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Preface GK Publications has been the "publisher of choice" to aspirants preparing for Staff Selection Commissions' coveted Junior Engineer recruitment examination. Our SSC JE series for this exam has been segmented into study guides and practice papers in Hindi and English modes for Civil, Electrical and Mechanical streams. With last year's exam going online, we put in effort to help students prepare better with our online test series. Our web portal gave access to full length mock tests which imitated the actual exam and helped students get familiar with online exam environment & improved analytics. Our electrical stream study guide is vastly popular among candidates taking up Paper II Conventional of SSC JE. The study guide comprises of numerous practice questions, adhering to the exam pattern, as laid out by SSC. Complete solutions to these questions have been given towards end to help students assess the actual level of their preparation. This book has solved papers of last 12 years, from 2007 to 2018 with exhaustive explanatory solutions to all questions. This valuable practice material help students become acquainted with exam pattern and have better understanding, study and self-practice to face Paper - II with confidence. Timing while practicing with this textbook will also escalate their speed and always accuracy. We hope that the book meets expectations of future Junior Electrical Engineers. Due care has been taken to keep the book error free. However, any and all erroneous statements and feedback to improve the book are always welcome. Wishing a prestigious career to all students,

A ll the Best! Team GKP

Contents 

Exam Pattern



Syllabus

SOLVED PAPERS  2018 (Conducted on 29.12.2019)

1 – 12

 2017

1 – 16

 2016

1 – 16

 2015

1 – 14

 2014

1–9

 2013

1–8

 2012

1–9

 2011

1–5

 2010

1 – 13

 2009

1 – 11

 2008

1 – 14

 2007

1 – 10

 Practice Paper-1

1–8

 Practice Paper-2

1 – 12

Exam Pattern Papers

Subject

Paper – I Objective Type (Computer Based Test)

(i) General Intelligence & Reasoning (ii) General Awareness (iii) General Engineering

Maximum Marks

Duration

50 50 100

2 Hours

300

2 Hours

Paper – II Conventional Type (Paper Based Test)

General Engineering

Cut-offs Marks Paper I Electrical 2017 2016 2015 Category Cut-off Candidates Cut-off Candidates Cut-off Candidates Marks Available Marks Available Marks Available SC 120.00 235 99.00 369 114.75 261 ST 114.50 139 94.50 176 105.50 191 OBC 133.25 413 109.50 1094 125.25 888 OH 113.00 18 87.00 40 100.00 39 HH 83.50 18 54.00 36 80.50 15 UR 136.25 599 115.00 725 131.00 407 Total 1422 2440 1801

Cut-off Marks 102.50 93.75 109.50 93.00 69.00 117.50

2014 Candidates Available 243 123 1068 27 25 649 2135

Paper II Electrical Category Cut-off Marks SC 268.50 ST 265.50 OBC 299.00 OH 303.50 HH 247.50 UR 164.75 Total

2017 2016 2015 Candidates Cut-off Candidates Cut-off Available Marks Available Marks 44 210.50 169 50 21 206.50 81 50 110 245.75 285 62 132 205.25 15 40 3 126.00 16 40 9 285.75 610 107 319 1176

2014 Cut-off Candidates Marks Available 149 108 126 55 180 471 124 16 64 10 215 190* 850

Syllabus Basic Concepts Concepts of resistance, inductance, capacitance, and various factors affecting them. Concepts of current, voltage, power, energy and their units.

Circuit Law Kirchhoff’s law, Simple Circuit solution using network theorems.

Magnetic Circuit Concepts of flux, mmf, reluctance, Different kinds of magnetic materials, Magnetic calculations for conductors of different configuration e.g. straight, circular, solenoidal, etc. Electromagnetic induction, self and mutual induction.

AC Fundamentals Instantaneous, peak, R.M.S. and average values of alternating waves, Representation of sinusoidal wave form, simple series and parallel AC Circuits consisting of R.L. and C, Resonance, Tank Circuit. Poly Phase system – star and delta connection, 3 phase power, DC and sinusoidal response of R-L and R-C circuit.

Measurement and Measuring Instruments Measurement of power (1 phase and 3 phase, both active and re-active) and energy, 2 wattmeter method of 3 phase power measurement. Measurement of frequency and phase angle. Ammeter and voltmeter (both moving oil and moving iron type), extension of range wattmeter, Multimeters, Megger, Energy meter AC Bridges. Use of CRO, Signal Generator, CT, PT and their uses. Earth Fault detection.

