SSC JE Electrical CBT 2: 50 Important Technical MCQs with Answers and Explanations
Preparing for the SSC JE Electrical CBT Paper 2 examination requires a strong understanding of core Electrical Engineering concepts, technical formulas, and numerical problem-solving techniques.
Regular practice with technical multiple-choice questions helps candidates revise important topics, identify weak areas, improve accuracy, and develop confidence.
In this article, Electric Sine presents 50 important SSC JE Electrical CBT 2 practice MCQs with correct answers and detailed explanations, covering Electrical Circuits, Electrical Machines, Power Systems, Measurements, Basic Electronics, Power Electronics, and Utilization of Electrical Energy.
These questions are useful for diploma students, B.Tech Electrical Engineering graduates, and government engineering job aspirants.
Note: These are original syllabus-based practice questions, not officially verified SSC JE previous-year questions.
SSC JE Electrical CBT Paper 2: Exam Overview
SSC JE Paper 2 evaluates candidates' technical knowledge in their selected engineering discipline.
Electrical Engineering candidates should prepare according to the official SSC JE syllabus and the examination notification applicable to their recruitment year.
The following 50 questions are arranged subject-wise for systematic preparation.
Part 1: Electrical Circuits and Network Theory – Questions 1 to 10
Q1. What is the equivalent resistance of two 6 Ω resistors connected in parallel?
A. 12 Ω
B. 6 Ω
C. 3 Ω
D. 1.5 Ω
Correct Answer: C. 3 Ω
Explanation:
For two resistors connected in parallel:
Equivalent Resistance = (R1 × R2) / (R1 + R2)
= (6 × 6) / (6 + 6)
= 36 / 12 = 3 Ω
Therefore, the equivalent resistance is 3 Ω.
Q2. Kirchhoff's Current Law is based on the conservation of:
A. Energy
B. Charge
C. Power
D. Magnetic flux
Correct Answer: B. Charge
Explanation:
Kirchhoff's Current Law states that the total current entering a junction equals the total current leaving it.
This law is based on the principle of conservation of electric charge.
Q3. In a purely inductive AC circuit, current:
A. Leads voltage by 90°
B. Lags voltage by 90°
C. Is in phase with voltage
D. Leads voltage by 180°
Correct Answer: B. Lags voltage by 90°
Explanation:
In a purely inductive AC circuit, current lags behind the applied voltage by 90 electrical degrees.
Q4. What is the power factor of a purely resistive AC circuit?
A. Zero
B. 0.5 lagging
C. Unity
D. 0.5 leading
Correct Answer: C. Unity
Explanation:
In a purely resistive AC circuit, voltage and current are in phase.
Power Factor = cos 0° = 1
Therefore, the power factor is unity.
Q5. At resonance in a series RLC circuit, impedance is:
A. Maximum
B. Minimum
C. Infinite
D. Purely capacitive
Correct Answer: B. Minimum
Explanation:
At resonance, inductive reactance equals capacitive reactance.
XL = XC
Both reactances cancel each other.
Therefore, total impedance becomes equal to resistance and reaches its minimum value.
Q6. The time constant of an RC circuit is:
A. R / C
B. C / R
C. RC
D. 1 / RC
Correct Answer: C. RC
Explanation:
The time constant of an RC circuit is:
Time Constant = Resistance × Capacitance
τ = R × C
Its unit is seconds.
Q7. A 10 Ω resistor carries a current of 2 A. Calculate the power dissipated.
A. 10 W
B. 20 W
C. 40 W
D. 80 W
Correct Answer: C. 40 W
Explanation:
Power = Current² × Resistance
P = I² × R
P = 2² × 10
P = 4 × 10 = 40 W
Therefore, the power dissipated is 40 W.
Q8. The condition for maximum power transfer in a DC resistive network is:
A. Load resistance is zero
B. Load resistance equals Thevenin resistance
C. Load resistance is infinite
D. Load resistance is twice the Thevenin resistance
Correct Answer: B. Load resistance equals Thevenin resistance
Explanation:
According to the maximum power transfer theorem, maximum power is delivered to the load when:
Load Resistance = Thevenin Resistance
RL = Rth
This condition applies to a linear resistive Thevenin network.
