,600 ftΒ³.

Stormwater Detention Storage5–10 min
Solve β†’
QUESTION06
E-110Water Resources Environmental Engineering

The Watershed That Sent 60.5 cfs, Not 100.8

The rainfall intensity and drainage area were correct, but the runoff coefficient was omitted. The design peak runoff was 60.5 cfs, not 100.8 cfs.

Rational Method Runoff5–10 min
Solve β†’
QUESTION07
E-109Construction Engineering

The Slab Order That Needed 80 ydΒ³, Not 74.1

The geometric concrete volume was 74.1 ydΒ³, but the purchase order had to include 8% waste. The correct order quantity was 80 ydΒ³.

Construction Quantity Takeoff5–10 min
Solve β†’
QUESTION08
E-108Construction Engineering

The Project With CPI 0.75, Not 1.33

Actual cost was divided by earned value, producing 1.33 and a false impression of good performance. CPI is EV/AC, so the project was only at 0.75.

Earned Value Management5–10 min
Solve β†’
QUESTION09
E-107Construction Engineering

The Schedule That Finished in 16 Days, Not 12

The shorter branch finished in 12 days, but project completion waited on the longer A-B-D-F path. The correct duration was 16 days.

Critical Path Method5–10 min
Solve β†’
QUESTION10
E-106Transportation Engineering

The PM Peak That Sent 199 Cars In, Not 306

The project generated 306 total trips in the PM peak hour, but only 65% were entering. The inbound design demand was about 199 veh/h.

Trip Generation Directional Split5–10 min
Solve β†’
QUESTION11
E-105Transportation Engineering

The Lane That Carried 1,800 veh/h, Not 1,227

The spacing was 132 ft per vehicle, but 45 mph was treated as 45 ft/s. Correct unit conversion gives density 40 veh/mi and flow 1,800 veh/h/lane.

Traffic Flow Fundamental Relation5–10 min
Solve β†’
QUESTION12
E-104Transportation Engineering

The Curve That Needed 833 ft, Not 1,190

The design speed was correct, but superelevation was omitted. Including both superelevation and side friction reduced the required curve radius to about 833 ft.

Horizontal Curve Radius5–10 min
Solve β†’
QUESTION13
E-103Geotechnical Engineering

The Footing That Held 614 kPa, Not 331

Using only the surcharge term gave 331 kPa. The width-dependent soil-weight term raised the gross ultimate bearing capacity to about 614 kPa.

Bearing Capacity5–10 min
Solve β†’
QUESTION14
E-102Geotechnical Engineering

The Soil at 7 m That Carried 94.8 kPa, Not 134

Total overburden at 7 m was 134 kPa, but pore water carried 39.2 kPa. The soil skeleton carried only 94.8 kPa effective stress.

Effective Stress5–10 min
Solve β†’
QUESTION15
E-101Geotechnical Engineering

The Soil Sample With e = 0.80, Not 0.50

Using wet density as dry density produced e=0.50. Correcting for 20% water content gave e=0.80 and 67.5% saturation.

Soil Phase Relations5–10 min
Solve β†’
QUESTION16
E-100Materials

The Stronger Material That Made the Heavier Tie Rod

Material A had the higher yield strength, but its density penalty was much larger. After sizing both rods to the same factor of safety, Material B produced the lighter design.

Material Selection Strength To Density5–10 min
Solve β†’
QUESTION17
E-099Materials

The Wire That Had 0.17 Ohm, Not 170

The resistivity formula was correct, but 2 mmΒ² was treated as 2Γ—10^-9 mΒ² instead of 2Γ—10^-6 mΒ². The wire resistance was 0.17 ohm.

Electrical Resistivity5–10 min
Solve β†’
QUESTION18
E-098Materials

The 2 m Bar That Grew 2.3 mm, Not 1.15

The thermal strain was correct, but it was applied as if the bar were only 1 m long. The 2 m bar expanded 2.3 mm.

Linear Thermal Expansion5–10 min
Solve β†’
QUESTION19
E-097Materials

The Alloy That Was 40% Beta, Not 57%

Dividing the overall composition by the beta composition gave 57%, but phase fractions come from tie-line segment lengths. The beta fraction was 40%.

Lever Rule Phase Fractions5–10 min
Solve β†’
QUESTION20
E-096Engineering Economics

The Risky Project Worth $52,000, Not $35,000

The outcomes were not equally likely. Probability-weighting gives an expected value of $52,000, not the simple average of $35,000.

