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SSC JE Electrical Paper-II Practice Paper (2026 Exam Pattern, Solved Set)

SSC Junior Engineer (JE) • 2026 • 100 Questions • Free practice with solutions

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Showing 100 Questions with Step-by-Step Solutions
Q1Electrical Engineering (SSC JE)
Easy
A 6 Ω and a 3 Ω resistor in parallel are connected in series with a 4 Ω resistor across a 12 V supply. The current drawn from the supply is
A. 3 A
B. 1 A
C. 4 A
D. 2 A
View answer and explanation
Correct Option: D
6 Ω ∥ 3 Ω = 2 Ω. Total = 2 + 4 = 6 Ω. I = 12/6 = 2 A.
Q2Electrical Engineering (SSC JE)
Easy
Resistors of 20 kΩ and 30 kΩ are connected in series across a 100 V supply. The voltage across the 30 kΩ resistor is
A. 50 V
B. 40 V
C. 75 V
D. 60 V
View answer and explanation
Correct Option: D
V = 100 × 30 / (20 + 30) = 60 V.
Q3Electrical Engineering (SSC JE)
Easy
A 2 kW heater runs for 3 hours every day for 30 days. At ₹6 per unit, the monthly cost of the energy is
A. ₹540
B. ₹180
C. ₹1,080
D. ₹2,160
View answer and explanation
Correct Option: C
Energy = 2 × 3 × 30 = 180 kWh. Cost = 180 × 6 = ₹1,080.
Q4Electrical Engineering (SSC JE)
Hard
A battery of emf 12 V and internal resistance 0.5 Ω supplies a 5.5 Ω load. The terminal voltage is
A. 11.5 V
B. 11 V
C. 10.5 V
D. 12 V
View answer and explanation
Correct Option: B
I = 12/(0.5 + 5.5) = 2 A. Terminal voltage = 12 − 2 × 0.5 = 11 V.
Q5Electrical Engineering (SSC JE)
Easy
Capacitors of 6 μF and 3 μF are connected in series. The equivalent capacitance is
A. 18 μF
B. 2 μF
C. 4.5 μF
D. 9 μF
View answer and explanation
Correct Option: B
C = (6 × 3)/(6 + 3) = 2 μF.
Q6Electrical Engineering (SSC JE)
Easy
If the distance between the plates of a parallel plate capacitor is doubled, keeping everything else the same, the capacitance
A. becomes half
B. remains the same
C. becomes four times
D. doubles
View answer and explanation
Correct Option: A
C = εA/d, so C is inversely proportional to the plate separation d.
Q7Electrical Engineering (SSC JE)
Hard
A resistor has a resistance of 10 Ω at 20 °C and a temperature coefficient of 0.004 per °C at 20 °C. Its resistance at 70 °C is
A. 12 Ω
B. 14 Ω
C. 10.2 Ω
D. 13 Ω
View answer and explanation
Correct Option: A
Rt = R0 [1 + α (T − T0)] = 10 × [1 + 0.004 × 50] = 12 Ω.
Q8Electrical Engineering (SSC JE)
Hard
A current of 5 A flows for 2 minutes through a 10 Ω resistor. The heat produced is
A. 3 kJ
B. 30 kJ
C. 600 J
D. 300 kJ
View answer and explanation
Correct Option: B
H = I²Rt = 25 × 10 × 120 = 30 000 J.
Q9Electrical Engineering (SSC JE)
Medium
The resistivity of annealed copper at 20 °C is about
A. 1.7 × 10⁻¹¹ Ω·m
B. 1.7 × 10⁻² Ω·m
C. 1.7 × 10⁻⁵ Ω·m
D. 1.7 × 10⁻⁸ Ω·m
View answer and explanation
Correct Option: D
Copper has one of the lowest resistivities among common metals (silver is slightly lower).
Q10Electrical Engineering (SSC JE)
Medium
The maximum permissible operating temperature of Class A insulation is
A. 105 °C
B. 155 °C
C. 180 °C
D. 130 °C
View answer and explanation
Correct Option: A
Class A (cotton, silk, paper impregnated) is 105 °C, Class B 130 °C, Class F 155 °C and Class H 180 °C.
Q11Electrical Engineering (SSC JE)
Medium
The specific gravity of the electrolyte in a fully charged lead-acid cell is about
A. 1.8
B. 1.5
C. 1.2 to 1.28
D. 1.0
View answer and explanation
Correct Option: C
It falls toward 1.1 as the cell discharges, because sulphuric acid is used up and water forms. This is why a hydrometer shows the state of charge.