Q9. What is the SI unit of electrical conductance?
A. Ohm
B. Siemens
C. Henry
D. Farad
Correct Answer: B. Siemens
Explanation:
Electrical conductance is the reciprocal of electrical resistance.
Conductance = 1 / Resistance
G = 1 / R
The SI unit of conductance is Siemens (S).
Q10. In a balanced star-connected three-phase system, line voltage is:
A. Equal to phase voltage
B. Three times phase voltage
C. √3 times phase voltage
D. Half the phase voltage
Correct Answer: C. √3 times phase voltage
Explanation:
In a balanced star connection:
Line Voltage = √3 × Phase Voltage
VL = √3 × Vph
The line current is equal to the phase current.
Part 2: Electrical Machines – Questions 11 to 20
Q11. Maximum efficiency of a transformer occurs when:
A. Copper loss is zero
B. Iron loss equals copper loss
C. Iron loss is twice copper loss
D. The transformer operates at no load
Correct Answer: B. Iron loss equals copper loss
Explanation:
For a transformer operating at constant voltage and frequency, maximum efficiency occurs when variable copper loss equals constant iron loss.
Copper Loss = Iron Loss
This is an important transformer efficiency condition.
Q12. What is the EMF equation of a transformer?
A. E = 2.22 × f × N × Φm
B. E = 4.44 × f × N × Φm
C. E = 1.11 × f × N × Φm
D. E = 6.28 × f × N × Φm
Correct Answer: B. E = 4.44 × f × N × Φm
Explanation:
For sinusoidal magnetic flux, the RMS induced EMF is:
E = 4.44 × f × N × Φm
Where:
E = RMS induced EMF in volts
f = Supply frequency in Hz
N = Number of turns
Φm = Maximum magnetic flux in weber
Q13. The slip of a three-phase induction motor at synchronous speed is:
A. 1
B. 0.5
C. 0
D. 2
Correct Answer: C. 0
Explanation:
Slip = (Synchronous Speed − Rotor Speed) / Synchronous Speed
s = (Ns − Nr) / Ns
At synchronous speed:
Ns = Nr
Therefore:
Slip = 0
Q14. A four-pole induction motor operates on a 50 Hz supply. Calculate its synchronous speed.
A. 750 rpm
B. 1000 rpm
C. 1500 rpm
D. 3000 rpm
Correct Answer: C. 1500 rpm
Explanation:
Synchronous Speed = (120 × Frequency) / Number of Poles
Ns = (120 × 50) / 4
Ns = 6000 / 4
Ns = 1500 rpm
Q15. What is the main purpose of a starter in a DC motor?
A. Increase motor frequency
B. Limit starting armature current
C. Eliminate field flux
D. Increase supply voltage
Correct Answer: B. Limit starting armature current
Explanation:
At the moment of starting, the back EMF of a DC motor is zero.
Since armature resistance is low, a very high starting current can flow.
A starter limits this current and protects the motor.
Q16. In a DC shunt motor, what generally happens to speed when field flux decreases?
A. Speed decreases
B. Speed increases
C. Speed becomes zero
D. Speed remains exactly constant
Correct Answer: B. Speed increases
Explanation:
The approximate speed relation of a DC motor is:
Speed ∝ (Applied Voltage − Armature Voltage Drop) / Field Flux
When field flux decreases, motor speed generally increases, assuming other operating conditions remain comparable.
Q17. The open-circuit test of a transformer is mainly used to determine:
A. Full-load copper loss
B. Core loss and no-load parameters
C. Short-circuit current at rated voltage
D. Winding insulation resistance
Correct Answer: B. Core loss and no-load parameters
Explanation:
During the open-circuit test, the transformer draws a small no-load current.
Copper loss is relatively small.
