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English SSC CGL Tier I papers. Free 60-minute mocks, 100 questions, 200 marks; +2 correct and −0.5 incorrect. Source: SSC response sheets preserved by SSC Portal. The marked archived keys have not been independently revalidated against final SSC keys; worked explanations are pending review. A/B/C/D correspond to source options 1/2/3/4.

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Question 1 · Electrical Engineering (SSC JE) · Electrical Machines
A 4-pole DC generator has a wave winding with 500 conductors and a flux of 0.02 Wb per pole. At 1200 rpm, the generated emf is
Choose your answer for question 1
Answer and explanation

Correct answer: C

E = Φ Z N P / (60 A). For wave winding A = 2: E = 0.02 × 500 × 1200 × 4 / (60 × 2) = 400 V.

Question 2 · Electrical Engineering (SSC JE) · Electrical Machines
A DC motor is supplied at 220 V and takes an armature current of 20 A. The armature resistance is 0.5 Ω. The back emf is
Choose your answer for question 2
Answer and explanation

Correct answer: C

Eb = V − Ia Ra = 220 − 20 × 0.5 = 210 V.

Question 3 · Electrical Engineering (SSC JE) · Electrical Machines
The speed of a DC shunt motor can be increased above its normal speed by
Choose your answer for question 3
Answer and explanation

Correct answer: D

N ∝ Eb/Φ, so weakening the field raises the speed (field control).

Question 4 · Electrical Engineering (SSC JE) · Electrical Machines
A DC series motor should never be started without a load because
Choose your answer for question 4
Answer and explanation

Correct answer: B

With no load, the armature current and the flux are both small, and the speed rises to a very high value.

Question 5 · Electrical Engineering (SSC JE) · Electrical Machines
The emf equation of a transformer is
Choose your answer for question 5
Answer and explanation

Correct answer: B

E_rms = 4.44 f N Φm, where Φm is the maximum core flux.

Question 6 · Electrical Engineering (SSC JE) · Electrical Machines
A transformer has a primary voltage of 11 kV and a secondary voltage of 400 V. If the primary has 1100 turns, the secondary has
Choose your answer for question 6
Answer and explanation

Correct answer: C

N2 = N1 × V2/V1 = 1100 × 400/11 000 = 40 turns.

Question 7 · Electrical Engineering (SSC JE) · Electrical Machines
The open circuit test on a transformer gives
Choose your answer for question 7
Answer and explanation

Correct answer: C

At no load, the current is small, so copper loss is negligible and the input power is the core loss.

Question 8 · Electrical Engineering (SSC JE) · Electrical Machines
The efficiency of a transformer is maximum when
Choose your answer for question 8
Answer and explanation

Correct answer: C

Iron loss is constant and copper loss varies with the square of the load. The efficiency peaks when the two are equal.

Question 9 · Electrical Engineering (SSC JE) · Electrical Machines
An autotransformer is most economical compared to a two-winding transformer when the transformation ratio is
Choose your answer for question 9
Answer and explanation

Correct answer: A

With K = V1/V2 (greater than 1), the copper saved is a fraction 1/K of that in the two-winding transformer. The saving is therefore greatest when K is close to 1.

Question 10 · Electrical Engineering (SSC JE) · Electrical Machines
An induction motor runs at 1440 rpm on a 50 Hz supply with 4 poles. The slip is
Choose your answer for question 10
Answer and explanation

Correct answer: C

Ns = 120 × 50/4 = 1500 rpm. s = (1500 − 1440)/1500 = 0.04.

Question 11 · Electrical Engineering (SSC JE) · Electrical Machines
The synchronous speed of a 6-pole machine on a 50 Hz supply is
Choose your answer for question 11
Answer and explanation

Correct answer: A

Ns = 120 f / P = 120 × 50 / 6 = 1000 rpm.

Question 12 · Electrical Engineering (SSC JE) · Electrical Machines
The frequency of the rotor current of an induction motor at a slip of 0.04 on a 50 Hz supply is
Choose your answer for question 12
Answer and explanation

Correct answer: C

fr = s f = 0.04 × 50 = 2 Hz.

Question 13 · Electrical Engineering (SSC JE) · Electrical Machines
Star-delta starting of a three-phase induction motor reduces the starting line current to
Choose your answer for question 13
Answer and explanation

Correct answer: D

In star, the phase voltage is 1/√3 of the line voltage, so the line current falls to 1/3.

Question 14 · Electrical Engineering (SSC JE) · Electrical Machines
An over-excited synchronous motor operates at
Choose your answer for question 14
Answer and explanation

Correct answer: C

An over-excited motor supplies reactive power to the system, so it acts like a capacitor (synchronous condenser).

Question 15 · Electrical Engineering (SSC JE) · Electrical Machines
The voltage regulation of an alternator is highest for a load of
Choose your answer for question 15
Answer and explanation

Correct answer: A

For a lagging load, armature reaction is demagnetising, and the terminal voltage falls the most.

Question 16 · Electrical Engineering (SSC JE) · Electrical Machines
A single-phase induction motor is not self-starting because the stator field is
Choose your answer for question 16
Answer and explanation

Correct answer: B

A pulsating field can be resolved into two equal and opposite rotating fields that give no net torque at standstill.

Question 17 · Electrical Engineering (SSC JE) · Electrical Machines
A universal motor can operate on
Choose your answer for question 17
Answer and explanation

Correct answer: C

It is a series-wound motor, whose torque direction does not change when the supply polarity reverses.

Question 18 · Electrical Engineering (SSC JE) · Electrical Machines
Interpoles in a DC machine are provided to
Choose your answer for question 18
Answer and explanation

Correct answer: D

They produce a flux that induces an emf in the coil under commutation which opposes the reactance voltage.

Question 19 · Electrical Engineering (SSC JE) · Electrical Machines
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
Choose your answer for question 19
Answer and explanation

Correct answer: B

Ia = 40 + 2 = 42 A. E = V + Ia Ra = 220 + 42 × 0.1 = 224.2 V.

Question 20 · Electrical Engineering (SSC JE) · Electrical Machines
A DC motor delivers 10 kW at 1000 rpm. The shaft torque is about
Choose your answer for question 20
Answer and explanation

Correct answer: C

T = P / ω = 10 000 × 60 / (2π × 1000) = 95.5 N·m.

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