Q12Electrical Engineering (SSC JE)
Medium
In a network with b branches and n nodes, the number of independent KVL (loop) equations is
A. n − 1
B. b + n − 1
C. b − n
D. b − n + 1
View answer and explanation
Correct Option: D
There are n − 1 independent KCL equations and b − (n − 1) = b − n + 1 independent loop equations.
Q13Electrical Engineering (SSC JE)
Hard
A series circuit with R = 30 Ω, XL = 70 Ω and XC = 30 Ω is connected to a 100 V ac supply. The current is
A. 1 A
B. 10 A
C. 3.3 A
D. 2 A
View answer and explanation
Correct Option: D
X = 70 − 30 = 40 Ω. |Z| = √(30² + 40²) = 50 Ω. I = 100/50 = 2 A.
Q14Electrical Engineering (SSC JE)
Hard
A load has V = 200∠0° V and I = 10∠−30° A. The reactive power is
A. 500 VAr
B. 2000 VAr
C. 1000 VAr
D. 1732 VAr
View answer and explanation
Correct Option: C
Q = V I sin φ = 200 × 10 × sin 30° = 1000 VAr. The real power is 200 × 10 × cos 30° = 1732 W.
Q15Electrical Engineering (SSC JE)
Hard
A parallel resonant circuit has L = 1 mH, C = 1 μF and a coil resistance of 10 Ω. The dynamic impedance at resonance is
A. 10 Ω
B. 31.6 Ω
C. 100 Ω
D. 1000 Ω
View answer and explanation
Correct Option: C
Zd = L / (C R) = 10⁻³ / (10⁻⁶ × 10) = 100 Ω.
Q16Electrical Engineering (SSC JE)
Easy
In a series RC circuit being charged from a dc supply, the capacitor voltage reaches what fraction of the final value after one time constant?
A. 50 percent
B. 63.2 percent
C. 36.8 percent
D. 86.5 percent
View answer and explanation
Correct Option: B
v(t) = V (1 − e^(−t/τ)). At t = τ this is V (1 − 0.368) = 0.632 V.
Q17Electrical Engineering (SSC JE)
Medium
For a reciprocal two-port network, the ABCD parameters satisfy
A. AD − BC = 0
B. AD − BC = 1
C. AD + BC = 1
D. AD = BC
View answer and explanation
Correct Option: B
Reciprocity requires the determinant of the transmission matrix to be one.
Q18Electrical Engineering (SSC JE)
Hard
A series RLC circuit has R = 10 Ω, L = 1 mH and C = 1 μF. The quality factor is about
A. 31.6
B. 0.316
C. 3.16
D. 316
View answer and explanation
Correct Option: C
Q = (1/R) √(L/C) = (1/10) × √(10⁻³/10⁻⁶) = 31.62/10 = 3.16.
Q19Electrical Engineering (SSC JE)
Easy
In a balanced three-phase four-wire system, the current in the neutral wire is
A. zero
B. √3 times the line current
C. three times the line current
D. equal to the line current
View answer and explanation
Correct Option: A
The three phase currents are equal in magnitude and 120° apart, so their phasor sum is zero.
Q20Electrical Engineering (SSC JE)
Easy
A balanced delta-connected load has a phase current of 10 A. The line current is
A. 17.3 A
B. 30 A
C. 10 A
D. 5.77 A
View answer and explanation
Correct Option: A
In delta IL = √3 Iph = 1.732 × 10 = 17.3 A.
Q21Electrical Engineering (SSC JE)
Hard
In a two-wattmeter test on a balanced load, the readings are 3 kW and 1 kW (both positive). The power factor is about
A. 0.5
B. 0.756
C. 0.6
D. 0.866
View answer and explanation
Correct Option: B
tan φ = √3 (W1 − W2)/(W1 + W2) = 1.732 × 2/4 = 0.866, so φ = 40.9° and cos φ = 0.756.
Q22Electrical Engineering (SSC JE)
Hard
Two coils with L1 = 1 H, L2 = 2 H and mutual inductance M = 0.5 H are connected in series aiding. The total inductance is
A. 4 H
B. 3 H
C. 2 H
D. 3.5 H
View answer and explanation
Correct Option: A
Ltotal = L1 + L2 + 2M = 1 + 2 + 1 = 4 H. In series opposing it would be 2 H.
Q23Electrical Engineering (SSC JE)
Medium
For maximum power transfer in an ac circuit with a complex source impedance Zs, the load impedance should be
A. the complex conjugate of Zs
B. zero
C. the reciprocal of Zs
D. equal to Zs
View answer and explanation
Correct Option: A
This cancels the reactance and makes the load resistance equal to the source resistance.