Therefore, the measured input power approximately represents the transformer core loss.
Q18. The short-circuit test of a transformer is primarily used to determine:
A. Core loss at rated voltage
B. Equivalent series impedance and copper loss
C. Magnetizing current at no load
D. Turns ratio only
Correct Answer: B. Equivalent series impedance and copper loss
Explanation:
In the short-circuit test, a reduced voltage is applied to circulate the required test current.
Core loss is small because the applied voltage is low.
The measured power approximately represents copper loss at the test current.
The test also helps determine equivalent series impedance.
Q19. Which motor is generally preferred when nearly constant speed is required over varying loads?
A. DC series motor
B. DC shunt motor
C. Universal motor
D. Single-phase repulsion motor
Correct Answer: B. DC shunt motor
Explanation:
A DC shunt motor has approximately constant field flux.
Therefore, it provides relatively good speed regulation when load changes.
Q20. A synchronous motor normally operates at:
A. A speed determined by slip
B. Synchronous speed
C. Half synchronous speed
D. A speed greater than synchronous speed under all loads
Correct Answer: B. Synchronous speed
Explanation:
A synchronous motor runs at synchronous speed when synchronized with the supply.
Synchronous Speed = (120 × Frequency) / Number of Poles
Ns = (120 × f) / P
Its steady-state speed is determined by supply frequency and the number of poles.
Part 3: Power Systems and Protection – Questions 21 to 30
Q21. What is the primary purpose of a protective relay?
A. Generate electrical power
B. Detect abnormal conditions and initiate protective action
C. Increase transmission voltage
D. Improve insulation resistance
Correct Answer: B. Detect abnormal conditions and initiate protective action
Explanation:
A protective relay monitors electrical system conditions.
When a fault or abnormal condition is detected, it can send a trip command to the circuit breaker.
This helps isolate faulty sections of the power system.
Q22. Which device is designed to interrupt fault current in a power system?
A. Isolator alone
B. Circuit breaker
C. Potential transformer
D. Surge capacitor
Correct Answer: B. Circuit breaker
Explanation:
A circuit breaker can interrupt normal load current and fault current within its rated capacity.
An isolator is primarily intended for isolation and is not a substitute for a fault-interrupting circuit breaker.
Q23. What is the main purpose of earthing electrical equipment?
A. Increase operating voltage
B. Provide a protective fault-current path and reduce shock risk
C. Increase power consumption
D. Reduce supply frequency
Correct Answer: B. Provide a protective fault-current path and reduce shock risk
Explanation:
Protective earthing connects exposed conductive parts to the earthing system.
It helps reduce dangerous touch voltages and supports the operation of protective devices under fault conditions.
Q24. Which conductor type is widely used for overhead transmission lines?
A. Silver
B. ACSR
C. Nichrome
D. Tungsten
Correct Answer: B. ACSR
Explanation:
ACSR stands for Aluminium Conductor Steel Reinforced.
It combines the electrical conductivity of aluminium with the mechanical strength provided by steel reinforcement.
Q25. Why are capacitor banks connected to inductive loads?
A. To increase reactive power demand from the supply
B. To improve power factor
C. To increase supply frequency
D. To eliminate active power consumption
Correct Answer: B. To improve power factor
Explanation:
Inductive loads draw lagging reactive power.
Capacitor banks supply leading reactive power, reducing the net reactive power demand from the supply.
This improves power factor and can reduce line current.
Q26. The Ferranti effect is associated with:
A. A decrease in receiving-end voltage under light load
B. Receiving-end voltage exceeding sending-end voltage under light load
C. Transformer copper loss
D. DC motor commutation
Correct Answer: B. Receiving-end voltage exceeding sending-end voltage under light load
Explanation:
In long AC transmission lines, line capacitance can cause the receiving-end voltage to become higher than the sending-end voltage during light-load or no-load conditions.
This phenomenon is known as the Ferranti effect.
