Advanced Electrical interview questions

125 questions with answers (EEE-301 to EEE-425). Try each one aloud before reading the answer.

Circuit Theory

EEE-301 · A 415 V, 50 Hz three-phase plant draws 100 kW at 0.7 PF lagging. What capacitor bank raises PF to 0.95?

Q_c = P (tan φ₁ − tan φ₂) = 100 (1.020 − 0.329) ≈ 69.2 kVAR. For delta-connected capacitors, each phase gives 23.1 kVAR at 415 V, so C = Q/(ωV²) = 23 070/(314.2 × 415²) ≈ 426 µF per phase.

EEE-302 · A series RLC circuit has R = 10 Ω, L = 0.1 H and C = 10 µF. Find f₀, Q and bandwidth.

f₀ = 1/(2π√(LC)) = 1/(2π × 10⁻³) ≈ 159 Hz. Q = (1/R)√(L/C) = (1/10)√10⁴ = 10. Bandwidth = f₀/Q ≈ 15.9 Hz.

EEE-303 · A 12 V source with 4 Ω series resistance feeds a 6 Ω shunt resistor across the output terminals. What load gives maximum power and how much?

V_th = 12 × 6/10 = 7.2 V and R_th = 4 ∥ 6 = 2.4 Ω. Choose R_L = 2.4 Ω, giving P_max = V_th²/(4R_th) = 51.84/9.6 = 5.4 W.

EEE-304 · A 100 µF capacitor charges from 10 V through 1 kΩ. What is its voltage after 0.2 s?

τ = RC = 0.1 s. V = 10 (1 − e^(−t/τ)) = 10 (1 − e⁻²) ≈ 8.65 V.

EEE-305 · A 100 V DC source is switched onto R = 10 Ω and L = 0.5 H in series. Find the current after 50 ms.

τ = L/R = 0.05 s. i = (V/R)(1 − e^(−t/τ)) = 10 (1 − e⁻¹) ≈ 6.32 A. Final current is 10 A after about 5τ.

EEE-306 · How do you find the neutral shift of an unbalanced star load with no neutral wire?

Use Millman's theorem: V_N = (V_a Y_a + V_b Y_b + V_c Y_c)/(Y_a + Y_b + Y_c). Each phase load voltage is then V_phase − V_N, so lightly loaded phases see overvoltage.

EEE-307 · What are symmetrical components and why are they used?

Any unbalanced three-phase set decomposes into positive, negative and zero sequence sets. With a = 1∠120°, V₀ = (V_a + V_b + V_c)/3, V₁ = (V_a + aV_b + a²V_c)/3 and V₂ = (V_a + a²V_b + aV_c)/3. They simplify unsymmetrical fault and unbalance analysis.

EEE-308 · A voltage has a 230 V fundamental, 11.5 V 5th harmonic and 8 V 7th harmonic. What is its THD?

THD = √(V₅² + V₇²)/V₁ = √(132.25 + 64)/230 = 14.0/230 ≈ 6.1%.

EEE-309 · A rectifier load has displacement PF 0.98 and current THD of 80%. What is the true power factor?

True PF = displacement PF × distortion factor = 0.98/√(1 + 0.8²) = 0.98/1.281 ≈ 0.77. Capacitors cannot fix the distortion part; filtering is needed.

EEE-310 · In the two-wattmeter method, W₁ = 8 kW and W₂ = 4 kW on a balanced load. Find total power and power factor.

P = 12 kW. tan φ = √3 (8 − 4)/(8 + 4) = 0.577, so φ = 30° and PF = 0.866.

EEE-311 · A balanced star load of 10∠36.87° Ω per phase is fed at 400 V line. Find line current and total power.

V_ph = 400/√3 = 230.9 V, so I = 23.1 A. P = 3 I² R = 3 × 23.1² × 8 ≈ 12.8 kW, matching √3 × 400 × 23.1 × 0.8.

Electromagnetics

EEE-312 · A core has mean length 0.5 m, μᵣ = 2000 and a 1 mm air gap, with B = 1 T. Find the MMF needed.

Core: H = B/(μ₀μᵣ) = 398 A/m, giving 199 AT. Gap: H = B/μ₀ = 795 800 A/m, giving 796 AT. Total ≈ 995 AT, so the 1 mm gap needs four times the MMF of the 0.5 m core.

