BPSC AEO Electrical Main Exam Mock Test – Part 1 (Circuits & Network Analysis, Q1–20)

Derive the expression for RMS and average values of sinusoidal current and voltage.
For the series RLC circuit, R = 10 Ω, L = 0.1 H, C = 100 μF, applied voltage V = 220 V, 50 Hz:
a) Calculate current amplitude, phase angle, and power factor.
b) Draw phasor diagram.
Derive mesh and nodal analysis equations for a general 3-node resistive network.
Define resonance in series and parallel circuits. Derive expressions for resonant frequency.
A two-port network has Z-parameters: Z11 = 4 Ω, Z12 = 2 Ω, Z21 = 2 Ω, Z22 = 3 Ω. Determine input impedance when load ZL = 6 Ω.
Using superposition theorem, find the current through a 10 Ω resistor in a circuit with two voltage sources 20 V and 10 V.
Derive Thevenin and Norton equivalent circuits. Apply to find current through a load resistor 5 Ω in a network.
In a balanced three-phase star-connected load, line voltage 400 V, Z = 10 + j5 Ω. Calculate line current, phase current, power, and power factor.
Draw phasor diagram for the above three-phase circuit and determine reactive power.
Derive AC power formulas: instantaneous, active, reactive, and apparent power.
Determine maximum power transfer condition for AC circuits with complex impedance and derive expression.
A DC voltage source V = 100 V is applied across a series R-L circuit with R = 10 Ω, L = 0.1 H.
a) Solve for transient current i(t) after switch-on.
b) Find time constant and steady-state current.
Derive impedance and admittance matrices for a two-port network.
Define power factor improvement and derive condition for capacitor rating in a lagging load.
Explain phasor relationships in series and parallel RLC circuits. Solve for voltage drops across each element.
A voltage source V = 120 V, 50 Hz, is applied across a parallel RLC circuit: R = 30 Ω, L = 0.2 H, C = 100 μF. Determine current through each branch, total current, and power factor.
Derive relation between line and phase quantities in star and delta connections.
Using nodal analysis, solve for node voltages in a 3-node resistive network with given voltage sources.
Derive expressions for resonant impedance, current, and voltage in series RLC circuit at resonance.
For a DC circuit with multiple voltage sources and resistors, use superposition to determine current through a given resistor.

Part 2 – Electrical Machines (Q21–40)

Derive the torque equation of a DC motor and calculate the torque for a shunt motor with: V = 220 V, Ia = 10 A, field flux Φ = 0.05 Wb.
A DC shunt motor has R_a = 0.5 Ω, R_f = 100 Ω, V = 220 V, no-load speed = 1000 rpm.
a) Determine armature current and speed when delivering rated torque.
b) Plot torque-speed characteristics.
Derive the speed-torque characteristics of a series, shunt, and compound DC motor. Explain differences.
A 3-phase, 50 Hz, 440 V, star-connected induction motor has Z1 = 0.5 + j1 Ω per phase. Calculate:
a) Full-load current
b) Power factor
c) Rotor copper loss
Explain slip, synchronous speed, and rotor frequency in an induction motor. Solve a numerical with N_s = 1500 rpm, slip s = 3%.
Derive equivalent circuit of induction motor and explain how rotor parameters are referred to the stator.
A synchronous generator delivers 500 kVA, 0.8 pf lagging, 400 V line voltage. Reactance X_d = 10 Ω per phase. Determine excitation voltage required to maintain terminal voltage.
Derive the induced EMF equation of a 3-phase alternator and explain voltage regulation by synchronous impedance method.
A 3-phase alternator, 50 Hz, 400 V, delivers 100 kW at 0.8 pf lag. Calculate:
a) Armature current
b) Active and reactive power
c) Efficiency assuming given losses
Derive OC and SC tests for transformers and determine equivalent circuit parameters from test data.
A single-phase transformer, 220/110 V, 5 kVA, has R1 = 0.5 Ω, X1 = 1 Ω, R2 = 0.2 Ω, X2 = 0.8 Ω. Calculate:
a) Efficiency at full load, 0.8 pf lag
b) Voltage regulation
Derive the relationship between primary and secondary current, voltage, and impedance in an ideal transformer.
A DC series motor delivers 5 kW at 220 V. Determine current and torque if series field has 0.5 Ω and armature 0.2 Ω.
Explain starting methods of DC and AC motors, including star-delta, autotransformer, and rheostat starters, with numerical illustration.
A 3-phase induction motor draws 20 A at 400 V, 50 Hz. Efficiency = 90%, slip = 3%. Determine:
a) Output power
b) Rotor copper loss
c) Air-gap power
Derive torque-slip characteristic of a 3-phase induction motor and determine maximum torque and corresponding slip.
A synchronous motor operating at 440 V, 50 Hz, takes 20 A at 0.8 pf lag. Determine:
a) Mechanical power developed
b) Reactive power supplied or absorbed
c) Effect of over-excitation
Explain power factor correction using synchronous condensers and capacitor banks, with numerical example.
A DC shunt generator has open-circuit voltage 250 V at 1000 rpm. Find terminal voltage at 100 A load, R_a = 0.5 Ω, R_f = 100 Ω.
Explain parallel operation of alternators. Derive conditions for voltage, frequency, and phase sequence matching.

Part 3 – Power Systems & Protection (Q41–60)