Q24Electrical Engineering (SSC JE)
Easy
A 12 V source in series with a 4 Ω resistor is converted to its Norton equivalent. The Norton current is
A. 0.33 A
B. 3 A
C. 12 A
D. 48 A
View answer and explanation
Correct Option: B
IN = VTh / RTh = 12/4 = 3 A, with the 4 Ω in parallel.
Q25Electrical Engineering (SSC JE)
Medium
Millman’s theorem is used to find
A. the maximum power transferred
B. the current in a series circuit
C. the phase angle of a load
D. the voltage at the common node of parallel branches containing sources and resistances
View answer and explanation
Correct Option: D
V = Σ(Vk/Rk) / Σ(1/Rk) reduces several parallel branches to a single equivalent source.
Q26Electrical Engineering (SSC JE)
Hard
The peak value of a full-wave rectified sine wave is 157 V. Its average value is about
A. 100 V
B. 50 V
C. 111 V
D. 157 V
View answer and explanation
Correct Option: A
Vavg = 2Vm/π = 2 × 157/3.1416 = 99.95 ≈ 100 V.
Q27Electrical Engineering (SSC JE)
Hard
A coil of 200 turns is wound on a core whose reluctance is 4 × 10⁵ AT/Wb. Its self-inductance is
A. 1.0 H
B. 0.2 H
C. 0.01 H
D. 0.1 H
View answer and explanation
Correct Option: D
L = N²/S = 200² / (4 × 10⁵) = 40 000 / 400 000 = 0.1 H.
Q28Electrical Engineering (SSC JE)
Easy
A conductor 0.4 m long carries 10 A at right angles to a field of 0.5 T. The force on it is
A. 5 N
B. 2 N
C. 0.2 N
D. 20 N
View answer and explanation
Correct Option: B
F = B I L = 0.5 × 10 × 0.4 = 2 N.
Q29Electrical Engineering (SSC JE)
Easy
The energy lost per cycle per unit volume due to hysteresis is equal to
A. the coercive force
B. the area of the B-H loop
C. the peak value of B
D. the residual flux density
View answer and explanation
Correct Option: B
The area of the loop (in J/m³) is the energy dissipated as the material is taken once through the cycle.
Q30Electrical Engineering (SSC JE)
Easy
The eddy current loss in a core varies as
A. the cube of the lamination thickness
B. the inverse of the lamination thickness
C. the lamination thickness
D. the square of the lamination thickness
View answer and explanation
Correct Option: D
Pe ∝ f² Bm² t². Thin laminations are used to reduce this loss.
Q31Electrical Engineering (SSC JE)
Hard
A DC shunt generator delivers 40 A at 220 V. The field current is 2 A and the armature resistance is 0.1 Ω. The generated emf is
A. 216 V
B. 224.2 V
C. 228 V
D. 220 V
View answer and explanation
Correct Option: B
Ia = 40 + 2 = 42 A. E = V + Ia Ra = 220 + 42 × 0.1 = 224.2 V.
Q32Electrical Engineering (SSC JE)
Hard
A DC motor delivers 10 kW at 1000 rpm. The shaft torque is about
A. 10 N·m
B. 9.55 N·m
C. 95.5 N·m
D. 955 N·m
View answer and explanation
Correct Option: C
T = P / ω = 10 000 × 60 / (2π × 1000) = 95.5 N·m.
Q33Electrical Engineering (SSC JE)
Hard
A DC shunt motor on a 220 V supply with Ra = 0.4 Ω runs at 1000 rpm taking 25 A. When the armature current rises to 50 A with the same flux, the speed becomes about
A. 952 rpm
B. 905 rpm
C. 1000 rpm
D. 1050 rpm
View answer and explanation
Correct Option: A
Eb1 = 220 − 10 = 210 V and Eb2 = 220 − 20 = 200 V. N2 = 1000 × 200/210 = 952 rpm.
Q34Electrical Engineering (SSC JE)
Easy
Swinburne’s test cannot be used for
A. a DC shunt generator
B. a DC series motor
C. a DC shunt motor
D. a separately excited DC motor
View answer and explanation
Correct Option: B
The test needs the machine to run at no load with a constant flux. A series motor would race at no load.
Q35Electrical Engineering (SSC JE)
Hard
The secondary voltage of a transformer falls from 240 V at no load to 230 V at full load. The percentage regulation is about
A. 4.35 percent
B. 10 percent
C. 2.2 percent
D. 4.17 percent
View answer and explanation
Correct Option: D
Regulation of a transformer is taken with respect to the no-load voltage: (240 − 230)/240 = 4.17 percent.
Q36Electrical Engineering (SSC JE)
Hard
A 100 kVA transformer has an iron loss of 1 kW and a full-load copper loss of 1.5 kW. Its efficiency at full load and unity power factor is about
A. 97.0 percent
B. 98.5 percent
C. 96.2 percent
D. 97.6 percent
View answer and explanation
Correct Option: D
Output = 100 kW. Losses = 2.5 kW. η = 100 / 102.5 = 0.9756.