EEE-313 · Why are air gaps deliberately added to inductor cores?

A gap dominates reluctance, so inductance becomes nearly independent of core permeability. This linearizes L, stores energy in the gap and prevents saturation under DC bias, at the cost of more turns and fringing loss.

EEE-314 · A transformer designed for 50 Hz runs at 60 Hz at the same voltage. How do core losses change?

B_m ∝ V/f falls to 5/6. Eddy loss ∝ B²f² stays about the same. Hysteresis loss ∝ B^1.6 f falls to (5/6)^1.6 × 1.2 ≈ 0.90, so total core loss drops slightly.

EEE-315 · What is the skin depth of copper at 50 Hz and why does it matter?

δ = √(2ρ/(ωμ)) = √(2 × 1.72 × 10⁻⁸/(314 × 4π × 10⁻⁷)) ≈ 9.3 mm. Solid conductors much thicker than about 2δ waste copper, so large busbars use hollow, flat or stranded segmented conductors.

EEE-316 · Two parallel busbars 0.2 m apart carry a 50 kA peak short-circuit current. What force acts per metre?

F/l = μ₀I²/(2πd) = 2 × 10⁻⁷ × (5 × 10⁴)²/0.2 = 2500 N/m. Supports must be designed for peak asymmetrical fault current.

EEE-317 · How is reluctance torque produced?

A ferromagnetic rotor aligns to minimize reluctance. Torque T = ½ i² dL/dθ, independent of current direction. Switched reluctance and synchronous reluctance motors use it.

EEE-318 · What energy is stored in a 1 mm air gap of 10 cm² area at B = 1 T?

Energy density w = B²/(2μ₀) ≈ 3.98 × 10⁵ J/m³. Gap volume is 10⁻⁶ m³, so energy ≈ 0.40 J.

EEE-319 · How can magnetic fields from heavy-current cables be reduced near sensitive equipment?

Lay phases close together in trefoil or twisted formation so fields cancel, keep phase and neutral together, increase distance, and use high-permeability shielding at power frequency.

EEE-320 · What risks demagnetizing permanent magnets in a PM machine?

High temperature lowers coercivity, especially in NdFeB, and fault or overload currents produce opposing armature MMF. Together they can push the operating point below the knee of the demagnetization curve, causing permanent flux loss.

EEE-321 · Why is finite element analysis used in electrical machine design?

Magnetic circuits with saturation, slotting, fringing and complex geometry cannot be solved accurately by hand. FEA gives flux distribution, inductance, torque ripple, losses and local saturation for design optimization.

Transformers

EEE-322 · A 100 kVA transformer has 1 kW iron loss and 1.5 kW full-load copper loss. Find full-load and maximum efficiency at 0.8 PF.

Full load: η = 80/(80 + 2.5) ≈ 96.97%. Maximum efficiency at x = √(1/1.5) = 0.816, i.e. 81.6 kVA: output 65.3 kW, losses 2 kW, η ≈ 97.0%.

EEE-323 · A transformer has 1% resistance and 5% reactance. Find approximate regulation at 0.8 PF lagging and leading.

Regulation ≈ ε_r cos φ ± ε_x sin φ. Lagging: 0.8 + 3.0 = 3.8%. Leading: 0.8 − 3.0 = −2.2%, meaning the secondary voltage rises on load.

EEE-324 · A 500 kVA (5%) and a 250 kVA (4%) transformer share a 750 kVA load. How is the load shared?

Share is proportional to S/Z_pu: 500/0.05 = 10 000 and 250/0.04 = 6250. Transformer 1 carries 750 × 10 000/16 250 ≈ 462 kVA; transformer 2 carries 288 kVA, a 15% overload despite total capacity matching load.

EEE-325 · Find the secondary fault current of a 1000 kVA, 415 V, 5% impedance transformer on an infinite source.

S_sc = 1000/0.05 = 20 MVA. I_sc = 20 × 10⁶/(√3 × 415) ≈ 27.8 kA. Real values are lower once source impedance is included.

EEE-326 · How does transformer differential protection avoid tripping on inrush current?

Inrush is rich in second harmonic while internal fault current is not. The relay uses harmonic restraint, blocking when the second-harmonic ratio exceeds a setting, and uses percentage bias for CT mismatch and tap changes.

