SSC JE Electrical Question paper 2017 Solved

Ques 1. Two 10 ohm resistance is connected in parallel, their effective resistance will be

  1. 5
  2. 10
  3. 7
  4. None of the above

Answer.1. 5 ohm


Since the resistance are connected in parallel, therefore, their effective resistance will be

R1 x R2 / R1 + R2

100/20 = 5 ohm


Ques 2. In a 3-phase half wave rectifier, if the input phase voltage is 200 V, the PIV required for each diode will be

  1. 400 V
  2. 346 V
  3. 370 V
  4. 200 V

Answer.2. 346 V


Peak inverse voltage or PIV of diode = Maximum value of secondary voltage

= 200 x √3 = 346 V


Ques 3. For 100 turns as primary, 1000 V, 100 turns secondary transformer will have an output power of?

  1. 1000 V
  2. 500 V
  3. 250 V
  4. 100 V

Answer.4. 100 V


Transformer ratio is given as

= 1000/100 = 1000/Vs

= Vs = 100 Volt


Ques 4. Two couple coil of L1 = 0.8 H and L2 = 0.2 H have a coupling coefficient of K = 0.9. The mutual inductance M is

  1. 0.144 H
  2. 0.23 H
  3. 0.36 H
  4. 0.43 H

Answer.3. 0.36


Mutual inductance of two coupling coil is

M = K√L1L2

= 0.9√0.16

= 0.36 H


Ques 5. Transformer efficiency compared to other motor is

  1. Higher
  2. Remain same
  3. Lower
  4. None of these

Answer.1. Higher


  • .As Transformer is a static device, it does not have any rotating parts hence the mechanical losses are absent which increases the efficiency of the transformer.


Ques 6. Hall effect transducer with hall coefficient KH = -1 x 10-8 is required to measure a magnetic field of 10,000 gauss. A 2 mm bismuth slab is used as the transducer with a current of 3A. The output voltage of the transducer will be

  1. -15 x 10 -6 V
  2. -7.5 x 10 -6 V
  3. 10 x 10 -6 V
  4. -22.5 x 10 -6 V

Answer.1. -15 x 10 -6 V


Hall effect Output Voltage

= EH = KH IB / T

Where I is current

T is Thickness

B = Magnetic field = 10000 gauss = 10000 x 10-5 Wb/m2


Ques 7. Which power plant has the minimum running cost?

  1. Coal-based
  2. Hydel
  3. Nuclear
  4. None of these

Answer.2. Hydel


Operating cost of the hydro plant is low because of prime fuel, pressurised water comes almost free.


Ques 8. For power measurement of the three-phase circuit by two wattmeter method, when the value of power factor is less than 0.5 lagging.

  1. One of the wattmeters will read zero
  2. Both give the same reading
  3. One of the wattmeter connections will have to be reversed
  4. Pressure coil of the wattmeter will become ineffective.

Answer.3. One of the wattmeter connections will have to be reversed


Some point must be noted for two wattmeter method! READ THE POINT CAREFULLY

  1. If phase angle Φ = 0°, the readings of the two wattmeters are equal.
  2. If phase angle Φ < 60°, the readings of both the wattmeters are positive and the total power is the sum.
  3. If phase angle Φ = 60°, the reading P2 of the wattmeter W3 is zero.
  4. If phase angle Φ> 60°, i.e., if the power factor of the load is less than 0.5, the reading of wattmeter W2 is negative. In such case, the connection of either the current coil or the potential coil has to be reversed so that the pointer may deflect in the positive direction. The reading obtains after reversal of coil should then be taken as negative while calculating the power factor or the total power.

Now in the above question, the value of power factor is less than 0., therefore, one of the wattmeter connections will have to be reversed.


Ques 9. Which efficiency is more in case of cell

  1. Watt-hour efficiency
  2. Ampere-hour Efficiency
  3. Output Power efficiency
  4. None of the these

Answer.2. Ampere-hour Efficiency


The amp-hour is a unit of battery energy capacity hence Ampere hour efficiency is more in case of the cell and ampere-hour efficiency is always greater than watt-hour efficiency


Ques 10.  A 1000 ohms/V meter is used to measure a resistance on 150 V scale. The meter resistance is

  1. 150 KΩ
  2. 1K
  3. 6.67 ohms
  4. 0.001 Ohms

Answer.1. 150 KΩ


Resistance of Voltmeter = sensitivity in ohm × voltage in volts

= 1000 x 150 = 150 KΩ

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18 Comments on "SSC-JE-2016-2017-previous-year-electrical-question-paper-with-solution"

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Ques 93. The insulation resistance of a cable of length 10 Km is 1 MΩ. Its resistance for 50 Km length will be
1 MΩ
0.5 MΩ
0.2 MΩ
None of these
Hide Explanation
Answer.2. 0.5 MΩ
Insulation resistance of the cable
R α 1/Length
R2/R1 = L1/L2
Hence, insulation resistance for 50 km length
R2 = (1 x 10)/50
= 0.5 MΩ but ans is 0.2MEGA OHAM

Chetan gusain

35 The hysteresis coefficient is defined as the ratio of loss of energy per cycle to the field intensity or the intensity of electromagnetic stress in the material. Hence, the low hysteresis coefficient leads to reduce the hysteresis loss in that magnetic material.

Answer might be B

Chetan gusain

75 ….. answer C typing error

Chetan gusain

75 answer A


Question no 87’s answer is 80 % or 100%


In Q50 how the answer is all while sp. resistance depend uponmaterial and temp.

Kiran kumar

Explanation of question. 36 please.

Is there any app recording to your website?

This site is very helpful to me. Thank you so much.

Please provide GATE papers alsso


Ye exam paper jan2018 Wala he ya march 2017


Q24 answer is option 4

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