Q27. Buchholz protection is commonly provided for:
A. Dry-type transformers
B. Oil-immersed transformers with conservator tanks
C. DC generators only
D. Underground cables
Correct Answer: B. Oil-immersed transformers with conservator tanks
Explanation:
A Buchholz relay is a gas-operated protective device.
It is commonly installed between the main tank and conservator of suitable oil-immersed transformers.
It detects gas accumulation and abnormal oil movement caused by certain internal faults.
Q28. A current transformer is primarily used for measuring:
A. High voltage
B. Alternating current
C. Frequency only
D. Insulation resistance
Correct Answer: B. Alternating current
Explanation:
A current transformer reduces high AC current to a suitable proportional secondary current.
It is used with measuring instruments and protective relays.
Q29. Corona loss in overhead transmission lines occurs mainly due to:
A. Air ionization around conductors
B. Transformer hysteresis
C. Armature reaction
D. Battery discharge
Correct Answer: A. Air ionization around conductors
Explanation:
When the electric field around a transmission conductor becomes sufficiently high, the surrounding air can become ionized.
This causes corona discharge and associated power loss.
Q30. The load factor of a power system is:
A. Maximum Demand / Average Load
B. Average Load / Maximum Demand
C. Connected Load / Average Load
D. Maximum Demand / Connected Load
Correct Answer: B. Average Load / Maximum Demand
Explanation:
Load Factor = Average Load / Maximum Demand
It indicates how uniformly electrical demand is distributed over a specified period.
A higher load factor generally indicates more consistent utilization of the electrical system.
Part 4: Electrical Measurements and Instrumentation – Questions 31 to 38
Q31. A moving-iron instrument can measure:
A. DC only
B. AC only
C. Both AC and DC
D. Frequency only
Correct Answer: C. Both AC and DC
Explanation:
Moving-iron instruments operate using the magnetic effect of electric current.
They can measure both AC and DC quantities.
Q32. A conventional permanent-magnet moving-coil instrument is normally used for:
A. DC measurement
B. Direct AC measurement without a rectifier
C. High-frequency measurement only
D. Power factor measurement only
Correct Answer: A. DC measurement
Explanation:
A PMMC instrument responds to unidirectional current.
For AC measurement, an appropriate rectifier or conversion arrangement is required.
Q33. Which instrument is used to measure insulation resistance?
A. Wattmeter
B. Megohmmeter
C. Frequency meter
D. Tachometer
Correct Answer: B. Megohmmeter
Explanation:
A megohmmeter is used to measure high electrical resistance, especially insulation resistance.
It is commonly used for testing electrical cables, motors, and installations.
Q34. A wattmeter measures:
A. Electrical energy
B. Active electrical power
C. Reactive energy only
D. Frequency
Correct Answer: B. Active electrical power
Explanation:
A wattmeter measures active or real electrical power.
The SI unit of active power is the watt (W).
Electrical energy, on the other hand, is measured using an energy meter.
Q35. Why should the secondary winding of a current transformer not be open-circuited while the primary carries current?
A. Supply frequency will increase
B. Dangerous high voltage may develop across the secondary
C. The transformer becomes a DC source
D. Primary resistance becomes zero
Correct Answer: B. Dangerous high voltage may develop across the secondary
Explanation:
An open-circuited CT secondary may develop a dangerously high voltage.
This can cause insulation damage and create a serious electric shock hazard.
Therefore, CT secondary circuits must be handled using appropriate safety procedures.
Q36. Which bridge is commonly used to measure an unknown medium resistance?
A. Wheatstone bridge
B. Schering bridge
C. Maxwell bridge
D. Wien bridge
Correct Answer: A. Wheatstone bridge
Explanation:
The Wheatstone bridge uses four resistance arms.
At balance, the unknown resistance can be determined using the ratios of the other resistances.
It is commonly used for medium-resistance measurement.
Q37. The total active power in a three-phase, three-wire system can be measured using:
A. Only one ammeter
B. Two wattmeters
C. One voltmeter
D. One frequency meter
Correct Answer: B. Two wattmeters
Explanation:
The two-wattmeter method is used to measure total active power in a three-phase, three-wire system.