Derive the ABC-D parameters for a short and medium transmission line. Explain their significance with a numerical example.
A 3-phase, 50 Hz, 220 kV transmission line has R = 0.5 Ω/phase, X = 1 Ω/phase, sending end voltage 220 kV, receiving end load 100 MW at 0.8 pf lag. Determine:
a) Sending end current
b) Sending end voltage
c) Line losses and efficiency
Explain voltage regulation of transmission lines using the synchronous condenser method and compute numerical example.
Derive equations for series and shunt compensation of transmission lines. Solve for capacitor rating to improve power factor from 0.8 lag to 0.95 lag.
Explain per-unit system in power system calculations. Convert a transformer of 5 MVA, 33/11 kV to per-unit system and calculate per-unit impedances.
A three-phase fault occurs on a transmission line. Line-to-line voltage = 220 kV, Z_line = 10 + j20 Ω/phase. Calculate fault current using per-unit method.
Explain load flow analysis using Gauss-Seidel method. Solve a 3-bus system with given bus voltages and impedances.
Explain reliability in power systems. Calculate probability of supply continuity for a system with two parallel feeders.
Derive short-circuit current equations for:
a) Three-phase fault
b) Line-to-ground fault
Include numerical calculation with given network parameters.
Explain economic load dispatch and solve a numerical for two generators with cost functions C1 = 0.1 P1² + 20 P1, C2 = 0.08 P2² + 25 P2, total load 300 MW.
Explain protection schemes of transformers:
a) Differential protection
b) Buchholz relay
c) Overcurrent protection
Include example calculations.
A transmission line of length 100 km has R = 0.2 Ω/km, X = 0.5 Ω/km. Calculate:
a) Total line impedance
b) Voltage drop for a load of 50 MW at 0.8 pf lag
Explain protective relays: overcurrent, distance, differential. Solve a numerical for relay settings in a 33 kV feeder.
Explain circuit breaker selection for a given line: 33 kV, 100 MVA short-circuit capacity. Compute required interrupting capacity.
Derive equations for load sharing between parallel transformers. Solve a numerical for 2 transformers of 5 MVA and 3 MVA supplying 6 MW load.
Explain voltage drop and losses in three-phase lines. Compute efficiency and regulation for a line supplying 50 MW at 0.85 pf lag.
Derive expressions for critical clearing time of protective devices. Solve a numerical for a given line and fault.
Explain reactive power control using:
a) Shunt capacitors
b) Series capacitors
c) Synchronous condensers
Include numerical examples.
Explain bus bar arrangements in substations: single bus, double bus, ring bus. Solve a numerical to determine bus fault current.
A 3-phase, 220 kV system experiences line-to-ground fault on phase A. Z_line = 10 + j20 Ω/phase, load = 50 MW, pf = 0.8 lag. Calculate:
a) Fault current
b) Voltage at other phases
c) Current through protective relay

Part 4 – Electrical Measurements & Instrumentation (Q61–75)

Derive the operating principle of moving coil instruments and calculate deflection for a given current and coil parameters.
Explain moving iron instruments. Solve a numerical to determine current or voltage measurement for a given scale and spring constant.
A single-phase energy meter has constant Kh = 7.2 Wh/imp. Calculate:
a) Number of revolutions for a given load of 500 W over 2 hours
b) Accuracy if voltage varies by ±5%
Derive the current transformer (CT) ratio formula and solve a numerical for:
a) CT secondary current
b) Burden calculation
Derive the potential transformer (PT) ratio and voltage error. Solve a numerical with given primary and secondary voltage, load, and excitation.
Explain errors in instrument transformers:
a) Ratio error
b) Phase displacement
Include numerical calculation for CT/PT with given parameters.
Derive the two-wattmeter method for measuring power in a 3-phase system. Solve a numerical for load P = 50 kW, pf = 0.8 lag, line voltage = 400 V.
Explain DC and AC bridges (Wheatstone, Maxwell, Hay, Schering). Solve a numerical for unknown resistance or inductance.
Derive operation of Q-meter for measuring L and Q of an inductor. Solve a numerical for given frequency and voltage readings.
Derive the operation of oscilloscope, including horizontal and vertical deflection equations. Solve a numerical to determine peak voltage from screen deflection.
Explain strain gauge working principle. Solve a numerical to determine force or stress given gauge factor, resistance change, and applied strain.
Explain thermocouple and RTD for temperature measurement. Solve a numerical to find temperature for a given EMF or resistance change.
Derive loading effect of voltmeters and ammeters. Solve a numerical for voltage drop and measurement error.
Explain digital multimeter (DMM) operation for measuring AC/DC voltage, current, and resistance. Solve a numerical for accuracy calculation.
Explain instrument calibration and error calculation. Solve a numerical for percentage error given observed and true values.

Part 5 – Control Systems & Power Electronics (Q76–100)

Derive the transfer function of a DC motor for speed and torque. Solve a numerical to find speed response for a step input voltage.
Explain open-loop and closed-loop control systems. Solve a numerical to determine system response and error constants.
Derive time response of first-order systems (step and ramp inputs). Solve a numerical for time constant and settling time.
Derive time response of second-order systems. Solve a numerical to determine overshoot, rise time, and settling time for given ζ and ω_n.
Explain stability of control systems. Use Routh-Hurwitz criterion to check stability of a given characteristic equation.
Derive root-locus for a given transfer function. Solve a numerical to find gain for desired pole locations.
Draw Bode plot for a given transfer function and determine gain margin and phase margin. Solve numerical example.
Explain state-space representation and derive state equations for a given electrical system (R-L-C circuit).
Explain feedback control of DC and AC drives. Solve a numerical for speed regulation using feedback.
Derive PID controller equations. Solve a numerical to determine control action for given error signal.
Explain thyristor operation (SCR). Solve a numerical to calculate firing angle for a given load and output voltage.
Derive AC voltage controller output equations. Solve a numerical to determine RMS output voltage for given firing angle and load.
Explain single-phase and three-phase inverters. Solve a numerical to determine RMS output voltage and frequency.
Derive DC-DC converter equations for buck, boost, and buck-boost converters. Solve numerical to find output voltage for given duty ratio.
Explain chopper operation with continuous and discontinuous conduction modes. Solve a numerical to calculate average output voltage and current.
Explain phase-controlled rectifier operation (single-phase). Solve numerical to calculate average and RMS voltage for given firing angle.
Derive voltage and current equations of step-up and step-down choppers. Solve numerical for load voltage and ripple.
Explain UPS system working principle. Solve a numerical to determine backup time and battery rating for given load.
Explain SMPS operation and efficiency. Solve a numerical to determine output voltage and current given input voltage and duty cycle.
Explain thyristor-based AC regulators. Solve a numerical to calculate output voltage waveform and RMS value.
Explain PWM inverter operation and derive equations for fundamental and harmonic components. Solve numerical for RMS voltage.
Explain three-phase bridge rectifier operation with R-L load. Solve numerical to calculate average and RMS load voltage and current.
Explain phase-locked loop (PLL) operation. Solve a numerical for lock range and capture range.
Explain synchronous motor control using power electronics. Solve numerical to determine armature current and excitation voltage for given load.
Solve a combined system problem: A DC motor fed through a chopper drives an R-L load. Determine average armature current, torque, and speed response for given supply voltage, load, and duty ratio.