Electrical Machines (a)

D .C. Machine – Construction, Basic Principles of D.C. motors and generators, their characteristics, speed control and starting of D.C. Motors. Method of braking motor, Losses and efficiency of D.C. Machines. (b) 1 phase and 3 phase transformers – Construction, Principles of operation, equivalent circuit, voltage regulation, O.C. and S.C. Tests, Losses and efficiency. Effect of voltage, frequency and wave form on losses. Parallel operation of 1 phase /3 phase transformers. Auto transformers. (c) 3 phase induction motors, rotating magnetic field, principle of operation, equivalent circuit, torque-speed characteristics, starting and speed control of 3 phase induction motors. Methods of braking, effect of voltage and frequency variation on torque speed characteristics. Fractional Kilowatt Motors and Single Phase Induction Motors: Characteristics and applications.

Synchronous Machines Generation of 3-phase e.m.f. armature reaction, voltage regulation, parallel operation of two alternators, synchronizing, control of active and reactive power. Starting and applications of synchronous motors.

Generation, Transmission and Distribution Different types of power stations, Load factor, diversity factor, demand factor, cost of generation, inter-connection of power stations. Power factor improvement, various types of tariffs, types of faults, short circuit current for symmetrical faults. Switchgears – rating of circuit breakers, Principles of arc extinction by oil and air, H.R.C. Fuses, Protection against earth leakage / over current, etc. Buchholtz relay, Merz-Price system of protection of generators & transformers, protection of feeders and bus bars. Lightning arresters, various transmission and distribution system, comparison of conductor materials, efficiency of different system. Cable – Different type of cables, cable rating and derating factor.

Estimation and Costing Estimation of lighting scheme, electric installation of machines and relevant IE rules. Earthing practices and IE Rules.

Utilization of Electrical Energy Illumination, Electric heating, Electric welding, Electroplating, Electric drives and motors.

Basic Electronics Working of various electronic devices e.g. P N Junction diodes, Transistors (NPN and PNP type), BJT and JFET. Simple circuits using these devices.

Solved Paper 2018 (Conducted on 29.12.2019)

General Engineering Paper II Electrical 1. (a) Determine the unknown currents through and voltages across the resistances in the circuit of figure-1. (15) 2 2A

(c) The number of turns in two coupled coils is 600 and 1700, respectively. When a current of 6 A flows in the second coil, the total magnetic flux produced in this coil is 0.8 mWb, and the flux that links with the first coil is only 0.5 mWb. Calculate,



4

The field lines make an angle of 60° with the normal to the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (15)

Figure-1 (b) The resistance of a transmission line is 126 at 20°C. Determine the resistance of the line at – 35°C. The temperature coefficient of the material of transmission line is 0.00426 at 0°C. (15)

(i) The self-inductances of the two coils, (ii) The coefficient of coupling, and (iii) The coefficient of mutual inductance (3 × 10)

(c) Two heaters A and B are connected in parallel across a supply voltage. They produce 500 Kcal in 20 minutes and 1000 Kcal in 10 minutes, respectively. The resistance of heatr A is 10.

3. (a) Differentiate between absolute and secondary instruments. (10)

(i) Calculate the resistance of heater B. (15)

(c) A 50 A meter movement with an internal resistance of 1 K is to be used as a dc voltmeter of range 50 V. Calculate

(ii) If the two heaters are connected in series across the same supply voltage, how much heat will be produced in 5 minutes. (15) 2. (a) Determine the current through the 7 resistance in the network of figure-2. (15) 5 57V

3 4

42V 25V

7 6

70V

4V

Figure-2 (b) A circular coil of 30 turns and radius 8 cm carrying a current of 6 A is suspended vertically in a uniform horizontal magnetic field of 1.0 T.

(b) What is the basic difference between indicating instruments and integrating instruments ? (10)

(i) the multiplier resistance needed, and (ii) the voltage multiplying factor.