Expected Monetary Value5–10 min
Solve β†’
QUESTION21
E-095Engineering Economics

The Process Choice That Broke Even at 10,000 Units, Not 5,000

The fixed-cost difference was correct, but break-even depends on the difference in variable costs. The correct crossover is 10,000 units.

Break Even Analysis5–10 min
Solve β†’
QUESTION22
E-094Engineering Economics

The Upgrade Worth $3,675, Not $25,000

Undiscounted cash flows suggested a $25,000 gain; discounted NPV is about $3,675.

Present Worth5–10 min
Solve β†’
QUESTION23
E-093Safety Health Environment

The Relief Valve That Saw 480 N, Not 560

The vessel pressure was 700 kPa absolute, but the valve disk also had atmospheric pressure acting on the opposite side. Net pressure force depended on the 600 kPa differential.

Pressure Difference Force On Valve5–10 min
Solve β†’
QUESTION24
E-092Safety Health Environment

The Source That Fell to 30, Not 60

Eighteen hours represented three half-lives, not two. A 240-unit rate therefore fell to 30, not 60.

Radioactive Decay Half Life5–10 min
Solve β†’
QUESTION25
E-091Safety Health Environment

The Confined Space That Needed 18 mΒ³/min, Not 1,080

Six air changes per hour sounded like 1,080 mΒ³/min when the hour-to-minute conversion was skipped. The required flow was 18 mΒ³/min.

Confined Space Ventilation Air Changes5–10 min
Solve β†’
QUESTION26
E-090Quality

The 5% Defect Lot That Was Accepted 73.6% of the Time, Not 95%

A 5% defective process did not imply a 95% lot-acceptance probability. Under an n=20, c=1 plan, acceptance required observing zero or one defective item.

Single Sampling Plan Binomial Acceptance5–10 min
Solve β†’
QUESTION27
E-089Quality

The Part That Lost $45, Not $60

Quality loss did not grow linearly with deviation from target. Under the Taguchi quadratic model, a 3-unit deviation caused a $45 loss.

Taguchi Quality Loss5–10 min
Solve β†’
QUESTION28
E-088Quality

The Process With Cp 1.67 That Still Had Cpk 0.67

The process spread fit comfortably inside the specification width, but the mean had shifted toward the USL. Cp looked strong while Cpk exposed the centering problem.

Process Capability Cp Cpk5–10 min
Solve β†’
QUESTION29
E-087Quality

The Sample Mean That Crossed 53, Even Though 53.5 Looked Safe

The individual-process spread looked wide enough to contain 53.5, but an xΜ„ chart uses the standard error of the sample mean. The upper control limit was 53.0.

Xbar Control Chart Known Sigma5–10 min
Solve β†’
QUESTION30
E-086Facilities Supply Chain

The Higher-Rate Distribution Center That Cost $10,000 Less

Site A had the higher simple average freight rate, but most demand was on its low-cost lane. Demand-weighted annual cost made Site A $10,000 cheaper.

Demand Weighted Transportation Cost5–10 min
Solve β†’
QUESTION31
E-085Facilities Supply Chain

The Cell That Needed 4 Machines, Not 3

Three machines looked sufficient at 100% theoretical availability. At the planned 80% utilization, the cell required four machines.

Capacity Analysis Number Of Machines5–10 min
Solve β†’
QUESTION32
E-084Facilities Supply Chain

The Shorter Layout That Moved More Material

Layout A had the shorter average distance, but its busiest material-flow pair was much farther apart. Weighted load-distance made Layout B the better design.

Load Distance Facility Layout5–10 min
Solve β†’
QUESTION33
E-083Manufacturing Service Production Systems

The Inventory Policy That Ordered 707 Units, Not 10,000

Ordering the full annual demand once minimized ordering frequency, not total inventory cost. The EOQ was about 707 units.

Economic Order Quantity5–10 min
Solve β†’
QUESTION34
E-082Manufacturing Service Production Systems

The Forecast That Moved to 109, Not 130

The latest actual demand was 130 units, but exponential smoothing with Ξ±=0.30 did not replace the forecast with 130. The next forecast became 109.

Single Exponential Smoothing5–10 min
Solve β†’
QUESTION35
E-081Manufacturing Service Production Systems

The Assembly Line That Needed 4 Stations, Not 3.6

The arithmetic produced 3.6 stations, but a production line cannot operate with a fraction of a station. The theoretical minimum was 4.