BPSC AEO Electrical Main Exam – Part 1 Solutions (Q1–20)

Q1. RMS and Average Values of Sinusoidal Voltage and Current

For sinusoidal waveform $v(t) = V_m \sin \omega t$ v(t)=Vmsinωt:

RMS Value:

$V_{rms} = \sqrt{\frac{1}{T}\int_0^T v^2(t) dt} = \frac{V_m}{\sqrt{2}}$ Vrms=T1∫0Tv2(t)dt=2Vm

Average Value (over half cycle):

$V_{avg} = \frac{1}{T/2}\int_0^{T/2} v(t) dt = \frac{2 V_m}{\pi}$ Vavg=T/21∫0T/2v(t)dt=π2Vm

Q2. Series RLC Circuit

Given: $R = 10 \Omega$ R=10Ω, $L = 0.1 H$ L=0.1H, $C = 100 \mu F$ C=100μF, $V = 220 V$ V=220V, $f = 50 Hz$ f=50Hz

Reactances:

$X_L = 2 \pi f L = 2 \pi (50)(0.1) = 31.42 \ \Omega$ XL=2πfL=2π(50)(0.1)=31.42 Ω $X_C = \frac{1}{2 \pi f C} = \frac{1}{2 \pi (50)(100 \times 10^{-6})} \approx 31.83 \ \Omega$ XC=2πfC1=2π(50)(100×10−6)1≈31.83 Ω

Impedance:

$Z = R + j(X_L – X_C) = 10 + j(31.42 – 31.83) = 10 – j0.41$ Z=R+j(XL−XC)=10+j(31.42−31.83)=10−j0.41 $|Z| = \sqrt{10^2 + (-0.41)^2} \approx 10.008 \ \Omega$ ∣Z∣=102+(−0.41)2≈10.008 Ω

Current:

$I = \frac{V}{Z} = \frac{220}{10.008} \approx 21.99 \ A$ I=ZV=10.008220≈21.99 A

Phase angle:

$\phi = \arctan \frac{X_L – X_C}{R} = \arctan \frac{-0.41}{10} \approx -2.35^\circ \ (\text{current leads})$ ϕ=arctanRXL−XC=arctan10−0.41≈−2.35∘ (current leads)

Phasor diagram: Voltage across R, L, and C can be drawn with L leading current, C lagging.

Q3. Mesh and Nodal Analysis

Mesh analysis: $\sum V = 0$ ∑V=0 around loops
Nodal analysis: $\sum I_{into \ node} = 0$ ∑Iinto node=0
Write equations for a 3-node resistive network:

$(V_1 – V_2)/R_1 + V_1/R_2 + … = 0$ (V1−V2)/R1+V1/R2+…=0

Solve using simultaneous equations.

(Numerical depends on actual network)

Q4. Resonance in Series and Parallel Circuits

Series resonance: $X_L = X_C$ XL=XC

$f_r = \frac{1}{2 \pi \sqrt{LC}}$ fr=2πLC1

Parallel resonance: $X_L = X_C$ XL=XC (admittances equal)

$f_r = \frac{1}{2 \pi \sqrt{LC}}$ fr=2πLC1

At resonance: impedance is minimum in series, maximum in parallel.

Q5. Two-Port Network Input Impedance

Given: $Z_{11} = 4$ Z11=4, $Z_{12} = Z_{21} = 2$ Z12=Z21=2, $Z_{22} = 3$ Z22=3, Load $Z_L = 6 \Omega$ ZL=6Ω $Z_{in} = Z_{11} – \frac{Z_{12} Z_{21}}{Z_{22} + Z_L} = 4 – \frac{2 \cdot 2}{3 + 6} = 4 – \frac{4}{9} = 3.556 \ \Omega$ Zin=Z11−Z22+ZLZ12Z21=4−3+62⋅2=4−94=3.556 Ω

Q6. Superposition Theorem

Consider each voltage source individually, replace others with short circuit.
Solve current through 10 Ω resistor for each source.
Sum algebraically to get total current.

(Numerical depends on circuit values)

Q7. Thevenin & Norton Equivalents

Remove load resistor.
Vth = open-circuit voltage across terminals.
Rth = resistance seen from terminals with independent sources turned off.
In = Vth / (Rth + RL)

(Numerical solved accordingly)

Q8. Balanced Star Load

Given: V_L = 400 V, Z = 10 + j5 Ω

Phase voltage: $V_{ph} = V_L / \sqrt{3} = 230.94 V$ Vph=VL/3=230.94V
Phase current: $I_{ph} = V_{ph}/Z = 230.94 / \sqrt{10^2 + 5^2} = 20.63 A$ Iph=Vph/Z=230.94/102+52=20.63A
Line current: $I_L = I_{ph} = 20.63 A$ IL=Iph=20.63A
Power: $P = \sqrt{3} V_L I_L \cos \phi = \sqrt{3} \cdot 400 \cdot 20.63 \cdot (10/\sqrt{125}) \approx 143 kW$ P=3VLILcosϕ=3⋅400⋅20.63⋅(10/125)≈143kW

Q9. Phasor Diagram

Draw V_phase, V_R, V_L, V_C
Phase angle: $\phi = \arctan(X/R) = 26.57^\circ$ ϕ=arctan(X/R)=26.57∘

Q10. AC Power Formulas

Instantaneous: $p(t) = v(t) i(t) = V_m I_m \sin \omega t \sin (\omega t – \phi)$ p(t)=v(t)i(t)=VmImsinωtsin(ωt−ϕ)
Active: $P = V_{rms} I_{rms} \cos \phi$ P=VrmsIrmscosϕ
Reactive: $Q = V_{rms} I_{rms} \sin \phi$ Q=VrmsIrmssinϕ
Apparent: $S = V_{rms} I_{rms}$ S=VrmsIrms

Q11. Maximum Power Transfer (AC) $Z_L = Z_{th}^* \ (\text{complex conjugate of Thevenin impedance})$ ZL=Zth∗ (complex conjugate of Thevenin impedance)

Maximum power delivered: $P_{max} = \frac{|V_{th}|^2}{4 R_{th}}$ Pmax=4Rth∣Vth∣2

Q12. DC RL Transient

Given: V = 100 V, R = 10 Ω, L = 0.1 H $i(t) = \frac{V}{R}(1 – e^{-Rt/L}) = 10 (1 – e^{-100 t / 0.1}) = 10 (1 – e^{-1000 t}) A$ i(t)=RV(1−e−Rt/L)=10(1−e−100t/0.1)=10(1−e−1000t)A

Time constant: $\tau = L/R = 0.01 s$ τ=L/R=0.01s

Steady-state current: $I = V/R = 10 A$ I=V/R=10A

Q13. Two-Port Network Matrices

Z-matrix: $V = Z I$ V=ZI
Y-matrix: $I = Y V$ I=YV
Relations: $Y = Z^{-1}$ Y=Z−1, etc.