(2 × 10)

(d) In a moving coil instrument, the coil has a length of 5 cm, a width of 4 cm and 80 turns. The magnetic flux density in the air gap is 0.1 Wb/m2. The hair spring provides a controlling torque of 0.5 × 10–7 Nm/degree deflection of the coil. What current will be required to give a deflection of 60°? (20) 4. (a) A single phase 100 MVA, 132 KV/ 220 KV, 50 Hz transformer (ideal) is connected to 200 KV supply system. The secondary side of transformer is connected to a load of (300 + j 400) ohms. If the number of turns in low voltage (LV) side is 1000, find :

2

SOLVED PAPER 2018

A 200 m

(i) Turn ratio

300 m

300 m

(ii) Secondary side voltage (iii) Number of turns on the high voltage side (iv) The maximum value of core flux (v) Primary (source) and secondary (load) currents (vi) Power supplied by source

(6 × 5)

(b) A four-pole dc machine having wave winding has 294 conductors in armature. Find the following : (i) Flux per pole to generate 230 V when rotating at 1500 rpm (ii) Torque at this flux when rated armature current of 120 A is flowing. (2 × 15) 5. (a) What are the advantages of gas turbine plant over steam turbine plant? (15) (b) A single phase distributor fed at end a is loaded as shown in figure-3. The loop resistance and ractance per km are 0.3 and 0.15 respectively. Determine the voltage drop at the far end. (15)

p.f.

50A 0.707

60A 1.0

40A 0.8

Figure-3 (c) What are the various diagnostic techniques used for monitoring the health of various equipments on the system in a power grid? (15) (d) What do you mean by plant capacity factor? Describe the plant use factor. (15) 6. (a) Describe and differentiate the processes of direct resistance heating and indirect resistance heating. (15) (b) Determine the efficiency of a high frequency induction furnace which takes 10 minutes to melt 1.815 Kg of aluminum, the input to the furnace being 5 KW and the initial temperatures 15°C. (15) (c) If the input to an amplifier is 12 V and the output is 6 V, and the input and output impedances are equal, determine the dB gain of the amplifier. (15) (d) What is the main advantage of a comonemitter configuration over the common-base configuration of a bipolar junction transistor? (15)

SOLVED PAPER 2018

3

EXPLANATIONS 1. (a)

A 2A

2

B I1

(b) As given that,

C

4

R1 = 126 

I2

T1 = 20°C



T2 = –35°C 0 = 0.00426

F E D Here total current of 2A is distributed in branch 'BF' and 'BD'.

As we know that

R2 R1 = 1  0 T2 1   0 T1

Now in order to find the current passing through branch 'BE' and 'BD' we have to apply current division rule. According to current division rule, a parallel circuit act as a current divider as the current divides in all the parallel branches and voltage remains the same across them.

126 R2 = 1  0.00426  20 1  0.00426  (–35)

126  (0.8509)  R2 =   1.0852  = 98.8 

(c)

Heater A A

 6  I1 =   ×2 46 =

6 ×2 10

B Heater B

= 1.2 A and

 4  I2 =   ×2 46

=

8 10

I1

10

I2

x

= 0.8 A As given that The current through '2' resistor is '2A' Current through '4' resistor is 1.2 A and current through '6' resistor is 0.8 A and voltage across '2' resistor :

Let the resistance of heater 'B' be 'x' ohm. H1 = I12 Rt where 'H' is in Joule

V = RI

'I' is in Ampere

= 2 × 2 = 4V

'R' in Ohm

Voltage across '4' resistor

and t is time in sec.

V = RI = 4 × 1.2 = 4.8 V and voltage across '6' resistor V = RI = 6 × 0.8 = 4.8 V Also resistor '4' and '6' are in parallel. Therefore voltage remain same in parallel branch

500 10  20 1 Calorie = 4.2 Joule

I2 =

Now, 

500 × 1000C = ?

 500 K calorie = 500 × 1000 × 4.2 Joule 42 10 = 50000 × 42 = 21 × 105 Joule

= 500 × 1000 ×

4

SOLVED PAPER 2018

and time t = 20 min

(ii) New when two heater are connected in series we get

= 20 × 60 sec

10

= 1200 sec

2.5

Now applying formula H1 = I12 Rt 21 × 105 = I12 × 10 × 1200 

I

21  105 I = 1200  10

50 7 V

2 1

21  10 I12 = 12

I=

2

Now

2

 50 7  =   × (12.5) × 300  12.5 

2100 I = 12 I12 = 175



I1 = 5 7 A





= 10  5 7 V = 50 7 V Now in parallel voltage remains same  Voltage of heater VB = 50 7 V H2 = I R 2 t 2

2. (a)

3 I1 I2 5

2 2

42 × 105 = I 22  R 2  600 42 × 105 = VB × I2 × 600 42 × 105 = ( 50 7 ) × 600I2 I2 = I2 = I2 =



2500  7  125  300 12.5  125

2500  70  300 125 = 14 × 3 × 104 J = 42 × 104 Joule Hence when two resistors are connected in series across same supply voltage for 5 minuts heat produced is 42 × 104 Joule.