EEE-327 · What does dissolved gas analysis of transformer oil reveal?

Hydrogen suggests partial discharge, acetylene suggests arcing, ethylene suggests high-temperature thermal faults, and CO/CO₂ indicate paper insulation degradation. Trends and gas ratios are used for diagnosis.

EEE-328 · How would you size a transformer feeding mostly non-linear loads?

Harmonic currents increase winding eddy losses roughly with h², causing overheating. Use a K-factor rated transformer or derate a standard one, and consider a delta winding to trap triplen harmonics.

EEE-329 · How does hot-spot temperature affect transformer life?

Insulation aging is exponential with temperature; for standard paper it roughly doubles for each 6 °C increase. Sustained overloads that raise hot-spot temperature therefore shorten life sharply.

EEE-330 · What is a zigzag earthing transformer used for?

It creates a neutral point on an ungrounded or delta system. It has low zero-sequence impedance, allowing earth fault current for protection, and high positive-sequence impedance, so it draws little normal current.

EEE-331 · What capacity does an open-delta (V-V) bank provide?

Two single-phase transformers in open delta supply 57.7% of the original delta-delta rating, or 86.6% of the combined rating of the two units used.

EEE-332 · A transformer's Buchholz relay trips on energization. How do you proceed?

Do not re-energize. Collect and analyze the gas, perform DGA, insulation resistance, winding resistance, turns ratio and magnetizing current tests. Check whether trapped air from recent filling was the cause before restoring service.

DC Machines

EEE-333 · A 220 V shunt motor with R_a = 0.5 Ω runs at 1000 rpm drawing I_a = 40 A. Find speed at I_a = 20 A with constant flux.

E_b1 = 220 − 20 = 200 V and E_b2 = 220 − 10 = 210 V. N₂ = 1000 × 210/200 = 1050 rpm.

EEE-334 · For that motor, what starting resistance limits current to 80 A?

Total resistance = 220/80 = 2.75 Ω, so external resistance = 2.75 − 0.5 = 2.25 Ω.

EEE-335 · If flux in that motor is reduced to 80% at constant load torque, find the new current and speed.

T ∝ ΦI_a, so I_a = 40/0.8 = 50 A. E_b = 220 − 25 = 195 V. N = 1000 × (195/200)/0.8 ≈ 1219 rpm.

EEE-336 · A 4-pole lap-wound DC generator has 500 conductors, Φ = 0.02 Wb and runs at 1200 rpm. Find the EMF.

For lap winding A = P = 4. E = PΦNZ/(60A) = 4 × 0.02 × 1200 × 500/(60 × 4) = 200 V.

EEE-337 · A 220 V shunt motor draws 42 A. R_sh = 110 Ω, R_a = 0.5 Ω and rotational losses are 600 W. Find efficiency.

I_sh = 2 A, so I_a = 40 A. Losses: armature 800 W, field 440 W, rotational 600 W, total 1840 W. Output = 9240 − 1840 = 7400 W, so η ≈ 80.1%.

EEE-338 · A series motor's current rises from 20 A to 30 A. How does torque change below saturation?

Φ ∝ I, so T ∝ I². Torque rises by (30/20)² = 2.25 times.

EEE-339 · What is the Ward-Leonard system?

A motor-generator set supplies variable armature voltage to a separately excited DC motor, giving smooth speed control in both directions and regeneration. It is costly and largely replaced by thyristor or IGBT drives.

EEE-340 · How do compound generators correct voltage drop?

Series field flux rises with load and offsets the I_a R_a drop and armature reaction. Flat-compounded machines give equal no-load and full-load voltage; over-compounded machines also compensate feeder drop.

EEE-341 · What is the reactance voltage in commutation and how is it neutralized?

During commutation, coil current reverses from +I to −I in time T_c, inducing e = L × 2I/T_c that delays reversal and causes sparking. Interpoles inject an opposing EMF, and brush resistance aids resistance commutation.

EEE-342 · How does a brushless DC motor differ from a brushed DC motor?

A BLDC motor has PM rotor and wound stator, with electronic commutation from Hall sensors or back-EMF sensing. It removes brush wear and sparking, has better efficiency and cooling, but needs an inverter and controller.

Induction Machines

EEE-343 · A 4-pole, 50 Hz induction motor runs at 1440 rpm. Find slip and rotor frequency.