Q14. Power Factor Improvement

Required capacitor:

$Q_c = P (\tan \phi_1 – \tan \phi_2)$ Qc=P(tanϕ1−tanϕ2)

Q15. Phasor Relations

Series RLC: $V = V_R + V_L – V_C$ V=VR+VL−VC
Parallel RLC: $I = I_R + I_L + I_C$ I=IR+IL+IC

Q16. Parallel RLC Circuit

Branch currents:

$I_R = V / R, \ I_L = V / jX_L, \ I_C = V / (-jX_C)$ IR=V/R, IL=V/jXL, IC=V/(−jXC)

Total current: phasor sum.
Power factor: $\cos \phi = P/S$ cosϕ=P/S

Q17. Line and Phase Relations

Star: $V_L = \sqrt{3} V_{ph}, I_L = I_{ph}$ VL=3Vph,IL=Iph
Delta: $V_L = V_{ph}, I_L = \sqrt{3} I_{ph}$ VL=Vph,IL=3Iph

Q18. Nodal Analysis Example

Write KCL at each node. Solve simultaneous equations.

Q19. Series Resonance Quantities

Resonant frequency: $f_r = 1/(2 \pi \sqrt{LC})$ fr=1/(2πLC)
Impedance: minimum, current: maximum

Q20. DC Circuit Superposition

Consider each source separately, replace others with short.
Calculate current through resistor for each source. Sum algebraically.

Q21. Torque Equation of a DC Motor

Electromagnetic torque:

$T = \frac{P \Phi I_a}{2 \pi N_s} \quad \text{or} \quad T = k \Phi I_a$ T=2πNsPΦIaorT=kΦIa

Where $I_a$ Ia = armature current, $\Phi$ Φ = flux per pole.

Given: V = 220 V, I_a = 10 A, Φ = 0.05 Wb

Assume constant $k = 1$ k=1 for simplicity: $T = k \Phi I_a = 0.05 \cdot 10 = 0.5 \text{ N·m (per unit adjustment)}$ T=kΦIa=0.05⋅10=0.5 N\cdotpm (per unit adjustment)

(For real motor, use correct units including pole pairs and speed.)

Q22. DC Shunt Motor – Armature Current and Speed

Given: R_a = 0.5 Ω, R_f = 100 Ω, V = 220 V

Field current:

$I_f = \frac{V}{R_f} = \frac{220}{100} = 2.2 \text{ A}$ If=RfV=100220=2.2 A

Armature current at rated torque: depends on mechanical load;

$I_a = \frac{T}{k \Phi} = \text{as per torque}$ Ia=kΦT=as per torque

Speed:

$N = N_0 \left(1 – \frac{I_a R_a}{V}\right)$ N=N0(1−VIaRa)

Torque-speed curve: linear for shunt motor; slope depends on R_a.

Q23. Speed-Torque Characteristics

Shunt motor: constant speed, torque proportional to armature current.
Series motor: speed varies inversely with load, high starting torque.
Compound motor: combination; nearly constant speed under load.

Q24. 3-Phase Induction Motor – Current & Power Factor

Given: Z1 = 0.5 + j1 Ω/phase, 3-phase, 50 Hz, 440 V

Phase voltage (star):

$V_{ph} = V_L / \sqrt{3} = 440 / 1.732 \approx 254 V$ Vph=VL/3=440/1.732≈254V

Phase current:

$I_{ph} = \frac{V_{ph}}{Z} = \frac{254}{\sqrt{0.5^2 + 1^2}} = \frac{254}{1.118} \approx 227 A$ Iph=ZVph=0.52+12254=1.118254≈227A

Power factor:

$\cos \phi = \frac{R}{|Z|} = \frac{0.5}{1.118} \approx 0.447$ cosϕ=∣Z∣R=1.1180.5≈0.447

Q25. Slip, Synchronous Speed, Rotor Frequency

Synchronous speed:

$N_s = \frac{120 f}{P} = 1500 \text{ rpm (given)}$ Ns=P120f=1500 rpm (given)

Slip:

$s = \frac{N_s – N_r}{N_s} = 0.03$ s=NsNs−Nr=0.03

Rotor frequency:

$f_r = s f = 0.03 \cdot 50 = 1.5 \text{ Hz}$ fr=sf=0.03⋅50=1.5 Hz

Q26. Induction Motor Equivalent Circuit

Stator: R1 + jX1
Rotor (referred to stator): R2’/s + jX2’
Magnetizing branch: jXm in parallel
Total impedance: $Z_{total} = R_1 + jX_1 + (R_2’/s + jX_2′) || jX_m$ Ztotal=R1+jX1+(R2′/s+jX2′)∣∣jXm

Q27. Synchronous Generator Excitation Voltage

Given: S = 500 kVA, V_L = 400 V, pf = 0.8 lag, X_d = 10 Ω

Line current:

$I_L = \frac{S}{\sqrt{3} V_L} = \frac{500,000}{1.732 \cdot 400} \approx 721.7 A$ IL=3VLS=1.732⋅400500,000≈721.7A

Phase voltage: V_phase = V_L / √3 = 400 / 1.732 ≈ 231 V
Internal EMF:

$E_a = \sqrt{V^2 + (I_a X_d)^2 + 2 V I_a X_d \sin \theta}$ Ea=V2+(IaXd)2+2VIaXdsinθ

Plug values to get E_a.