V = RI

Now

H= =

Now voltage of heater A

Now

A

H = I2Rt

2 1



50 7

12.5 

42  105



57V I 3

(I1–I2) 42V 1



(I2–I3) 2

25V



7  102

3I1 + 4I1 – 4I2 = 67

I2 = 20 7A VB = R2 × I2 50 7 = R2 × 20 7 50 7



R2 =



R2 = 2.5 

20 7

Hence resistance of heater 'B' is 2.5 

4V

Applying KVL in 100 loop 1 –42 + 3I1 + 4(I1 – I2) – 25 = 0

7 × 20

3

70V

50 7  600

5 7 6

7

 7I1 – 4I2 = 67

...(i)

Applying KVL in loop 2 we get, 25 – 4(I1 – I2) + 5I2 + 57 + 6(I2 – I3) + 70 = 0 25 – 4I1 + 4I2 + 5I2 + 57 + 6I2 – 6I3 + 70 = 0 –4I1 + 15I2 – 6I3 = – 152 4I1 – 15I2 + 6I3 = 150 Applying KVL in loop 3 we get, –70 – 6(I2 – I3) + 7I3 – 4 = 0 –70 6I2 + 6I3 + 7I3 – 4 = 0

...(ii)

SOLVED PAPER 2018

13I3 – 6I2 = 74

...(iii)

Solving equations (i), (ii) and (iii) we get

 7I  67  13I3 – 6  1  = 74 4   52I3 – 42I1 + 402 = 296

(i) Self inductance of second coil L2 =

N 2 2 I2

1700  0.8 × 10–3 6 = 0.226 H

=

52I3 – 42I1 = –106

...(iv)

So, self inductance of first coil :

7I  67  and 4I1 – 15  1  + 6I3 = 152 4  

L 1 = L2 ×

16I1 – 105I1 + 1005 + 24I3 = 608

N12 N 22

= 0.226 ×

–89I1 + 24I3 = 608 – 1005 –89I1 + 24I3 = – 397

...(v)

Solving (iv) and (v) we get

(600)2 (1700)2

= 0.028 H (ii) The cofficient of coupling (K)

I3 = 2, I2 = –8 and I1 = 5 Hence current flowing through 7 resistor is 2A

=

21 2

0.5  103 = 0.625 0.8  10 3 (iii) The cofficient of mutual induction (M) =

(b) Given data, N = 30 r = 8 cm

= K L1 L1

I=6A

= 0.625 0.028  0.226 = 0.05 H

B = 1T  = 60° Magnitude of counter torque that must be applied to prevent the coil from turning is given by  = NIAB sin  Now area of circular ring (A) = r2 A = (0.08)2 

5

30  3.14  (0.08)2  6  1  3 ° = 2   3.13 Nm

(c)

A Given data, 1 = 0.5 mwb 2 = 0.8 mwb N1 = 600 N2 = 1700

B C

D

3. (a) There are two types of Electrical Measuring Instruments: 1. Absolute Instruments : these instruments give the values of the quantity that has to be measured in terms of physical constants and their deflection only. They do not need to be calibrated and do not need any comparison with other standard instruments. Examples: Tangent Galvanometer, Absolute electrometer, and Raleigh current balance. 2. Secondary instruments : these are instruments whose output is measured to give the value of the quantity. The quantity to be measured is determined by the deflection value of these instruments. These instruments need to be calibrated against an absolute instrument. Examples : ammeter, voltmeter, amperehour meter, wattmeter etc. Absolute instruments are used in laboratories as standardizing instruments, whereas secondary instruments are used in everyday work.

6

SOLVED PAPER 2018

Main difference between absolute and secondary instrument are : Absolute Instrument 1. These give magnitude of quantity in terms physical constants of instruments. 2. Calibration is not required.

3. Measurement is time consuming as of tedious calc ulations. 4. Very rarely used in practical applications. 5. Absolute instrum ents are used in laboratories as standardizing instruments.