Q37Electrical Engineering (SSC JE)
Hard
For the transformer of 1 kW iron loss and 1.5 kW full-load copper loss, the maximum efficiency occurs at a load of about
A. 75 percent of full load
B. 100 percent of full load
C. 66.7 percent of full load
D. 81.6 percent of full load
View answer and explanation
Correct Option: D
x = √(Pi / Pcu,fl) = √(1/1.5) = 0.816.
Q38Electrical Engineering (SSC JE)
Medium
For the parallel operation of two three-phase transformers, it is essential that they have
A. the same phase sequence and zero relative phase displacement
B. the same kVA rating
C. the same cooling system
D. the same number of tappings
View answer and explanation
Correct Option: A
Without matching phase displacement, a large circulating current would flow between them even at no load.
Q39Electrical Engineering (SSC JE)
Medium
The phase displacement between the primary and secondary line voltages of a Dyn11 transformer is
A. 120°
B. 30°
C. 60°
D. 0°
View answer and explanation
Correct Option: B
In the vector group Dyn11 the low voltage side leads the high voltage side by 30° (11 o’clock).
Q40Electrical Engineering (SSC JE)
Medium
In an induction motor, the slip at which the maximum torque occurs is equal to
A. X2 / R2
B. R2 × X2
C. R2 / X2
D. R2 + X2
View answer and explanation
Correct Option: C
sm = R2/X2, where R2 is the rotor resistance and X2 the standstill rotor reactance. The value of the maximum torque is independent of R2.
Q41Electrical Engineering (SSC JE)
Easy
The input to the rotor of an induction motor is 10 kW and the slip is 0.04. The rotor copper loss is
A. 0.04 kW
B. 0.4 kW
C. 9.6 kW
D. 4 kW
View answer and explanation
Correct Option: B
Rotor copper loss = s × rotor input = 0.04 × 10 = 0.4 kW. The mechanical power developed is 9.6 kW.
Q42Electrical Engineering (SSC JE)
Easy
The starting torque of a slip-ring induction motor can be improved by
A. removing the slip rings
B. adding external resistance in the rotor circuit
C. reducing the supply voltage
D. increasing the supply frequency
View answer and explanation
Correct Option: B
The extra resistance raises the rotor power factor at standstill and, within limits, the starting torque. It is cut out as the motor speeds up.
Q43Electrical Engineering (SSC JE)
Easy
A 50 Hz alternator running at 300 rpm has how many poles?
A. 24
B. 40
C. 10
D. 20
View answer and explanation
Correct Option: D
P = 120 f / Ns = 120 × 50 / 300 = 20.
Q44Electrical Engineering (SSC JE)
Medium
In the V-curve of a synchronous motor (armature current against field current), the armature current is minimum at
A. 0.5 power factor lagging
B. zero power factor leading
C. unity power factor
D. zero power factor lagging
View answer and explanation
Correct Option: C
To the left of the minimum the motor is under-excited (lagging) and to the right over-excited (leading).
Q45Electrical Engineering (SSC JE)
Medium
Damper windings in a synchronous machine are provided to
A. prevent hunting and provide starting torque for motors
B. reduce the stator copper loss
C. raise the efficiency at light loads
D. increase the field flux
View answer and explanation
Correct Option: A
The bars carry induced currents that oppose any oscillation of the rotor about synchronous speed.
Q46Electrical Engineering (SSC JE)
Easy
Among single-phase induction motors, the one with the highest starting torque is the
A. capacitor-start motor
B. reluctance motor
C. resistance split-phase motor
D. shaded-pole motor
View answer and explanation
Correct Option: A
A large capacitor in the starting winding gives a phase shift of nearly 90°, producing starting torque of around 3 to 4 times the full-load value.
Q47Electrical Engineering (SSC JE)
Easy
A stepper motor has 200 steps per revolution. Its step angle is
A. 1.8°
B. 0.9°
C. 18°
D. 3.6°
View answer and explanation
Correct Option: A
Step angle = 360°/200 = 1.8°.
Q48Electrical Engineering (SSC JE)
Medium
Because of armature reaction in a DC generator, the magnetic neutral axis shifts
A. in the direction of rotation
B. by exactly 90° electrical
C. against the direction of rotation
D. not at all
View answer and explanation
Correct Option: A
In a generator the brushes must be moved forward (in the direction of rotation) to the new neutral position. In a motor they move backward.