Line Balancing Cycle Time Minimum Workstations5–10 min
Solve β†’
QUESTION36
E-080Solid Hazardous Waste

The Mixed Waste Batch That Was 5.67%, Not 4.5%

Averaging 8% and 1% gave 4.5%, but the two waste streams had different masses. The correct mixed concentration was 5.67% by mass.

Waste Mixture Component Mass Balance5–10 min
Solve β†’
QUESTION37
E-079Solid Hazardous Waste

The Incinerator That Achieved 99.9%, Not 99%

An outlet of 0.025 kg/hr from a 25 kg/hr hazardous constituent feed means only 0.1% remains. The destruction/removal efficiency is 99.9%.

Destruction Removal Efficiency5–10 min
Solve β†’
QUESTION38
E-078Solid Hazardous Waste

The Landfill That Produced 12,000 mΒ³ of Leachate, Not 22,000

Precipitation minus evapotranspiration was not enough. Runoff and storage change reduced the annual leachate estimate to 12,000 mΒ³.

Landfill Water Balance Leachate5–10 min
Solve β†’
QUESTION39
E-077Solid Hazardous Waste

The Landfill That Lasted 3.42 Years, Not 4.11

Ignoring daily cover made the landfill appear to last 4.11 years. Including 20% cover-volume use reduced the life to about 3.42 years.

Landfill Airspace And Cover Volume5–10 min
Solve β†’
QUESTION40
E-076Air Quality Control

The Two Collectors That Removed 94%, Not 150%

A 70% cyclone followed by an 80% baghouse did not provide 150% removal. The second device treated only what escaped the first, giving 94% overall removal.

Series Particle Collection Efficiency5–10 min
Solve β†’
QUESTION41
E-075Air Quality Control

The Room That Stabilized at 6.5 mg/mΒ³, Not 0.2

Dividing generation by room volume produced a concentration rise rate. The steady indoor concentration was controlled by ventilation and reached 6.5 mg/mΒ³.

Well Mixed Indoor Air Steady State5–10 min
Solve β†’
QUESTION42
E-074Air Quality Control

The Scrubber That Left 4 lb/hr, Not 36

The scrubber removed 36 lb/hr, but that was not the stack emission. Only 4 lb/hr remained after 90% control.

Emission Factor And Control Efficiency5–10 min
Solve β†’
QUESTION43
E-073Water Wastewater

The Contact Tank That Had Only 21 Effective Minutes

The hydraulic detention time was 30 minutes, but baffling reduced the effective contact time to 21 minutes. The CT value was 31.5, not 45.

Disinfection Effective Contact Time Ct5–10 min
Solve β†’
QUESTION44
E-072Water Wastewater

The Clarifier System That Needed 12,000 ftΒ², Not 6,000

Splitting the flow between two clarifiers did not halve the total required area. The system still needed 12,000 ftΒ² in service.

Clarifier Surface Overflow Rate5–10 min
Solve β†’
QUESTION45
E-071Water Wastewater

The Treatment Plant That Removed 4,379 lb/day, Not 210

The concentration drop was 210 mg/L, but that was not a mass removal rate. At 2.5 MGD, the plant removed about 4,379 lb/day of BOD.

Mass Loading And Removal Efficiency5–10 min
Solve β†’
QUESTION46
E-070Chemical Reaction Engineering

The 70% Conversion That Produced Only 50% Yield

Seventy percent of A reacted, but only fifty percent of the feed became the desired product B. Conversion was not the same as yield.

Conversion Yield Selectivity Parallel Reactions5–10 min
Solve β†’
QUESTION47
E-069Chemical Reaction Engineering

The Rate Constant That Increased Sixfold at 330 K

A 30 K temperature rise did not cause a 10% kinetic change. The Arrhenius relation increased the rate constant by a factor of about 6.19.

Arrhenius Temperature Dependence5–10 min
Solve β†’
QUESTION48
E-068Chemical Reaction Engineering

The CSTR That Needed 600 L, Not 367

The PFR-sized volume looked efficient, but the well-mixed CSTR needed 600 L to reach the same 60% conversion.

First Order Cstr Design Equation5–10 min
Solve β†’
QUESTION49
E-067Chemical Reaction Engineering

The Batch Reactor That Reached 63%, Not 100%

The shortcut kt=1 suggested 100% conversion after 5 minutes. The first-order batch model gave only 63.2%.