Secondary Instrument 1. These give reading directly of the quantity at the time of measurement. 2. Calibration with absolute instruments is required time to time as per requirements. 3. Measurement is quick because of direct measureement. 4. Very widely used in practic al applications. 5. Secondary instruments a re us ed in everyday work.

(b) Secondary Instruments: Definition : They are defined as the instruments that give a ready measure of the quantities with the help of graduated scales. 

The value of the electrical quantity to be measured is determined from the deflection of these instruments.



With an absolute instrument these instruments are calibrated. ie, the secondary instruments are compared with known standards or absolute instruments.



2. Recording Instruments : A continuous record of variations of the electrical quantity over a long period of time is given by these instruments. It has a moving system which carries an inked pen which rests tightly on a graph chart. Example : Graphic recorders and galvanometer recorders are the examples of these instruments. 3. Integrating instruments : The total amount of either electricity or electrical energy supplied over a period of time is measured by these instruments. Example : Ampere hour meters, watt-hour meters, energy meters are the few examples of these instruments. The basic difference between indicating instruments and integrating instruments are : Indicating Instruments That indicate the magnitude of an electrical quantity at the time when it is being measured. Their indications are given by a pointer moving over calibrated dial

3. (c) (i)

1. Indicating instruments 2. Recording instruments 3. Integrating instruments 1. Indicating Instruments : The value of the electrical quantity is indicated by these instruments at the time when it is being measured. Pointers moving over the scale give the indication. Example : Ordinary Ammeters, Voltmeters and wattmeters are the examples of these instruments.

Iin = 50 A Rin = 1 k V = 50 V

Multiplier resistance (Rs) = ? New multiplier resistance

All the electrical measuring instruments belong to this category.

There are three categories of secondary instruments.

Integrating Instruments That measure we by a set of dials and pointer either the total amount of electrical energy or total quantity of electrical energy supplied to a circuit in a given time.

(Rs) = V – Rin Iin

50    1000  =  6 50  10   = (106 – 1000) = 103(103 – 1) = 999 k (ii) Voltage multiplying factor is given by (m) m= 1 

Rs R in

 999k  m=1 +    1k  = 1 + 999 = 1000

 Voltage multiplier factor = 1000

SOLVED PAPER 2018

(d) Given data,

N1 =

l = 5 cm = 0.05 m b = 4 cm = 0.04 m

220  1000 132 N1  1667

Magnetic flux density in air gap (B) = 0.1 Wb/m2

(iv) Maximum core flux (m) :

Controlling torque spring constant

E1 = 4.44 m × f × N1

= 0.5 × 10 Nm/degree

200 kV = 4.44 m × 50 × 1667

–7

I = ? (deflection of 60°)

 200  1000  m =   4.44  50  1667  = 0.54 Wb

By the basic law electromagnet torque, deflection torque is given by T d = NBAI T d = 80 × 0.1 × 0.05 × 0.04 × I ...(i)

Now Tc for 60°

m = 0.54 Wb (v) Primary and Secondary current  120  103  I2 =    300  i400 

= 0.5 × 10–7 × 60 5 × 60 × 10–7 = 30 × 10–7 10 Now at steady state position

=

=

30 × 10–7 = 80 × 0.1 × 0.05 × 0.04 × I

  I2 I1 =    Turn Ratio 

and,

–7

30  10 80  0.1  0.05  0.04 = 187.5 × 10–6 = 187.5 A

I=

= 24053.13 5   3

220 kV

132 kV

(300 + 400 j)

200 kV

4. (a)

I1 = 144 53.13 (vi) Power supply by source : P = VI1 = 200 kV × 14453.13

1000 turn

220  (i) Turn Ratio (a) =   132  5 = 3 (ii) Secondery voltage



V1 = 5 V2 3

3 3 = 5 5 = 120 kV

V2 = V1 

(iii) Turn in High voltage side : 

120  103 50053.13

I2 = 240 53.13

T c = Td



V1  N2 V2

=

N = 80 turns



7

V N1 = 1 V2 N2

= (17280041.16 + 23039.96 J) = (17280.04 + J23.039) kVA So, active power = 17280.04 kW Reactive power = 23.039 kVAR 4. (b) As given that (i) No of conductors in armature (z) = 294 P = 4, A = 2 Ea = 230 V N = 1500 As we know that Ea = =

zNP 60A Ea  60A zNP