N_s = 1500 rpm, s = 60/1500 = 0.04 and f_r = sf = 2 Hz.

EEE-344 · That motor takes 50 kW with 2 kW stator losses and 0.5 kW friction and windage. Find output and efficiency.

P_ag = 48 kW, rotor copper loss = sP_ag = 1.92 kW, P_m = 46.08 kW. Output = 45.58 kW, so η ≈ 91.2%.

EEE-345 · Find the electromagnetic torque for an air-gap power of 48 kW in a 4-pole, 50 Hz machine.

ω_s = 2π × 1500/60 = 157.1 rad/s. T = P_ag/ω_s ≈ 306 N·m.

EEE-346 · A motor's DOL starting current is 6 × FLC and starting torque 1.8 × FLT. What does star-delta starting give?

Both fall to one-third: starting current 2 × FLC and torque 0.6 × FLT. Star-delta suits only loads needing less than about 0.6 × FLT to break away.

EEE-347 · What does a 65% autotransformer tap do to starting current and torque?

Motor current falls to 0.65 of DOL and line current and torque fall to 0.65² ≈ 0.42 of DOL values.

EEE-348 · If R₂′ = 0.2 Ω and X₂′ = 1 Ω, at what slip does maximum torque occur and how would you get maximum torque at starting?

s_max = R₂′/X₂′ = 0.2. For maximum torque at s = 1, total rotor resistance must equal X₂′, so add 0.8 Ω external resistance per phase (slip-ring motor).

EEE-349 · What is field-oriented (vector) control?

It transforms stator currents into a rotating d-q frame aligned with rotor flux, controlling flux (i_d) and torque (i_q) independently like a separately excited DC motor. It gives fast torque response and full torque at zero speed.

EEE-350 · A VFD-driven motor overheats at low speed with constant torque load. Why, and what is the fix?

The shaft-mounted fan slows, reducing cooling while current stays near rated. Fit a separately powered blower, derate the motor or choose an inverter-duty motor rated for the speed range.

EEE-351 · What causes bearing currents in VFD-fed motors and how are they mitigated?

PWM common-mode voltage couples through stray capacitances to the shaft and discharges through bearings, causing fluting. Use insulated or ceramic bearings, shaft grounding rings, common-mode filters and symmetrical shielded cable.

EEE-352 · Why is supply voltage unbalance harmful to induction motors?

Unbalance produces negative-sequence voltage, and the low negative-sequence impedance draws large negative-sequence current that creates braking torque and rotor heating. Current unbalance is several times voltage unbalance, so motors must be derated.

EEE-353 · What limits a self-excited induction generator with capacitors?

It needs residual magnetism and capacitance above a minimum value. Voltage and frequency vary with speed and load, and it can collapse under heavy or inductive load, so it suits small isolated systems.

Synchronous Machines

EEE-354 · An alternator with X_s = 1.2 pu delivers 1 pu current at 0.8 PF lagging and 1 pu voltage. Find E, δ and regulation.

E = 1 + j1.2(0.8 − j0.6) = 1.72 + j0.96, so |E| ≈ 1.97 pu and δ ≈ 29.2°. Regulation = (1.97 − 1)/1 ≈ 97%.

EEE-355 · With E = 1.5 pu, V = 1 pu and X = 0.8 pu, find maximum power and the load angle at P = 1 pu.

P_max = EV/X = 1.875 pu. At P = 1 pu, sin δ = 1/1.875 = 0.533, so δ ≈ 32.2°.

EEE-356 · What is the swing equation?

(2H/ω_s) d²δ/dt² = P_m − P_e in per unit, where H is the inertia constant in MJ/MVA. It describes rotor angle dynamics after disturbances.

EEE-357 · What is the equal area criterion?

For a single machine on an infinite bus, the system is transiently stable if the decelerating area under the P-δ curve can equal the accelerating area. It gives the critical clearing angle for faults.

EEE-358 · What is the power equation of a salient-pole machine?

P = (EV/X_d) sin δ + (V²/2)(1/X_q − 1/X_d) sin 2δ. The second term is reluctance power, which exists even without excitation.

EEE-359 · A plant draws 500 kW at 0.8 lag. A 100 kW synchronous motor is added. What motor kVA brings the plant to unity PF?