First Order Batch Reactor Kinetics5–10 min
Solve β†’
QUESTION50
E-066Thermodynamics

The Methane Flame That Released 890 kJ per Mol

The reactant methane formation enthalpy was only -74.8 kJ/mol. The balanced combustion released nearly 890 kJ/mol.

Standard Heat Of Reaction5–10 min
Solve β†’
QUESTION51
E-065Thermodynamics

The Compressor That Heated the Gas to 446 K

No heater was attached, yet adiabatic reversible compression raised the ideal-gas temperature from 300 K to about 446 K.

Isentropic Compression Ideal Gas5–10 min
Solve β†’
QUESTION52
E-064Thermodynamics

The Liquid Mixture That Started Boiling at 64 kPa

A binary liquid did not boil when total pressure equaled either pure-component vapor pressure. Its bubble pressure was the composition-weighted sum.

Vapor Liquid Equilibrium Raoults Law5–10 min
Solve β†’
QUESTION53
E-063Material Energy Balances

The Heater That Needed 418 kW, Not 209

Heating the liquid required 209 kJ/kg. At 2 kg/s, that became 418 kW of heat transfer.

Steady Flow Energy Balance Sensible Heating5–10 min
Solve β†’
QUESTION54
E-062Material Energy Balances

The Evaporator That Removed 750 kg/h of Water

The feed was 1000 kg/h at 10 wt% solids. Concentrating to 40 wt% left only 250 kg/h liquid product.

Evaporator Nonvolatile Solids Balance5–10 min
Solve β†’
QUESTION55
E-061Material Energy Balances

The Mixer That Needed 100 kg/h of Concentrate

A 10 wt% stream and a 40 wt% stream had to produce 25 wt% product. The concentrate flow was 100 kg/h, not 25.

Steady State Component Balance5–10 min
Solve β†’
QUESTION56
E-060Mechanical Design Analysis

The 30 kW Shaft That Needed 239 NΒ·m of Torque

The shaft power was 30 kW at 1200 rpm. That meant about 239 NΒ·m torque, not 30 NΒ·m.

Power Transmission Torque Speed5–10 min
Solve β†’
QUESTION57
E-059Mechanical Design Analysis

The Thin-Wall Vessel That Saw 100 MPa Hoop Stress

At 2 MPa internal pressure, the vessel's hoop stress reached 100 MPaβ€”twice the longitudinal stress.

Thin Wall Pressure Vessel Stress5–10 min
Solve β†’
QUESTION58
E-058Mechanical Design Analysis

The Spring That Deflected 25.6 mm Under 500 N

The spring force was 500 N, but the deflection depended strongly on wire diameter, coil diameter, and active coils.

Helical Compression Spring Rate5–10 min
Solve β†’
QUESTION59
E-057Mechanical Design Analysis

The 50 mm Shaft That Saw 81.5 MPa of Shear

The torque was only 2 kNΒ·m, but the outer surface of the 50 mm shaft reached about 81.5 MPa shear stress.

Machine Element Torsional Stress5–10 min
Solve β†’
QUESTION60
E-056Fluid Mechanics

The Pump That Needed 14 kW to Deliver 9.81 kW to the Water

The water received 9.81 kW. A 70% efficient pump required about 14.0 kW of shaft input.

Pump Hydraulic Power Efficiency5–10 min
Solve β†’
QUESTION61
E-055Fluid Mechanics

The 50 m Pipe That Lost 20 kPa

The velocity stayed at 2 m/s, yet wall friction consumed 20 kPa of pressure over the pipe length.

Darcy Weisbach Head Loss5–10 min
Solve β†’
QUESTION62
E-054Fluid Mechanics

The Fast-Looking Oil Flow That Was Still Laminar

The oil was moving at 2 m/s, but its Reynolds number was only 1700.

Reynolds Number Flow Regime5–10 min
Solve β†’
QUESTION63
E-053Thermodynamics

The Refrigerator With a COP Greater Than One

The refrigerator removed 6 kW of heat using 2 kW of work. A COP of 3 was correct, not impossible.

Refrigerator Cop5–10 min
Solve β†’
QUESTION64
E-052Thermodynamics

The Turbine That Produced 1.2 MW From an Enthalpy Drop

The turbine power came from a 600 kJ/kg enthalpy drop, not from subtracting internal energies blindly.