Q49Electrical Engineering (SSC JE)
Medium
The ABCD constants of a short transmission line with series impedance Z are
A. A = 1, B = 0, C = Z, D = 1
B. A = D = 1, B = Z, C = 0
C. A = D = Z, B = C = 1
D. A = D = 0, B = 1, C = Z
View answer and explanation
Correct Option: B
With negligible shunt admittance, Vs = Vr + Z Ir and Is = Ir.
Q50Electrical Engineering (SSC JE)
Medium
The Ferranti effect refers to
A. the rise of receiving end voltage above the sending end voltage on a lightly loaded long line
B. the loss of power by corona
C. the unequal current distribution in a conductor
D. the fall of receiving end voltage under heavy loads
View answer and explanation
Correct Option: A
The charging current of the line capacitance flows through the line inductance and raises the voltage at the open end.
Q51Electrical Engineering (SSC JE)
Hard
A line has an inductance of 1.2 mH/km and a capacitance of 0.012 μF/km. The surge impedance is about
A. 1000 Ω
B. 316 Ω
C. 31.6 Ω
D. 100 Ω
View answer and explanation
Correct Option: B
Zc = √(L/C) = √(1.2 × 10⁻³ / 1.2 × 10⁻⁸) = √10⁵ = 316 Ω.
Q52Electrical Engineering (SSC JE)
Hard
A bus has a Thevenin reactance of 0.2 pu on a 100 MVA base. The short-circuit level at that bus is
A. 500 MVA
B. 100 MVA
C. 200 MVA
D. 20 MVA
View answer and explanation
Correct Option: A
Short-circuit MVA = base MVA / X pu = 100/0.2 = 500 MVA.
Q53Electrical Engineering (SSC JE)
Easy
In a balanced three-phase short circuit, the fault current is determined by
A. the zero sequence network only
B. the positive sequence network only
C. the negative sequence network only
D. all three sequence networks in series
View answer and explanation
Correct Option: B
A symmetrical fault has no negative or zero sequence components.
Q54Electrical Engineering (SSC JE)
Medium
The making current of a circuit breaker is related to its symmetrical breaking current by
A. approximately 0.5 times
B. equal to it
C. approximately 10 times
D. approximately 2.55 times
View answer and explanation
Correct Option: D
The peak of the first cycle of the asymmetrical fault current is about 2.55 times the rms symmetrical value, and the breaker must close on it without damage.
Q55Electrical Engineering (SSC JE)
Hard
An overcurrent relay has a plug setting of 125 percent on a 5 A coil and is fed from a 400/5 A CT. For a primary fault current of 2000 A, the plug setting multiplier is
A. 3.2
B. 8
C. 4
D. 5
View answer and explanation
Correct Option: C
CT ratio = 80, so the relay current = 2000/80 = 25 A. Pick-up = 1.25 × 5 = 6.25 A. PSM = 25/6.25 = 4.
Q56Electrical Engineering (SSC JE)
Medium
A major advantage of HVDC transmission over long distances is that it
A. has no reactive power loss and no skin effect
B. uses simple, low-cost terminal equipment
C. needs no converter stations
D. needs a larger right-of-way than ac
View answer and explanation
Correct Option: A
DC lines have no charging current, no stability limit from the line angle and carry power on fewer conductors. The converter stations are the main cost.
Q57Electrical Engineering (SSC JE)
Hard
A travelling wave on a line has a reflection coefficient at an open-circuited end of
A. −1
B. 2
C. 0
D. +1
View answer and explanation
Correct Option: D
The current reflects with opposite sign and the voltage doubles, so the voltage reflection coefficient is +1. At a short-circuited end it is −1.
Q58Electrical Engineering (SSC JE)
Medium
A Petersen coil is used to
A. neutralise the capacitive earth fault current
B. improve the power factor
C. protect against lightning
D. increase the fault current
View answer and explanation
Correct Option: A
It is a tuned reactor between the system neutral and earth. At resonance, it cancels the capacitive current at the fault, which extinguishes the arc.
Q59Electrical Engineering (SSC JE)
Hard
A two-part tariff has a charge of ₹200 per kW of maximum demand and ₹5 per kWh. A consumer with a maximum demand of 10 kW uses 2000 kWh in a month. The monthly bill (taking the demand charge per month) is
A. ₹12,000
B. ₹10,000
C. ₹12,500
D. ₹22,000
View answer and explanation
Correct Option: A
Demand charge = 200 × 10 = ₹2,000. Energy charge = 5 × 2000 = ₹10,000. Total = ₹12,000.
Q60Electrical Engineering (SSC JE)
Hard
A 100 MW plant operates at a capacity factor of 0.5 for a year of 8760 hours. The energy generated is
A. 876 000 MWh
B. 438 000 MWh
C. 43 800 MWh
D. 219 000 MWh
View answer and explanation
Correct Option: B
Energy = 100 × 8760 × 0.5 = 438 000 MWh.
