NCEES FE Other Disciplines · USA · NCEES
Original practice questions written to the published syllabus · not past papersCircuits, power, transformers, electronic devices and safe wiring: what a farm engineer needs before motors and sensors.
In the exams: NCEES FE Basic Electrical Engineering 6–9 questions; NCEES PE electrical circuits
NCEES FE Other Disciplines — Basic Electrical Engineering (6–9 questions)
Short calculations straight from the reference handbook: Ohm's and Kirchhoff's laws, series–parallel reduction, Thevenin/Norton equivalents and maximum power, RC/RL time constants, AC impedance and phasors, real/reactive/apparent power and power factor, three-phase power, and ideal transformer ratios. Speed with complex numbers on the calculator decides marks.
How our questions for NCEES FE Other Disciplines spread across the units of Basic Electrical & Electronics Engineering. Difficulty: 4 easy, 21 medium, 8 hard.
| DC circuits and network theorems | 5 | |
| AC circuits, phasors and power factor | 5 | |
| Single-phase and three-phase supply | 4 | |
| Transformers | 3 | |
| Operational amplifiers | 3 | |
| Digital logic | 3 | |
| House wiring and earthing | 3 | |
| Electrical safety | 3 | |
| Electrical measuring instruments | 2 | |
| Semiconductor devices and rectifiers | 2 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A sensor network has a Thevenin equivalent of 18 V in series with 3 Ω. A 6 Ω load is connected. What current does the load draw?
D. 2 A
I = V_Th/(R_Th + R_L) = 18/(3 + 6) = 2 A. Dividing by the load alone (3 A) ignores the source resistance; 6 A is the short-circuit (Norton) current; 2.4 A cannot be obtained from the stated 18 V, 3 Ω equivalent with a 6 Ω load.
A source has a Thevenin voltage of 18 V and Thevenin resistance of 3 Ω. What is the maximum power it can deliver to a resistive load, and at what efficiency?
D. 27 W at 50 % efficiency
Maximum power transfer occurs at R_L = R_Th = 3 Ω, giving P = V_Th²/(4R_Th) = 324/12 = 27 W. Equal power is lost inside the source, so efficiency is 50 %. 108 W is V²/R with the load shorted out — the power dissipated internally on short circuit, not load power.
Why can the superposition theorem not be used to find the power dissipated in a resistor by adding the powers due to each source acting alone?
D. Power varies with current squared, so it is not linear in the sources
Superposition relies on linearity: currents and voltages add. Power is I²R, and (I₁ + I₂)² ≠ I₁² + I₂², so adding powers is wrong. Find the total current by superposition first, then square it. Superposition is pointless with one source and it applies to currents as well as voltages.
A 230 V, 50 Hz single-phase borehole pump motor draws 10 A at 0.8 power factor lagging. What shunt capacitance raises the power factor to 0.95 lagging?
C. 46.6 µF
P = 230 × 10 × 0.8 = 1840 W; tan φ₁ = 0.75, tan φ₂ = 0.3287; Q_c = 1840 × (0.75 − 0.3287) = 775 var; C = Q_c/(ωV²) = 775/(314.16 × 230²) = 46.6 µF. 83 µF corrects to unity; 10729 µF divides by V instead of V²; 293 µF forgets the 2π in ω.
Why are three-phase induction motors preferred to single-phase ones for farm pumps above a few kilowatts?
B. Their three windings give a rotating field, so they self-start and deliver steady power
Three-phase currents in three windings 120° apart produce a constant rotating field, so the motor self-starts without an auxiliary winding and the total power does not pulsate. Induction motors run below synchronous speed (slip), still need overload protection, and must be earthed like any other metal-framed equipment.
A clamp meter is closed around both the phase and neutral conductors of a single-phase pump cable while the pump runs normally. What does it read?
B. Close to zero, because equal and opposite currents cancel
The clamp senses the net current through its jaws. Phase and neutral carry equal and opposite currents, so the net is zero except for any leakage to earth — which is how earth-leakage clamps find faults. To read load current, clamp one conductor only.
An inverting amplifier has R_f = 100 kΩ and R_i = 10 kΩ. What is the output for a +0.2 V input, assuming the op-amp stays within its output range?
B. −2.0 V
Gain = −R_f/R_i = −100/10 = −10; V_out = −10 × 0.2 = −2.0 V. +2.0 V ignores the inversion; ±2.2 V uses the non-inverting gain 1 + R_f/R_i = 11.
In an inverting amplifier with the non-inverting input grounded, why is the inverting input called a 'virtual ground'?
C. Feedback holds it at 0 V, yet no current flows from it to ground
With negative feedback and very high open-loop gain, the output drives the input difference to nearly zero, so the inverting input sits at ground potential. But because the input impedance is very high, no current flows into it; the input current all passes through R_f. It is not physically grounded.
In a TT earthing system, how are the consumer's exposed metal parts earthed?
B. Through the consumer's own earth electrode, separate from the supply earth
TT: the source neutral is earthed (first T) and the installation's exposed parts go to an independent local electrode (second T). A PEN conductor is TN-C/TN-C-S; a separate protective conductor from the source is TN-S; an isolated supply is IT.
An enclosure for a milking-parlour pump controller is rated IP55 under IEC 60529. What does this mean?
D. Dust-protected and protected against water jets
In the IP code the first digit is protection against solids (5 = dust-protected; 6 = dust-tight) and the second against water (5 = water jets; 7 = temporary immersion). The digits are not humidity or temperature ratings.
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