Existing reactive load = 500 tan 36.87° = 375 kVAR, so the motor must supply 375 kVAR leading. Motor kVA = √(100² + 375²) ≈ 388 kVA at about 0.26 PF leading.

EEE-360 · What happens on loss of excitation in a grid-connected generator?

The machine loses synchronism, runs above synchronous speed as an induction generator, absorbs large reactive power and overheats the rotor. An offset mho relay on the impedance plane detects it and trips the unit.

EEE-361 · What limits define a generator capability curve?

Stator current (circle about origin), field current heating (circle about −V²/X_s), prime mover power limit, stator end-region heating and steady-state stability in the leading (under-excited) region.

EEE-362 · What are subtransient, transient and synchronous reactances?

After a sudden short circuit, current is limited first by X_d″ (damper effects, few cycles), then X_d′ (field effect, about a second) and finally X_d. X_d″ < X_d′ < X_d, and X_d″ sets breaker duty.

EEE-363 · What happens if a generator is synchronized far out of phase?

Large voltage difference drives very high currents and transient shaft torques, possibly damaging the shaft, couplings and windings. Synch-check relays block closing outside angle, slip and voltage limits.

Power Generation

EEE-364 · A hydro site has Q = 50 m³/s, H = 100 m and overall efficiency 90%. What is the output?

P = ρgQHη = 1000 × 9.81 × 50 × 100 × 0.9 ≈ 44.1 MW.

EEE-365 · An 80 m diameter wind turbine sees 10 m/s wind with C_p = 0.45. Estimate its power.

A = π × 40² ≈ 5027 m². P = ½ρAv³C_p = 0.5 × 1.225 × 5027 × 1000 × 0.45 ≈ 1.39 MW. Doubling wind speed would raise available power eightfold, up to rated output.

EEE-366 · Units of 100 MW (4% droop) and 200 MW (5% droop) share a 30 MW load increase at 50 Hz. Find the frequency drop and sharing.

Stiffness: 100/0.04 = 2500 and 200/0.05 = 4000 MW per pu. Δf = 30/6500 = 0.00462 pu ≈ 0.23 Hz. Unit 1 picks up ≈ 11.5 MW and unit 2 ≈ 18.5 MW.

EEE-367 · What is the principle of economic load dispatch?

Without losses, total cost is minimized when all units operate at equal incremental cost dC/dP, within their limits. With transmission losses, incremental costs are multiplied by penalty factors.

EEE-368 · A thermal plant has a heat rate of 10 000 kJ/kWh. What is its efficiency?

η = 3600/heat rate = 3600/10 000 = 36%.

EEE-369 · What must you check when designing PV strings for an inverter?

Open-circuit voltage rises at low temperature, so the string's cold-weather V_oc must stay below inverter maximum DC voltage. The hot-weather MPP voltage must stay within the MPPT window, and string current within input limits.

EEE-370 · What is the difference between grid-following and grid-forming inverters?

Grid-following inverters use a PLL to lock to grid voltage and inject controlled current, so they need a strong grid. Grid-forming inverters set their own voltage and frequency, supporting weak grids and islands.

EEE-371 · How does high renewable penetration affect frequency stability?

Converter-connected sources add little natural inertia, so the rate of change of frequency after a trip rises. Remedies include fast frequency response from batteries, synthetic inertia, synchronous condensers and grid-forming control.

EEE-372 · What is black start capability?

The ability of a unit, such as hydro or a diesel-started gas turbine, to start without external grid supply. Black-start units energize transmission paths and restore the grid in islands that are later synchronized.

EEE-373 · A system has 100 MW maximum demand and supplies 438 GWh in a year. Find the load factor.

Average load = 438 000 MWh/8760 h = 50 MW. Load factor = 50/100 = 50%.

Transmission & Distribution

EEE-374 · What is the Ferranti effect?

At light or no load on a long line or cable, charging current through the series inductance makes the receiving-end voltage higher than the sending-end voltage. Shunt reactors are used to control it.

EEE-375 · What is corona discharge and how is it reduced?

Ionization of air around a conductor when the surface electric field exceeds air's breakdown strength. It causes power loss, audible noise and radio interference. Use larger diameter or bundled conductors and smooth hardware.

EEE-376 · How does HVDC compare with HVAC transmission?