Q61Electrical Engineering (SSC JE)
Medium
The on-load tap changer of a transformer is used to
A. change the frequency
B. increase the efficiency
C. regulate the voltage without interrupting the supply
D. reduce the iron loss
View answer and explanation
Correct Option: C
Taps in the winding are changed while the transformer carries load, using diverter and selector switches.
Q62Electrical Engineering (SSC JE)
Medium
In a single-core cable, the maximum electrostatic stress in the dielectric occurs
A. midway in the insulation
B. at the armour
C. at the sheath
D. at the surface of the conductor
View answer and explanation
Correct Option: D
The stress varies inversely with the radius, so it is highest at the conductor surface. Grading is used to even it out.
Q63Electrical Engineering (SSC JE)
Easy
The scale of a moving iron instrument is
A. logarithmic
B. cramped at the upper end
C. uniform throughout
D. non-uniform, cramped at the lower end
View answer and explanation
Correct Option: D
The deflection torque varies with the square of the current, so the scale is crowded at the low end.
Q64Electrical Engineering (SSC JE)
Hard
An energy meter has a constant of 600 rev/kWh. The disc makes 60 revolutions in 5 minutes. The load is
A. 2.4 kW
B. 0.6 kW
C. 12 kW
D. 1.2 kW
View answer and explanation
Correct Option: D
60 rev = 0.1 kWh in 5 min = 1/12 h. Power = 0.1 × 12 = 1.2 kW.
Q65Electrical Engineering (SSC JE)
Easy
A dc potentiometer is standardised using
A. a standard cell
B. a battery of unknown emf
C. a voltmeter
D. a decade resistance box
View answer and explanation
Correct Option: A
The working current is adjusted so the voltage drop along the wire matches the known emf of the standard cell, giving a known potential gradient.
Q66Electrical Engineering (SSC JE)
Easy
An LVDT is used to measure
A. resistance
B. temperature
C. displacement
D. frequency
View answer and explanation
Correct Option: C
A linear variable differential transformer converts the position of a core into an ac voltage proportional to the displacement.
Q67Electrical Engineering (SSC JE)
Easy
A thermocouple works on the principle of the
A. Peltier effect
B. Seebeck effect
C. Piezoelectric effect
D. Hall effect
View answer and explanation
Correct Option: B
A temperature difference between two junctions of dissimilar metals produces an emf (the Seebeck effect).
Q68Electrical Engineering (SSC JE)
Easy
A current transformer of ratio 200/5 is connected to an ammeter which reads 4 A. The primary current is
A. 4 A
B. 200 A
C. 160 A
D. 40 A
View answer and explanation
Correct Option: C
The ratio is 40, so I1 = 4 × 40 = 160 A.
Q69Electrical Engineering (SSC JE)
Hard
A 3½ digit digital voltmeter can display a maximum reading of
A. 1999
B. 3999
C. 9999
D. 999
View answer and explanation
Correct Option: A
The half digit can only be 0 or 1, so the display runs from −1999 to +1999.
Q70Electrical Engineering (SSC JE)
Hard
On the 2 V range, a 3½ digit digital voltmeter has a resolution of
A. 1 V
B. 1 mV
C. 0.1 mV
D. 10 mV
View answer and explanation
Correct Option: B
The full-scale display is 1.999 V with three decimals, so the smallest step is 0.001 V.
Q71Electrical Engineering (SSC JE)
Easy
An error due to wrong calibration of an instrument is a
A. random error
B. parallax error only
C. systematic error
D. gross error
View answer and explanation
Correct Option: C
It is repeatable and consistent in direction, and can be corrected by recalibration.
Q72Electrical Engineering (SSC JE)
Hard
A wattmeter has ratings of 150 V, 5 A and a full scale of 150 divisions. The wattmeter constant is
A. 25 W per division
B. 1 W per division
C. 0.2 W per division
D. 5 W per division
View answer and explanation
Correct Option: D
Full-scale power = 150 × 5 = 750 W. Constant = 750/150 = 5 W per division.
Q73Electrical Engineering (SSC JE)
Hard
A second order system has the characteristic equation s² + 4s + 16 = 0. The damping ratio is
A. 1
B. 0.25
C. 0.5
D. 2
View answer and explanation
Correct Option: C
ωn² = 16, so ωn = 4. 2ζωn = 4, so ζ = 4/(2 × 4) = 0.5.