Q28. Induced EMF of Alternator $E = 4.44 f N \Phi$ E=4.44fNΦ

f = frequency
N = number of turns per phase
Φ = flux per pole

Voltage regulation: $\%VR = \frac{E_0 – V}{V} \times 100$ %VR=VE0−V×100

Q29. 3-Phase Alternator Calculations

Given: P = 100 kW, pf = 0.8 lag, V_L = 400 V

Line current:

$I_L = \frac{P}{\sqrt{3} V_L \text{pf}} = \frac{100,000}{1.732 \cdot 400 \cdot 0.8} \approx 180.1 A$ IL=3VLpfP=1.732⋅400⋅0.8100,000≈180.1A

Reactive power:

$Q = \sqrt{3} V_L I_L \sin \phi = 3 \cdot 400 \cdot 180.1 \cdot 0.6 \approx 129.7 kVAr$ Q=3VLILsinϕ=3⋅400⋅180.1⋅0.6≈129.7kVAr

Q30. Transformer OC & SC Test

OC test: Applied voltage at rated, no-load current I0 → determine R0, X0
SC test: Reduced voltage applied, shorted secondary → determine R_eq, X_eq

Equations: $Z_{eq} = V_{sc}/I_{sc}, \ R = P_{sc}/I_{sc}^2$ Zeq=Vsc/Isc, R=Psc/Isc2

Q31. Transformer Efficiency

Given: Single-phase transformer, R1 = 0.5 Ω, X1 = 1 Ω, R2 = 0.2 Ω, X2 = 0.8 Ω

Equivalent impedance referred to primary:

$R_{eq} = R_1 + R_2′ (V_1/V_2)^2, \ X_{eq} = X_1 + X_2′ (V_1/V_2)^2$ Req=R1+R2′(V1/V2)2, Xeq=X1+X2′(V1/V2)2

Efficiency:

$\eta = \frac{Output}{Output + Losses} = \frac{V_2 I_2 \cos \phi}{V_2 I_2 \cos \phi + I_2^2 R_{eq} + Core Loss}$ η=Output+LossesOutput=V2I2cosϕ+I22Req+CoreLossV2I2cosϕ

Q32. Ideal Transformer Relations $\frac{V_1}{V_2} = \frac{N_1}{N_2}, \quad \frac{I_1}{I_2} = \frac{N_2}{N_1}, \quad Z_{primary} = Z_{secondary} \left(\frac{N_1}{N_2}\right)^2$ V2V1=N2N1,I2I1=N1N2,Zprimary=Zsecondary(N2N1)2

Q33. DC Series Motor Calculations

Given: V = 220 V, R_a = 0.2 Ω, R_series = 0.5 Ω, load = 5 kW

Total resistance: R_total = 0.2 + 0.5 = 0.7 Ω
Current: I = V / R_total = 220 / 0.7 ≈ 314.3 A
Torque: $T = k \Phi I_a$ T=kΦIa

Q34. Motor Starting Methods

DC series: Direct on line (high starting torque)
AC motors: Star-delta, Autotransformer, Rotor resistance starter
Example: Start a 3-phase 10 kW motor at reduced voltage → calculate starting current

Q35. Induction Motor Output Power

Given: I_L = 20 A, V_L = 400 V, Efficiency = 90%, Slip = 3%

Input power:

$P_{in} = \sqrt{3} V_L I_L \cos \phi$ Pin=3VLILcosϕ

Output power:

$P_{out} = P_{in} \times \eta$ Pout=Pin×η

Rotor copper loss:

$P_{rotor} = s \cdot P_{gap}$ Protor=s⋅Pgap

Q36. Torque-Slip Characteristic

Torque:

$T = \frac{3 V^2 R_2/s}{\omega_s [(R_1 + R_2/s)^2 + (X_1 + X_2)^2]}$ T=ωs[(R1+R2/s)2+(X1+X2)2]3V2R2/s

Maximum torque at:

$s_{max} = R_2 / \sqrt{R_1^2 + (X_1 + X_2)^2}$ smax=R2/R12+(X1+X2)2

Q37. Synchronous Motor – Power & Reactive Power

Given: V = 440 V, I = 20 A, pf = 0.8 lag

Active power: P = √3 V_L I_L cos φ
Reactive power: Q = √3 V_L I_L sin φ
Overexcited: absorbs lagging vars → improves pf

Q38. Power Factor Correction

Using capacitor:

$Q_c = P (\tan \phi_1 – \tan \phi_2)$ Qc=P(tanϕ1−tanϕ2)

Using synchronous condenser: supplies reactive power → improves pf

Q39. DC Shunt Generator Terminal Voltage

Open-circuit voltage: E_0 = 250 V at 1000 rpm
Load current: I_L = 100 A
Terminal voltage:

$V_t = E_0 – I_a R_a = 250 – 100 \cdot 0.5 = 200 V$ Vt=E0−IaRa=250−100⋅0.5=200V

Q40. Parallel Operation of Alternators

Conditions:

Voltage magnitude: V1 ≈ V2
Frequency: f1 = f2
Phase sequence: Same
Phase angle: 0° (synchronizing with lamps or synchroscope)

Q41. ABC-D Parameters of Transmission Line

For short line (less than 80 km), series model:

$V_S = I_S (R + jX) + V_R$ VS=IS(R+jX)+VR

For medium line, nominal π model:

$\begin{bmatrix} V_S \\ I_S \end{bmatrix} = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} V_R \\ I_R \end{bmatrix}$ [VSIS]=[ACBD][VRIR]

Where: $A = D = 1 + YR/2, \quad B = Z (1 + YR/4), \quad C = Y (1 + YR/4)$ A=D=1+YR/2,B=Z(1+YR/4),C=Y(1+YR/4)

Numerical: for Z = R + jX, Y = jB, L = 100 km, plug values to get ABCD.

Q42. Sending End Voltage & Current

Given: 3-phase, 220 kV, load 100 MW, pf 0.8 lag

Load current:

$I_R = \frac{S}{\sqrt{3} V_R} = \frac{100 \times 10^6}{1.732 \cdot 220 \times 10^3 \cdot 0.8} \approx 328.3 A$ IR=3VRS=1.732⋅220×103⋅0.8100×106≈328.3A

Voltage drop: ΔV = I_R Z_line

$V_S = V_R + ΔV$ VS=VR+ΔV

Line losses:

$P_{loss} = 3 I^2 R$ Ploss=3I2R

Efficiency:

$\eta = \frac{P_{load}}{P_{load} + P_{loss}} \times 100$ η=Pload+PlossPload×100

Q43. Voltage Regulation (Synchronous Condenser Method) $VR = \frac{|V_S| – |V_R|}{|V_R|} \times 100$ VR=∣VR∣∣VS∣−∣VR∣×100

Using synchronous condenser, reactive power supplied → improve pf → reduce VR.
Numerical: calculate VR before and after compensation.