Steady Flow Energy Equation5–10 min
Solve β†’
QUESTION65
E-051Thermodynamics

The Sealed Tank That Reached 150 kPa, Not 133

The gas temperature rose from 300 K to 450 K. The pressure did not scale with Celsius temperature.

Ideal Gas Constant Volume5–10 min
Solve β†’
QUESTION66
E-050Fluid Mechanics Hydraulics

The Channel That Carried 5.69 Cubic Meters per Second

The flow area alone was not enough. Hydraulic radius and slope set the channel discharge.

Manning Open Channel Flow5–10 min
Solve β†’
QUESTION67
E-049Fluid Mechanics Hydraulics

The Faster Pipe Section That Lost 16 kPa of Pressure

The pipe stayed level. Velocity increased from 2 to 6 m/s, and pressure fell from 300 to 284 kPa.

Bernoulli Equation5–10 min
Solve β†’
QUESTION68
E-048Fluid Mechanics Hydraulics

The Pipe That Halved Its Diameter and Quadrupled Its Velocity

The diameter fell by half. The velocity did not merely doubleβ€”it rose from 1.5 to 6.0 m/s.

Continuity Equation5–10 min
Solve β†’
QUESTION69
E-047Fluid Mechanics Hydraulics

The Tank Gauge That Read 58.9 kPa at 6 m

The water was motionless, but pressure still increased with depth.

Fluid Statics Hydrostatic Pressure5–10 min
Solve β†’
QUESTION70
E-046Structural Engineering

The Column That Buckled Near 497 kips

The steel strength was not the first limit. Slender-column instability governed near 497 kips.

Euler Buckling5–10 min
Solve β†’
QUESTION71
E-045Structural Engineering

The Beam Moment That Peaked at 75 kip-ft, Not 300

The total load was 30 kips. Multiplying by half the span produced the wrong bending moment.

Simply Supported Beam Maximum Moment5–10 min
Solve β†’
QUESTION72
E-044Structural Engineering

The Floor Beam That Was Loaded at 400 lb/ft, Not 50

The floor load was 50 psf. The beam did not carry 50 lb/ftβ€”it carried 400 lb/ft.

Tributary Area Load Conversion5–10 min
Solve β†’
QUESTION73
E-043Statics

The Truss Diagonal That Carried More Than the Support Reaction

Each support reaction was only 6 kN, yet each inclined top member carried about 8.49 kN.

Truss Method Of Joints5–10 min
Solve β†’
QUESTION74
E-042Statics

The Beam Reactions That Refused to Split 50-50

The total load was 28 kN, but the support reactions were not 14 kN each.

Rigid Body Equilibrium Support Reactions5–10 min
Solve β†’
QUESTION75
E-041Statics

The Two Cables That Each Carried 10 kN

A 10 kN load was shared by two cables, but each cable still carried 10 kN of tension.

Force Equilibrium Cable Tension5–10 min
Solve β†’
QUESTION76
E-040Computer Networks

The Router That Ignored the Shorter-Looking Route

Two routes matched the destination. The router chose the more specific prefix, not the visually simpler route.

Longest Prefix Match5–10 min
Solve β†’
QUESTION77
E-039Computer Networks

The 1500-Byte Packet That Did Not Arrive in 12 Microseconds

Serialization took only 12 microseconds. The fiber path added another 500 microseconds.

Network Delay Components5–10 min
Solve β†’
QUESTION78
E-038Computer Networks

The /27 Network That Could Not Fit 40 Devices

The subnet looked large enough from the address range. The usable host count said otherwise.

Ipv4 Subnetting Host Capacity5–10 min
Solve β†’
QUESTION79
E-037Communications

The Four-Channel FDM Link That Needed 95 kHz

Four 20 kHz channels looked like an 80 kHz jobβ€”until the guard bands were counted.

Frequency Division Multiplexing Bandwidth5–10 min
Solve β†’
QUESTION80
E-036Communications

The PCM Link That Needed 64 kbps, Not 32

The voice channel bandwidth was 4 kHz. The data link was sized at 32 kbps and immediately overflowed.

Pcm Bit Rate5–10 min
Solve β†’
QUESTION81
E-035Communications

The AM Signal Whose Envelope Pinched Shut

The carrier was 5 V. The message amplitude was 6 V. The envelope detector stopped reproducing the message cleanly.

Am Modulation Index5–10 min
Solve β†’
QUESTION82
E-034Control Systems

The Higher-Gain Loop That Crossed Into Instability

Increasing gain improved responseβ€”until the characteristic equation crossed the stability limit.