Q74Electrical Engineering (SSC JE)
Hard
The percentage peak overshoot of a standard second order system with ζ = 0.5 is about
A. 37 percent
B. 4.3 percent
C. 16.3 percent
D. 50 percent
View answer and explanation
Correct Option: C
Mp = exp(−πζ / √(1 − ζ²)) = exp(−1.814) = 0.163.
Q75Electrical Engineering (SSC JE)
Medium
For a unity feedback type-1 system, the steady state error for a unit step input is
A. zero
B. 1
C. 1/Kv
D. infinite
View answer and explanation
Correct Option: A
A type-1 system has an integrator, so it follows a step input exactly. The error for a ramp is 1/Kv.
Q76Electrical Engineering (SSC JE)
Hard
A system has the characteristic equation s³ + 2s² + 3s + 6 = 0. The system is
A. stable
B. overdamped
C. unstable
D. marginally stable
View answer and explanation
Correct Option: D
The Routh array has a zero row at s¹: (2 × 3 − 6)/2 = 0. The roots are −2 and ±j√3, so there are poles on the imaginary axis.
Q77Electrical Engineering (SSC JE)
Easy
A positive phase margin of a closed loop system indicates that the system is
A. unstable
B. marginally stable
C. overdamped
D. stable
View answer and explanation
Correct Option: D
Positive gain and phase margins in a minimum phase system mean the closed loop is stable.
Q78Electrical Engineering (SSC JE)
Medium
The Nyquist stability criterion is based on the number of encirclements of
A. the point +1 + j0 by the closed-loop response
B. the origin by the closed-loop response
C. the imaginary axis by the root locus
D. the point −1 + j0 by the open-loop frequency response
View answer and explanation
Correct Option: D
The criterion relates the encirclements of the critical point to the right half plane poles of the closed loop.
Q79Electrical Engineering (SSC JE)
Easy
The ripple factor of a full-wave rectifier with a resistive load is about
A. 0.482
B. 0.707
C. 0.121
D. 1.21
View answer and explanation
Correct Option: A
A half-wave rectifier has a ripple factor of 1.21 and a full-wave rectifier 0.482.
Q80Electrical Engineering (SSC JE)
Hard
A transistor with β = 100 is biased with a fixed base resistor RB = 470 kΩ from VCC = 10.7 V (VBE = 0.7 V). The collector current is about
A. 0.213 mA
B. 1.0 mA
C. 21.3 mA
D. 2.13 mA
View answer and explanation
Correct Option: D
IB = (10.7 − 0.7)/470 kΩ = 21.3 μA. IC = β IB = 100 × 21.3 μA = 2.13 mA.
Q81Electrical Engineering (SSC JE)
Easy
The output impedance of an ideal operational amplifier is
A. zero
B. 1 kΩ
C. equal to the input impedance
D. infinite
View answer and explanation
Correct Option: A
An ideal op-amp has infinite gain and input impedance, and zero output impedance.
Q82Electrical Engineering (SSC JE)
Easy
A non-inverting amplifier uses Rf = 90 kΩ and R1 = 10 kΩ. The voltage gain is
A. 9
B. −9
C. 11
D. 10
View answer and explanation
Correct Option: D
Gain = 1 + Rf/R1 = 1 + 9 = 10.
Q83Electrical Engineering (SSC JE)
Hard
The 4-bit 2’s complement representation of the decimal number −5 is
A. 1011
B. 0101
C. 1100
D. 1010
View answer and explanation
Correct Option: A
5 = 0101. Invert to get 1010 and add 1 to get 1011.
Q84Electrical Engineering (SSC JE)
Easy
A JK flip-flop with J = K = 1 and a clock pulse
A. holds its previous state
B. resets its output
C. sets its output
D. toggles its output
View answer and explanation
Correct Option: D
With both inputs high the output complements on each clock pulse, which is the basis of counters.
Q85Electrical Engineering (SSC JE)
Easy
The minimum number of flip-flops needed to build a mod-10 counter is
A. 5
B. 10
C. 4
D. 3
View answer and explanation
Correct Option: C
Ten states need ⌈log₂ 10⌉ = 4 flip-flops, and the counter is reset at count 10.
Q86Electrical Engineering (SSC JE)
Easy
The Boolean expression A + A′B simplifies to
A. A′ + B
B. AB
C. A
D. A + B
View answer and explanation
Correct Option: D
A + A′B = (A + A′)(A + B) = A + B.
Q87Electrical Engineering (SSC JE)
Hard
A single-phase fully controlled bridge rectifier has a peak supply voltage of 300 V and a firing angle of 60°. The average output voltage (continuous conduction) is about
A. 95.5 V
B. 165.4 V
C. 47.7 V
D. 191 V
View answer and explanation
Correct Option: A
Vdc = (2Vm/π) cos α = (600/π) × 0.5 = 95.5 V.