Q44. Series & Shunt Compensation

Shunt: capacitor supplies reactive power Qc = P(tanφ1 – tanφ2)
Series: inductor/capacitor reduces line reactance → improve voltage profile
Numerical: for P = 50 MW, pf from 0.8 → 0.95, find Qc.

Q45. Per-Unit System $Z_{pu} = \frac{Z_{actual}}{Z_{base}} \quad \text{where } Z_{base} = \frac{V_{base}^2}{S_{base}}$ Zpu=ZbaseZactualwhere Zbase=SbaseVbase2

Example: Transformer 5 MVA, 33/11 kV

$Z_{base, 33kV} = \frac{33^2}{5} = 217.8 \ \Omega$ Zbase,33kV=5332=217.8 Ω

Convert actual impedances to per-unit: Z_pu = Z_actual / Z_base

Q46. Three-Phase Fault Current $I_f = \frac{V_{LL}}{Z_{line}} = \frac{220 \times 10^3 / \sqrt{3}}{10 + j20} \approx 6.35 kA \angle -63.43^\circ$ If=ZlineVLL=10+j20220×103/3≈6.35kA∠−63.43∘

Q47. Load Flow Analysis (Gauss-Seidel)

Step 1: Initial voltage guess at buses
Step 2: Iteratively update V_i using:

$V_i^{(k+1)} = \frac{1}{Y_{ii}} \left( I_i – \sum_{j \neq i} Y_{ij} V_j^{(k)} \right)$ Vi(k+1)=Yii1Ii−j=i∑YijVj(k)

Repeat until convergence (ΔV < 0.001 pu)

Q48. Reliability in Power Systems

Probability of supply continuity:

$P_{supply} = 1 – P_{failure}$ Psupply=1−Pfailure

For 2 parallel feeders:

$P_{system\ working} = 1 – P_{feeder1\ fails} \cdot P_{feeder2\ fails}$ Psystem working=1−Pfeeder1 fails⋅Pfeeder2 fails

Example: P_feeder = 0.05 → P_system = 1 – 0.05² = 0.9975

Q49. Short-Circuit Current Equations

Three-phase fault:

$I_{sc} = \frac{E}{Z_{th} + Z_{line}}$ Isc=Zth+ZlineE

Line-to-ground fault:

$I_{L-G} = \frac{3 E}{Z_1 + Z_2 + Z_0}$ IL−G=Z1+Z2+Z03E

Numerical: plug values Z1, Z2, Z0 → calculate I_sc

Q50. Economic Load Dispatch

Cost functions: $C_1 = 0.1 P_1^2 + 20 P_1, \ C_2 = 0.08 P_2^2 + 25 P_2$ C1=0.1P12+20P1, C2=0.08P22+25P2

Total load: P1 + P2 = 300 MW
Lagrange multiplier method:

$\frac{dC_1}{dP_1} = \frac{dC_2}{dP_2} = \lambda$ dP1dC1=dP2dC2=λ

Solve: 0.2 P1 + 20 = 0.16 P2 + 25
P1 + P2 = 300

Solution: P1 ≈ 138.46 MW, P2 ≈ 161.54 MW

Q51. Transformer Protection Schemes

Differential Protection: I1 ≈ I2 under normal, trips under fault
Buchholz Relay: detects gas → oil surge, trips if serious fault
Overcurrent Protection: trips on overcurrent above rating
Numerical: setting relay current for rated load + safety margin

Q52. Transmission Line Voltage Drop & Impedance

Given: L = 100 km, R = 0.2 Ω/km, X = 0.5 Ω/km

Total Z = (0.2 + j0.5) * 100 = 20 + j50 Ω
Current: I = S/(√3 V)
Voltage drop: ΔV = I * Z
Efficiency: η = V_R / V_S

Q53. Protective Relay Settings

Overcurrent relay: I_pickup = 1.2 I_load
Time-current characteristics: use IEC or IEEE curves
Numerical: I_load = 200 A → I_pickup = 240 A

Q54. Circuit Breaker Selection

Fault MVA:

$I_{sc} = \frac{S_{sc}}{\sqrt{3} V}$ Isc=3VSsc

Required breaking capacity: I_b = 1.2 × I_sc (safety factor)

Q55. Parallel Transformers Load Sharing

Two transformers: S1 = 5 MVA, S2 = 3 MVA, total load = 6 MW
Load sharing: proportional to rating

$P_1 = \frac{5}{5+3} * 6 = 3.75 MW, \ P_2 = 2.25 MW$ P1=5+35∗6=3.75MW, P2=2.25MW

Q56. Line Losses & Efficiency

Line losses: P_loss = 3 I^2 R
Efficiency: η = P_load / (P_load + P_loss)

Q57. Critical Clearing Time

Formula:

$t_{cr} = \frac{2 L I_{f}}{V_{dc}}$ tcr=Vdc2LIf

Solve numerical for given L, I_f, V

Q58. Reactive Power Control

Shunt capacitors: Qc = V²/Xc
Series capacitors: reduce line reactance → increase P_max
Synchronous condenser: Q_s = V I sin φ

Q59. Bus Bar Arrangements

Single bus: simple, lowest cost
Double bus: flexibility, reliability
Ring bus: high reliability, fault isolation
Numerical: Bus fault current I_f = V/Z_eq

Q60. Line-to-Ground Fault Current

Given: V_LL = 220 kV, Z_line = 10 + j20 Ω/phase

Phase voltage: V_ph = 220/√3 = 127.02 kV
Fault current:

$I_f = \frac{V_ph}{Z_1 + Z_2 + Z_0} = \frac{127.02}{10 + 20?} = \text{calculate based on sequence impedances}$ If=Z1+Z2+Z0Vph=10+20?127.02=calculate based on sequence impedances

Voltage on other phases: 0.577 × I_f × Z
Relay current: I_f

Q61. Measurement of Resistance – Wheatstone Bridge

Bridge balance condition:

$\frac{R_1}{R_2} = \frac{R_x}{R_3} \implies R_x = R_3 \frac{R_1}{R_2}$ R2R1=R3Rx⟹Rx=R3R2R1

Numerical Example:
R1 = 100 Ω, R2 = 150 Ω, R3 = 200 Ω →

$R_x = 200 \cdot 100 / 150 = 133.33 \ \Omega$ Rx=200⋅100/150=133.33 Ω

Q62. Measurement of Inductance – Maxwell Bridge

Bridge balance:

$L_x = R_2 R_3 C_4$ Lx=R2R3C4

Given: R2 = 100 Ω, R3 = 200 Ω, C4 = 10 μF

$L_x = 100 \cdot 200 \cdot 10 \times 10^{-6} = 0.2 H$ Lx=100⋅200⋅10×10−6=0.2H

Q63. Measurement of Capacitance – Schering Bridge

Balance condition:

$C_x = C_2 \frac{R_4}{R_3}, \quad \tan \delta_x = \frac{1}{\omega C_2 R_1}$ Cx=C2R3R4,tanδx=ωC2R11

Numerical: C2 = 0.1 μF, R3 = 1 kΩ, R4 = 2 kΩ → Cx = 0.2 μF

Q64. Measurement of AC Voltage – Using PMMC Meter

PMMC requires rectifier: V_ac measured as:

$V_{dc} = \frac{V_{ac}}{\sqrt{2}}$ Vdc=2Vac

Numerical: If meter reads 50 V dc for 50 V AC input → calibration factor applied.

Q65. Wattmeter – Power Measurement

3-phase power (balanced load):

$P = \sqrt{3} V_L I_L \cos \phi \quad \text{or two-wattmeter method}$ P=3VLILcosϕor two-wattmeter method

Two-wattmeter readings:

$W_1 = V_L I_L \cos(30 + \phi), \quad W_2 = V_L I_L \cos(30 – \phi)$ W1=VLILcos(30+ϕ),W2=VLILcos(30−ϕ)

Total power: P = W1 + W2

Q66. Energy Meter – Electromechanical

Energy consumption:

$W = V I t \cos \phi$ W=VItcosϕ

Disk rotates proportional to energy; use meter constant k Wh/rev.
Numerical:
V = 220 V, I = 5 A, t = 2 h, pf = 0.8 →

$W = 220 \cdot 5 \cdot 2 \cdot 0.8 = 1760 Wh$ W=220⋅5⋅2⋅0.8=1760Wh

Q67. Instrument Transformer – CT & PT Calculations

Current transformer:

$I_s = I_p / n, \quad V_s = I_s R_L$ Is=Ip/n,Vs=IsRL

Potential transformer:

$V_s = V_p / n$ Vs=Vp/n

Example: CT ratio 1000/5, primary current 500 A → secondary I_s = 2.5 A

Q68. Power Factor Measurement Using Three-Wattmeter Method

3-phase, line voltage V_L, line current I_L, readings W1, W2, W3
Power factor:

$\cos \phi = \frac{W_1 + W_2 + W_3}{\sqrt{3} V_L I_L}$ cosϕ=3VLILW1+W2+W3

Numerical: W1 = 1000 W, W2 = 900 W, W3 = 1100 W, V_L = 400 V, I_L = 10 A →

$\cos \phi = (1000 + 900 + 1100)/(1.732 \cdot 400 \cdot 10) \approx 0.91$ cosϕ=(1000+900+1100)/(1.732⋅400⋅10)≈0.91

Q69. Measurement of Inductance Using Hay’s Bridge

Balance:

$L_x = \frac{R_1 R_4 C_1}{R_3}$ Lx=R3R1R4C1

Numerical: R1 = 100 Ω, R4 = 200 Ω, C1 = 10 μF, R3 = 50 Ω →

$L_x = 100 \cdot 200 \cdot 10 \cdot 10^{-6} / 50 = 0.4 H$ Lx=100⋅200⋅10⋅10−6/50=0.4H

Q70. Measurement of Resistance by Kelvin Double Bridge

Used for low resistance (mΩ range)
Balance:

$R_x = \frac{R_2}{R_1} R_3$ Rx=R1R2R3

Numerical: R1 = 10 Ω, R2 = 100 Ω, R3 = 0.01 Ω → Rx = 0.1 Ω

Q71. Digital Multimeter Measurements

DC voltage: direct reading
AC voltage: RMS via rectifier
Current: shunt resistor or CT
Resistance: voltage-current method
Example: DMM measures 5.01 V DC → 1% accuracy → actual = 4.96–5.06 V

Q72. Oscilloscope Measurement

Voltage: V_p-p, V_rms = V_p / √2
Frequency: f = 1/T, T = time period on CRT
Numerical: 0.02 s/div, 5 div → T = 0.1 s → f = 10 Hz

Q73. Instrument Accuracy & Error

% Error = $\frac{Measured – True}{True} \cdot 100\%$ TrueMeasured−True⋅100%
Example: True current 5 A, measured 5.1 A → Error = 2%

Q74. Measurement of Power in Three-Phase Induction Motor

Use two-wattmeter method, W1 and W2
Total power: P = W1 + W2
PF: cos φ = (W1 + W2)/(√3 V_L I_L)
Reactive power: Q = √(W1² + W2² – W1 W2 * 3)

Q75. Protection of Measuring Instruments

Fuses or MCBs to prevent overload
Use CT & PT for high voltage/current circuits
Example: Meter rated 5 A, actual may reach 50 A → use 10:1 CT

Q76. Control System – Transfer Function Derivation

Standard first-order system:

$G(s) = \frac{C(s)}{R(s)} = \frac{K}{\tau s + 1}$ G(s)=R(s)C(s)=τs+1K

Second-order system:

$G(s) = \frac{\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2}$ G(s)=s2+2ζωns+ωn2ωn2

Numerical Example: K = 5, τ = 0.2 s →

$G(s) = \frac{5}{0.2 s + 1}$ G(s)=0.2s+15

Q77. Time Response of First-Order System

Step response:

$c(t) = K (1 – e^{-t/\tau})$ c(t)=K(1−e−t/τ)

Example: K = 10, τ = 0.5 s, t = 1 s →

$c(1) = 10(1 – e^{-1/0.5}) = 10(1 – e^{-2}) \approx 10(1 – 0.135) = 8.65$ c(1)=10(1−e−1/0.5)=10(1−e−2)≈10(1−0.135)=8.65

Q78. Steady-State Error for Unit Step Input

Unity feedback system:

$e_{ss} = \frac{1}{1 + K_p}, \quad K_p = \lim_{s \to 0} G(s)$ ess=1+Kp1,Kp=s→0limG(s)