HVDC has no reactive power or skin effect, lower line losses over long distances, needs fewer conductors and can link asynchronous grids. Converter stations are expensive, so HVDC pays off beyond a break-even distance and for long submarine cables.

EEE-377 · Why is the per-unit system used?

It normalizes quantities to chosen bases, so values are similar across transformer voltage levels and transformer impedance is the same on both sides. It simplifies calculations and makes equipment of different ratings comparable.

EEE-378 · Convert 0.1 pu on a 50 MVA base to a 100 MVA base at the same voltage.

Z_new = Z_old × (MVA_new/MVA_old) × (kV_old/kV_new)² = 0.1 × 2 × 1 = 0.2 pu.

EEE-379 · An 11 kV short line with R = 2 Ω and X = 4 Ω per phase carries 100 A at 0.8 PF lagging. Find sending voltage, regulation and efficiency.

V_R = 6351 V per phase. V_S ≈ 6351 + 100(2 × 0.8 + 4 × 0.6) = 6751 V, i.e. 11.69 kV line, so regulation ≈ 6.3%. Losses = 3 × 100² × 2 = 60 kW against 1524 kW output, η ≈ 96.2%.

EEE-380 · Two 400 kV buses are linked by X = 100 Ω with δ = 30°. Find the power transfer.

P = V_S V_R sin δ/X = (400 × 400 × 0.5)/100 = 800 MW. Maximum at δ = 90° would be 1600 MW.

EEE-381 · Why is reactive power supplied locally rather than transmitted?

Reactive flow mainly depends on voltage magnitude difference, Q ≈ V(V_S − V_R)/X. Transmitting it requires large voltage gradients, raises current and losses, and uses up line capacity.

EEE-382 · How does a STATCOM differ from an SVC?

An SVC uses thyristor-switched or controlled reactors and capacitors, so its output falls with V². A STATCOM is a VSC whose current is nearly independent of voltage, giving better support during low-voltage events and faster response.

EEE-383 · How would you reduce technical losses on a distribution network?

Correct power factor, balance phases, reconductor overloaded sections, shorten LV runs by siting transformers at load centres, use higher distribution voltage and replace old transformers with low-loss units.

EEE-384 · Find the voltage drop for 32 A on a 50 m run of 10 mm² copper single-phase cable at 230 V.

Loop length 100 m, R = 0.0172 × 100/10 = 0.172 Ω. Drop = 32 × 0.172 ≈ 5.5 V ≈ 2.4% (resistive approximation).

Protection & Switchgear

EEE-385 · Explain differential protection.

It compares current entering and leaving a protected zone through matched CTs. Under normal load or external faults they balance; a mismatch above the threshold indicates an internal fault and trips the breakers. It is used for transformers, generators, busbars and cables.

EEE-386 · Find the operating time of an IEC standard inverse relay at PSM = 10 and TMS = 0.1.

t = TMS × 0.14/(PSM^0.02 − 1) = 0.1 × 0.14/(1.0471 − 1) ≈ 0.30 s.

EEE-387 · What determines the grading margin between overcurrent relays?

Breaker interruption time, relay overshoot, relay and CT errors, and a safety margin. Typical margins are around 0.3–0.4 s for electromechanical schemes and less for numerical relays.

EEE-388 · An 11 kV bus fed by a generator (0.2 pu) and transformer (0.1 pu) on 100 MVA base suffers a three-phase fault. Find fault current.

I_f = 1/0.3 = 3.33 pu. Base current = 100 × 10⁶/(√3 × 11 000) ≈ 5249 A, so I_f ≈ 17.5 kA (333 MVA).

EEE-389 · With Z₁ = Z₂ = 0.2 pu and Z₀ = 0.1 pu, compare single line-to-ground and three-phase fault currents.

SLG: I_f = 3E/(Z₁ + Z₂ + Z₀) = 3/0.5 = 6 pu. Three-phase: 1/0.2 = 5 pu. When Z₀ < Z₁, as near solidly earthed transformers, an earth fault exceeds the three-phase fault.

EEE-390 · How do you specify a protection CT to avoid saturation?

Its knee-point voltage must exceed the voltage needed to drive maximum fault current through CT winding, lead and relay burden, with margin for DC offset and remanence as required by the relay scheme.