Total Power = Wattmeter 1 Reading + Wattmeter 2 Reading
P = W1 + W2
The method can also be used for unbalanced loads under suitable measurement conditions.
Q38. A 2 kW electrical load operates for 5 hours. Calculate the energy consumed.
A. 2 kWh
B. 5 kWh
C. 10 kWh
D. 20 kWh
Correct Answer: C. 10 kWh
Explanation:
Electrical Energy = Power × Time
Energy = 2 kW × 5 hours
Energy = 10 kWh
Therefore, the energy consumed is 10 units of electricity.
Part 5: Basic Electronics and Power Electronics – Questions 39 to 44
Q39. A semiconductor diode normally conducts substantial current when it is:
A. Reverse biased
B. Forward biased
C. Unbiased
D. Connected to an open circuit
Correct Answer: B. Forward biased
Explanation:
A PN junction diode allows substantial current to flow under forward bias when the applied voltage and circuit conditions are suitable.
Under normal reverse bias, only a small leakage current flows.
Q40. What is the main application of a Zener diode?
A. Voltage regulation
B. Mechanical speed control
C. Magnetic flux measurement
D. Transformer cooling
Correct Answer: A. Voltage regulation
Explanation:
A Zener diode is designed to operate in the reverse-breakdown region.
When operated within its specified limits, it maintains an approximately constant voltage.
Therefore, it is commonly used for voltage regulation and reference circuits.
Q41. A full-wave rectifier is supplied with 50 Hz AC. What is the ripple frequency?
A. 25 Hz
B. 50 Hz
C. 100 Hz
D. 200 Hz
Correct Answer: C. 100 Hz
Explanation:
For a full-wave rectifier:
Ripple Frequency = 2 × Supply Frequency
Ripple Frequency = 2 × 50
Ripple Frequency = 100 Hz
Q42. An SCR can normally be turned ON by applying:
A. A suitable gate pulse while forward biased
B. A negative anode voltage only
C. A reverse gate pulse only
D. Zero anode current
Correct Answer: A. A suitable gate pulse while forward biased
Explanation:
SCR stands for Silicon Controlled Rectifier.
An SCR can be triggered into conduction using a suitable gate pulse when it is forward biased.
Once conducting, it normally remains ON until the anode current falls below the holding current, assuming normal operating conditions.
Q43. A DC chopper is mainly used to convert:
A. AC to DC
B. DC to controlled DC
C. DC to fixed-frequency AC
D. AC to AC without frequency change
Correct Answer: B. DC to controlled DC
Explanation:
A DC chopper is a power electronic converter.
It converts a fixed DC input into a controlled DC output by switching the supply ON and OFF at high speed.
Q44. What is the function of an inverter?
A. Convert DC into AC
B. Convert AC into DC
C. Convert mechanical energy into DC
D. Convert AC into mechanical energy
Correct Answer: A. Convert DC into AC
Explanation:
An inverter converts direct current into alternating current.
Inverters are widely used in solar power systems, UPS systems, and variable-frequency motor drives.
Part 6: Electrical Materials, Utilization and Estimating – Questions 45 to 50
Q45. Which material is commonly used as the filament in an incandescent lamp?
A. Copper
B. Tungsten
C. Aluminium
D. Lead
Correct Answer: B. Tungsten
Explanation:
Tungsten has a very high melting point.
It can operate at high temperatures and emit visible light.
Therefore, it is widely used as the filament material in conventional incandescent lamps.
Q46. What is the SI unit of illuminance?
A. Lumen
B. Lux
C. Candela
D. Weber
Correct Answer: B. Lux
Explanation:
Illuminance represents the luminous flux incident per unit area.
Illuminance = Luminous Flux / Area
1 lux = 1 lumen per square metre
Q47. Which electrical heating method uses eddy currents induced in a conducting workpiece?