Closed Loop Stability Routh5–10 min
Solve β†’
QUESTION83
E-033Control Systems

The Bode Plot That Lost 20 dB in One Decade

The low-frequency gain was 40 dB. One decade past the pole, the response was near 20 dB.

First Order Bode Pole5–10 min
Solve β†’
QUESTION84
E-032Control Systems

The Gain-10 Loop That Only Reached 0.909

The forward gain was 10. The closed-loop output did not become 10 times the reference.

Unity Feedback Closed Loop Response5–10 min
Solve β†’
QUESTION85
E-031Power Systems

The 8 Ohm Load That Became 128 Ohms

The secondary load was 8 Ξ©. The primary source did not see 8 Ξ©β€”or even 32 Ξ©.

Transformer Reflected Impedance5–10 min
Solve β†’
QUESTION86
E-030Power Systems

The Feeder That Delivered 228 V Instead of 240 V

The source measured 240 V. The load saw only 228 V because the feeder resistance was treated as one-way.

Distribution Voltage Drop5–10 min
Solve β†’
QUESTION87
E-029Power Systems

The Motor Load That Drew More Current Than the kW Suggested

The load was only 30 kW, but the feeder current was much higher than the kW-only calculation predicted.

Three Phase Power Factor5–10 min
Solve β†’
QUESTION88
E-028Electronics

The Op-Amp That Tried to Output -10 V on Β±5 V Rails

The gain equation requests -10 V, but the amplifier only has Β±5 V rails.

Op Amp Saturation5–10 min
Solve β†’
QUESTION89
E-027Electronics

The Transistor That Refused to Amplify 100Γ—

An NPN switch has VCC = 5 V, Rc = 100 Ξ©, Ξ² = 100, and Ib = 1 mA. Ξ²Ib predicts 100 mA, but with VCE(sat) β‰ˆ 0.2 V the collector circ

Bjt Saturation5–10 min
Solve β†’
QUESTION90
E-026Electronics

The Protection Diode That Starved the 5 V Sensor

The reverse-polarity protection worked, but the 5 V sensor would not start.

Diode Forward Voltage5–10 min
Solve β†’
QUESTION91
E-025Signal Processing

The Digital Filter That Lagged a Perfect Step

A perfect step arrived. The filter answered 2.0 instead of 10.

Digital Filter Difference Equation5–10 min
Solve β†’
QUESTION92
E-024Signal Processing

The 1 kHz Filter That Only Passed 70.7%

A 'passband' tone reached the cutoff and lost nearly 30% of its amplitude.

Analog Low Pass Cutoff5–10 min
Solve β†’
QUESTION93
E-023Signal Processing

The Vibration Sensor That Invented a 300 Hz Tone

The machine says 900 Hz. The sampled spectrum insists on 300 Hz.

Sampling Aliasing5–10 min
Solve β†’
QUESTION94
E-022Linear Systems

The Transfer Function That Settled at 5, Not 20

The numerator says 20. The unit-step output settles at 5. The model is working exactly as written.

Transfer Function Dc Gain5–10 min
Solve β†’
QUESTION95
E-021Linear Systems

The RLC Circuit That Peaked Instead of Filtering

Both reactive components looked like opposition. At resonance, they canceled and the current doubled the protection limit.

Series Rlc Resonance5–10 min
Solve β†’
QUESTION96
E-020Linear Systems

The Capacitor That Drew Too Much Current

The steady-state current is zero. At switch-on, the circuit violates its 5 mA startup limit.

Rc Transient Initial Condition5–10 min
Solve β†’
QUESTION97
E-019Circuit Analysis

The Voltage Divider That Collapsed Under Load

The divider measured exactly 6 V with no load. Connect one load, and the output falls to 4 V.

Voltage Divider Loading6–10 min
Solve β†’
QUESTION98
E-018Circuit Analysis

The Parallel Load That Tripped the Supply

Each branch looked safe. The supply still tripped the instant the second load was connected.

Parallel Circuits6–10 min
Solve β†’
QUESTION99
E-017Circuit Analysis

The Circuit That Shouldn’t Have Burned

A 12 V source is connected across a required 2 Ξ© load. The calculated and measured current is 6 A, but the resistor overheats and fails within seconds. Identify the missing engineering check, prove the root cause, and choose a safe correction.

Electrical Power5–10 min
Solve β†’
QUICK ANSWER

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