EEE-391 · How are distance protection zones typically set?

Zone 1 covers about 80% of the line instantaneously. Zone 2 covers about 120% with a time delay. Zone 3 covers further into adjacent lines as remote backup with a longer delay.

EEE-392 · What is breaker failure protection?

If a breaker receives a trip but current persists beyond a set time, the scheme trips all adjacent breakers that can feed the fault, usually through busbar protection tripping.

EEE-393 · How can arc-flash hazard be reduced in switchgear?

Lower clearing time with arc-flash relays (light plus current), zone-selective interlocking and maintenance mode settings. Use arc-resistant switchgear, remote racking and switching, and appropriate PPE based on incident energy studies.

EEE-394 · Why must busbar protection be very stable?

A busbar trip disconnects every circuit on the bus. The scheme must stay stable for heavy external faults even with CT saturation, using high-impedance differential or biased low-impedance numerical schemes, often with check zones.

EEE-395 · How is discrimination between an upstream fuse and a downstream device achieved?

The upstream fuse's pre-arcing I²t must exceed the downstream device's total let-through I²t at the prospective fault current. Manufacturer selectivity tables are used for the check.

Power Electronics

EEE-396 · Design a buck converter for 24 V to 12 V at 2 A, 100 kHz, with 0.4 A ripple current.

D = 0.5. L = (V_in − V_o)D/(f ΔI) = 12 × 0.5/(10⁵ × 0.4) = 150 µH. With C = 100 µF, ΔV = ΔI/(8fC) = 0.4/80 = 5 mV (ignoring ESR).

EEE-397 · A boost converter raises 12 V to 48 V at 100 W. Find duty cycle and input current.

D = 1 − V_in/V_o = 0.75. Input current = 100/12 ≈ 8.3 A for an ideal converter.

EEE-398 · Find the output of a three-phase full converter on 415 V at α = 30°.

V_dc = 1.35 × 415 × cos 30° ≈ 485 V.

EEE-399 · What happens when a fully controlled converter is fired beyond 90°?

Average output voltage becomes negative while current direction stays the same, so power flows from the DC side to AC (inversion). It needs a DC source such as a regenerating motor, and α must leave enough extinction margin to avoid commutation failure.

EEE-400 · How do you choose switching frequency in a converter?

Higher frequency shrinks inductors, capacitors and filters and improves response, but raises switching loss and EMI. Silicon IGBTs favour lower frequencies; SiC and GaN devices allow much higher frequencies efficiently.

EEE-401 · What harmonics does a six-pulse rectifier produce and how does twelve-pulse help?

A six-pulse bridge produces h = 6k ± 1 (5th, 7th, 11th, 13th). A twelve-pulse arrangement with two bridges fed via 30° phase-shifted windings cancels 5th and 7th, leaving 12k ± 1.

EEE-402 · A device dissipates 50 W with R_jc = 0.5, R_cs = 0.2 and R_sa = 0.8 °C/W at 40 °C ambient. Find T_j.

T_j = T_a + P ΣR = 40 + 50 × 1.5 = 115 °C. Check this against the device's maximum junction rating with margin.

EEE-403 · What are the advantages of space vector PWM over sinusoidal PWM?

SVPWM gives about 15% higher linear output voltage from the same DC bus, lower harmonic distortion and fewer switchings per cycle. It fits naturally with digital vector control.

EEE-404 · Why use an active front end instead of a diode front end in a drive?

An AFE draws near-sinusoidal current at controllable (near unity) power factor and allows regeneration to the grid. It costs more and needs a line filter.

EEE-405 · What features should an IGBT gate driver have?

Galvanic isolation, adequate peak gate current, negative turn-off bias, desaturation (short-circuit) detection with soft turn-off, Miller clamp, undervoltage lockout and short, low-inductance gate loops.

Electrical Measurements

EEE-406 · Extend a 1 mA, 100 Ω meter to read 1 A.

R_sh = I_m R_m/(I − I_m) = 0.1/0.999 ≈ 0.1001 Ω in parallel.

EEE-407 · Convert a 1 mA, 100 Ω meter into a 100 V voltmeter.

Total resistance = 100/0.001 = 100 kΩ, so series multiplier = 99.9 kΩ. Sensitivity is 1000 Ω/V.

EEE-408 · What do CT accuracy class designations such as 0.5 and 5P20 mean?