A. Resistance heating
B. Induction heating
C. Arc heating
D. Dielectric heating
Correct Answer: B. Induction heating
Explanation:
Induction heating uses an alternating magnetic field to induce electric currents in a conducting workpiece.
These currents produce heat due to electrical resistance.
Induction heating is widely used in industrial metal heating and processing.
Q48. Why is Nichrome commonly used in electrical heating elements?
A. Very low resistivity
B. High resistivity and good oxidation resistance
C. Zero temperature coefficient
D. Superconducting properties at room temperature
Correct Answer: B. High resistivity and good oxidation resistance
Explanation:
Nichrome has high electrical resistivity and good resistance to oxidation at elevated temperatures.
These properties make it suitable for electrical heating elements.
Q49. What is the main function of an MCB in a low-voltage electrical installation?
A. Improve power factor
B. Protect against overload and short-circuit current
C. Measure electrical energy
D. Maintain constant voltage
Correct Answer: B. Protect against overload and short-circuit current
Explanation:
MCB stands for Miniature Circuit Breaker.
An MCB automatically disconnects a circuit when excessive current flows due to overload or short-circuit conditions, according to its rated protective characteristics.
Q50. A 230 V, 1 kW single-phase resistive heater draws approximately how much current?
A. 2.17 A
B. 4.35 A
C. 8.70 A
D. 10 A
Correct Answer: B. 4.35 A
Explanation:
For a resistive heater:
Power = Voltage × Current
P = V × I
Current = Power / Voltage
I = 1000 / 230
I = 4.35 A (approximately)
Therefore, the heater draws approximately 4.35 A.
How to Prepare for SSC JE Electrical CBT Paper 2
Preparing for SSC JE Electrical Paper 2 requires systematic revision and consistent practice.
Here are some useful preparation strategies:
- Understand the official syllabus: Identify the technical subjects and important topics prescribed for the examination.
- Strengthen electrical fundamentals: Revise basic circuit laws, electrical machines, power systems, measurements, and electrical materials.
- Practise numerical questions: Focus on formulas, unit conversions, circuit calculations, and electrical machine problems.
- Solve previous-year questions: Use authentic SSC JE question papers to understand examination patterns and recurring concepts.
- Attempt mock tests: Practise under timed conditions to improve speed and accuracy.
- Analyse incorrect answers: Review every mistake and revise the related concept.
- Maintain a formula notebook: Keep important formulas and short notes for quick revision.
Frequently Asked Questions About SSC JE Electrical CBT 2
1. Are these actual SSC JE previous-year questions?
No. These are original practice questions based on important Electrical Engineering concepts relevant to SSC JE preparation.
2. Are these 50 questions enough for SSC JE Electrical Paper 2?
No. These questions are useful for technical revision, but candidates should prepare the complete official syllabus, practise verified previous-year questions, and attempt full-length mock tests.
3. Which technical subjects are important for SSC JE Electrical Paper 2?
Important areas include Electrical Circuits, Electrical Machines, Power Systems, Electrical Measurements, Electrical Materials, Basic Electronics, Utilization of Electrical Energy, and Estimating and Costing, subject to the applicable official syllabus.
4. How can I improve accuracy in SSC JE Electrical technical questions?
Revise fundamental concepts, practise numerical problems, understand why each answer is correct, and analyse mistakes after every practice test.
5. Where can I practise Electrical Engineering MCQs online?
Students and job aspirants can explore Electric Sine for Electrical Engineering objective questions, topic-wise learning resources, and available free mock tests.
Conclusion
SSC JE Electrical CBT Paper 2 preparation becomes more effective when candidates combine strong technical knowledge with regular question practice.
These 50 important Electrical Engineering MCQs help revise core concepts, strengthen numerical skills, and identify areas requiring additional preparation.
For better results, continue practising topic-wise questions, verified previous-year papers, and mock tests while following the latest official SSC JE examination guidelines.
Learn. Practice. Prepare.
Explore more Electrical Engineering study resources and free mock tests at Electric Sine.
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