Q88Electrical Engineering (SSC JE)
Hard
A step-up (boost) chopper has an input of 50 V and a duty ratio of 0.6. The output voltage is
A. 83.3 V
B. 200 V
C. 125 V
D. 30 V
View answer and explanation
Correct Option: C
Vo = Vin / (1 − D) = 50 / 0.4 = 125 V.
Q89Electrical Engineering (SSC JE)
Medium
In a voltage source inverter, feedback diodes across the switches are needed to
A. raise the output frequency
B. carry the reactive load current
C. protect against overvoltage on the supply
D. reduce the harmonics to zero
View answer and explanation
Correct Option: B
With an inductive load, the current continues to flow when the switches turn off. The diodes return the energy to the dc source.
Q90Electrical Engineering (SSC JE)
Easy
A TRIAC is equivalent to
A. two SCRs connected in series
B. two diodes in parallel
C. an SCR and a diode in series
D. two SCRs connected in antiparallel
View answer and explanation
Correct Option: D
It conducts in both directions and is triggered by a gate pulse of either polarity, so it is used for ac power control.
Q91Electrical Engineering (SSC JE)
Medium
A snubber circuit across an SCR protects it against
A. overvoltage on the gate
B. a high rate of rise of voltage (dv/dt)
C. a low holding current
D. a high rate of rise of current (di/dt)
View answer and explanation
Correct Option: B
The RC network slows the rise of the anode voltage, which would otherwise turn the device on falsely.
Q92Electrical Engineering (SSC JE)
Medium
In the V/f control of an induction motor, the ratio of voltage to frequency is kept constant in order to
A. keep the power factor at unity
B. increase the starting current
C. maintain a constant air gap flux
D. maintain a constant slip
View answer and explanation
Correct Option: C
The flux is proportional to V/f. Constant flux gives a constant torque capability over the speed range below the base speed.
Q93Electrical Engineering (SSC JE)
Hard
A room of 10 m × 6 m must be lit to 300 lux. The utilisation factor is 0.5 and the maintenance factor is 0.8. The total luminous flux required is
A. 22 500 lm
B. 90 000 lm
C. 18 000 lm
D. 45 000 lm
View answer and explanation
Correct Option: D
Flux = E A / (UF × MF) = 300 × 60 / (0.5 × 0.8) = 45 000 lm.
Q94Electrical Engineering (SSC JE)
Hard
A 10 kW heating element is designed to operate at 230 V. Its resistance is about
A. 5.29 Ω
B. 0.043 Ω
C. 52.9 Ω
D. 23 Ω
View answer and explanation
Correct Option: A
R = V²/P = 230² / 10 000 = 5.29 Ω.
Q95Electrical Engineering (SSC JE)
Easy
Regenerative braking in electric traction
A. is possible only in steam locomotives
B. needs no electrical machine
C. wastes the energy as heat in resistors
D. returns the kinetic energy of the train to the supply
View answer and explanation
Correct Option: D
The traction motors act as generators during braking and feed power back to the overhead line.
Q96Electrical Engineering (SSC JE)
Easy
According to Faraday’s first law of electrolysis, the mass of a substance deposited is proportional to
A. the square of the current
B. the temperature of the electrolyte
C. the voltage applied
D. the quantity of charge passed
View answer and explanation
Correct Option: D
m = Z Q, where Z is the electrochemical equivalent of the substance.
Q97Electrical Engineering (SSC JE)
Easy
The Fourier transform of a rectangular pulse is a
A. sinc function
B. delta function
C. step function
D. rectangular function
View answer and explanation
Correct Option: A
A pulse of width T has a spectrum proportional to sin(ωT/2)/(ω/2).
Q98Electrical Engineering (SSC JE)
Easy
Convolution of two signals in the time domain corresponds to which operation in the frequency domain?
A. division
B. multiplication
C. addition
D. differentiation
View answer and explanation
Correct Option: B
If y(t) = x(t) * h(t), then Y(ω) = X(ω) H(ω). This is why filtering is easier in the frequency domain.
Q99Electrical Engineering (SSC JE)
Easy
The Laplace transform of e^(−at) u(t) is
A. 1 / (s − a)
B. 1 / (s + a)
C. a / s
D. s / (s + a)
View answer and explanation
Correct Option: B
∫ e^(−at) e^(−st) dt from 0 to ∞ = 1 / (s + a).
Q100Electrical Engineering (SSC JE)
Hard
The energy of the signal x(t) = e^(−t) u(t) is
A. 2
B. 0.5
C. 0.25
D. 1
View answer and explanation
Correct Option: B
E = ∫ e^(−2t) dt from 0 to ∞ = 1/2.

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