Metering class 0.5 has 0.5% ratio error at rated current. Protection class 5P20 keeps composite error within 5% up to 20 times rated current. Metering CTs should saturate early to protect meters; protection CTs should not.

EEE-409 · What does a power quality analyzer measure?

RMS voltage and current, harmonics and THD, flicker, sags, swells, interruptions, transients, unbalance, frequency, and power, energy and power factor over time.

EEE-410 · How does an electronic energy meter measure energy?

It samples voltage and current with ADCs, multiplies instantaneous samples to get power and integrates over time. Current sensing uses shunts, CTs or Rogowski coils.

EEE-411 · What does a tan δ test indicate?

Tan δ is the ratio of resistive to capacitive current in insulation. A high or rising value with test voltage indicates moisture, contamination or aging in cables, bushings and windings.

EEE-412 · What is partial discharge measurement used for?

It detects small localized discharges in voids, surfaces or interfaces within HV insulation before breakdown. Electrical, acoustic and UHF methods are used for cables, GIS, transformers and machines.

EEE-413 · How many wattmeters are needed for an unbalanced four-wire load?

By Blondel's theorem, an n-wire system needs n − 1 wattmeters, so three are required for three-phase four-wire.

EEE-414 · Why do instrument transformer errors matter more at low power factor?

CT and VT phase angle errors shift the measured angle. Power error is roughly tan φ times the phase error, which grows large as PF falls.

EEE-415 · How do you measure winding resistance and correct for temperature?

Use a DC four-wire (Kelvin) method and wait for inductive current to stabilize. For copper, R₂ = R₁(235 + T₂)/(235 + T₁).

Electrical Safety, Wiring & Drives

EEE-416 · What maximum earth fault loop impedance allows a 32 A type B MCB to trip instantaneously at 230 V?

Type B trips magnetically at 5 × I_n = 160 A, so Z_s ≤ 230/160 ≈ 1.44 Ω before applying temperature and voltage correction factors.

EEE-417 · Estimate the resistance of a 3 m, 16 mm rod in 100 Ω·m soil.

R = (ρ/2πL)[ln(8L/d) − 1] = 5.31 × (ln 1500 − 1) ≈ 33.5 Ω. Parallel rods spaced well apart, longer rods or soil treatment reduce it.

EEE-418 · A centrifugal fan's speed is reduced to 80% with a VFD. How does power change?

Fan torque ∝ n² and power ∝ n³, so power falls to 0.8³ ≈ 51% of full-speed power.

EEE-419 · What happens when a VFD decelerates a high-inertia load quickly?

The motor regenerates and the DC bus voltage rises until an overvoltage trip. Use longer ramps, a braking chopper with resistor, a common DC bus or an active front end to return energy to the grid.

EEE-420 · A 50 kg·m² load at 1500 rpm stops in 10 s at constant deceleration. Size the braking energy and power.

ω = 157.1 rad/s, E = ½Jω² ≈ 617 kJ. Average power ≈ 61.7 kW and peak power at the start is about 123 kW.

EEE-421 · A motor runs 100 kW for 2 min, 50 kW for 4 min and is idle 4 min each cycle. Estimate rating by the RMS method.

P_rms = √((100² × 2 + 50² × 4)/10) = √3000 ≈ 55 kW. Check also peak torque capability and reduced cooling when idle.

EEE-422 · How do you mitigate harmonics from many VFDs in a plant?

Add AC line reactors or DC chokes, use 12- or 18-pulse front ends, active front ends, passive tuned filters or active harmonic filters. Check against the utility's limits at the point of common coupling.

EEE-423 · What is the correct procedure to prove a circuit dead before work?

Isolate and lock off, then test using an approved voltage tester: prove it on a known live source, test all conductors, and prove the tester again. Apply earths where required before touching.

EEE-424 · Why are IT (isolated) systems used in hospital operating theatres?

An isolating transformer with no earthed neutral means a first insulation fault does not trip the supply, keeping life-support equipment running. An insulation monitoring device alarms so the fault can be cleared.

EEE-425 · What factors govern the selection of a motor for a VFD application?

Load torque profile (constant or variable), speed range, cooling at low speed, insulation rating for PWM voltage spikes, bearing current protection, and whether field weakening above base speed is needed.

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