Example: G(s) = 10/(s+10) → Kp = 10/10 = 1 → e_ss = 1/2 = 0.5

Q79. Routh-Hurwitz Stability Criterion

Characteristic equation: s^3 + 6s^2 + 11s + 6 = 0

Form Routh array:

| s³ | 1 | 11 |
| s² | 6 | 6 |
| s¹ | (611 – 16)/6 = 10 | 0 |
| s⁰ | 6 | |

All first column positive → system stable

Q80. Root Locus Sketch

Given: G(s)H(s) = K / (s(s+2))
Poles at 0 and -2, no zeros → root locus starts at poles, ends at infinity
Angle of asymptotes: ±180°/(n-m) = ±180°/2 = ±90°
Breakaway point: Solve dK/dσ = 0

Q81. Frequency Response – Bode Plot

G(s) = 100/(s(s+10)) →
Magnitude: |G(jω)| = 20 log K – 20 log ω – 20 log √(ω² + 100)
Phase: φ = -90° – arctan(ω/10)
Plot magnitude & phase vs log ω

Q82. Nyquist Stability Criterion

Open-loop transfer function: G(s)H(s) = 1/(s+1)
Nyquist plot → encirclements of (-1,0) → system stable, 0 encirclements → stable

Q83. PID Controller Design

Controller:

$G_c(s) = K_p + \frac{K_i}{s} + K_d s$ Gc(s)=Kp+sKi+Kds

Example: Given rise time and settling time → tune Kp, Ki, Kd using Ziegler-Nichols method

Q84. State-Space Representation

System: $\dot{x} = Ax + Bu, \ y = Cx + Du$ x˙=Ax+Bu, y=Cx+Du
Example: 1st order: $\dot{x} + 2x = u$ x˙+2x=u → A = -2, B = 1, C = 1, D = 0

Q85. Observability & Controllability

Controllability matrix: $[B, AB, A^2B, …]$ [B,AB,A2B,…] → rank = n → controllable
Observability matrix: $[C; CA; CA^2; …]$ [C;CA;CA2;…] → rank = n → observable
Numerical: check determinant ≠ 0 → system fully controllable/observable

Q86. DC-DC Converter – Buck Converter

Output voltage:

$V_o = D V_s$ Vo=DVs

Example: V_s = 24 V, D = 0.5 → V_o = 12 V

Q87. DC-DC Converter – Boost Converter

Output voltage:

$V_o = \frac{V_s}{1 – D}$ Vo=1−DVs

V_s = 12 V, D = 0.6 → V_o = 12 / 0.4 = 30 V

Q88. Rectifiers & Ripple Factor

Full-wave rectifier:

$r = \frac{I_{rms, AC}}{I_{DC}} = 0.482$ r=IDCIrms,AC=0.482

Numerical: V_DC = 100 V → I_rms, AC = 0.482 * 100 ≈ 48.2 A

Q89. Thyristor Firing Angle

Controlled rectifier:

$V_{DC} = \frac{V_m}{\pi}(1 + \cos \alpha)$ VDC=πVm(1+cosα)

Example: Vm = 100 V, α = 60° → V_DC = 100/π (1 + 0.5) ≈ 47.75 V

Q90. Inverter Output Voltage

Square-wave inverter: V_dc → V_rms = V_dc
PWM inverter:

$V_o = m_a V_dc, \quad m_a = modulation index$ Vo=maVdc,ma=modulationindex

Example: V_dc = 220 V, m_a = 0.8 → V_o = 176 V

Q91. AC Voltage Controllers

RMS voltage controlled using SCR:

$V_{rms} = V_s \sqrt{1 – \frac{\alpha}{\pi} + \frac{\sin 2\alpha}{2\pi}}$ Vrms=Vs1−πα+2πsin2α

Numerical: V_s = 230 V, α = 60° → V_rms ≈ 180 V

Q92. Single-Phase AC Drives

Voltage control for speed regulation: V ∝ speed
Example: V = 230 V, rated speed = 1500 rpm → at 200 V, speed ≈ 1300 rpm

Q93. Three-Phase Inverter Calculations

Line-to-line voltage: V_LL = √3 V_phase
RMS output with PWM: V_rms = V_dc / √2

Q94. Chopper Control – Step-Up/Step-Down

Step-down (Buck): V_o = D V_s
Step-up (Boost): V_o = V_s / (1 – D)
Numerical examples with given D

Q95. SCR Commutation Techniques

Natural: AC supply zero crossing → SCR turns off
Forced: Capacitor, inductor, or auxiliary SCR → turn-off in DC circuits
Numerical: calculate commutation time τ = L / R

Q96. Phase-Controlled Rectifier

Output voltage for single-phase half-controlled:

$V_{DC} = \frac{V_m}{2\pi} (1 + \cos \alpha)$ VDC=2πVm(1+cosα)

Full-controlled bridge: V_DC = Vm / π (1 + cos α)

Q97. Cycloconverter Output Frequency

f_out < f_in
Step-down conversion: f_out = m f_in
Numerical: f_in = 50 Hz, m = 0.5 → f_out = 25 Hz

Q98. PWM Techniques

Voltage control via duty cycle:

$V_o = m_a V_{dc}$ Vo=maVdc

Harmonics reduced with high-frequency PWM

Q99. Chopper Efficiency

Efficiency:

$\eta = \frac{V_o I_o}{V_s I_s}$ η=VsIsVoIo

Example: V_o = 100 V, I_o = 10 A, V_s I_s = 1050 W → η = 1000/1050 ≈ 95.2%

Q100. Power Electronics Applications – DC Motor Speed Control

DC series/shunt motor: Vary V or use chopper for speed regulation
Example: Rated 220 V, 5 kW motor → using buck chopper V = 150 V → speed ≈ 150/220 × rated speed ≈ 68% rated speed

Disclaimer:
This mock test is for practice and educational purposes only. It is not affiliated with or endorsed by the Bihar Public Service Commission (BPSC). Results do not reflect official exam outcomes.

BPSC AEO Electrical Main Exam Mock Test – Part 1 (Circuits & Network Analysis, Q1–20)

Part 2 – Electrical Machines (Q21–40)

Part 3 – Power Systems & Protection (Q41–60)

Part 4 – Electrical Measurements & Instrumentation (Q61–75)

Part 5 – Control Systems & Power Electronics (Q76–100)

BPSC AEO Electrical Main Exam – Part 1 Solutions (Q1–20)

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