SSC JE electrical question paper explained 2016-2017

22 Comments / SSC previous year solved electrical papers / April 12, 2022

Ques 51. An op-amp has a common-mode gain of 0.01 and a differential mode gain of 105. Its common-mode rejection ratio would be

  1. 10-7
  2. 10-3
  3. 103
  4. 107

Answer.4. 107

Explanation:

CMMR = Differential mode gain / Common-mode gain = Ad/Ac

= 105/0.01

= 107

 

Ques 52. In which type of amplifier, the power output is highest possible?

  1. Common C
  2. Common B
  3. Common E
  4. Darlington Pair

Answer.4. Darlington Pair

Explanation:

Darlington pair

  • Darlington pair consists of two transistors in such a way that the current amplified by the first transistor is amplified further by the second one. This results in a very high level of current gain. The overall current gain of the Darlington pair is the product of the two individual transistors.
  • This means that if two transistors with modest current gains of 25 were used, then the overall current gain would be 25 × 25 = 625

 

Ques 53. A voltage source VAB = 4 sin ωt is applied to the terminal A and B of the circuit shown in the given figure. The diodes are assumed to be ideal. The impedance offered by the circuit across the terminal A and B is

Terminal A and B

  1. 5 kΩ
  2. 10 kΩ
  3. 15 kΩ
  4. 20 kΩ

Answer.2. 10 KΩ

Explanation:

In the first half of the cycle, Diode D1 conducts and the impedance offered will be 10 KΩ as shown in the below figure

Solution 53

In the negative half cycle, diode D2 conducts and impedance offer will be 10 KΩ

Solution 54

Hence equivalent impedance offered across terminals A and B is 10 KΩ

 

Ques 54.  Which of the following motor is best suitable for the hoist?

  1. 3 Phase Induction motor
  2. 1 Phase Induction motor
  3. 3 Phase synchronous motor
  4. Repulsion Motor

Answer.1. 3 Phase Induction motor

Explanation:

Crane and hoists require high starting torque during starting because they carry a heavy load. The induction motor has a starting torque of about 120 to 150 percent of full-load torque hence they are preferred for cranes and hoists.

 

Ques 55. The op-amp as shown in the figure has a gain of -100. The value of R1 is

op amp

  1. 10 KΩ
  2. 10.1 KΩ
  3. 20 KΩ
  4. 100 MΩ

Answer.1. 10 KΩ

Explanation:

OP-amp gain is given as

Gain = -(R2/R1)

-100 = -( 1 x 106/R1)

R1 = 1000Ω

R1 = 10 KΩ

 

Ques 56. Millman theorem is an extension of?

  1. Norton
  2. Thevenin
  3. Superposition
  4. Both 1 & 2

Answer.4. Both 1 & 2

Explanation:

Millman’s theorem provides an equivalent circuit that is Similar to Thevenin or Norton’s equivalent circuit. Hence we can say that Millman’s theorem is an extension of Thevenin’s or Norton’s theorem for the circuit containing the number of independent voltage or current sources respectively.

 

Ques 57. A tank circuit of RF amplifier has a coil of Q factor of 100. It is tuned to a frequency of 1 MHz. The bandwidth of the circuit is

  1. 1 kHz
  2. 10 kHz
  3. 100 kHz
  4. 1000 kHz

Answer.3. 10 KHz

Explanation:

Bandwidth = Frequency/Q Factor

= 1 x 106/100

=10 KHz

 

Ques 58. Diode resistance in case of forward biased in an ideal case

  1. Very very Low
  2. Zero
  3. Infinite
  4. Very High

Answer.2. Zero

Explanation:

  • In the ideal case diode resistance in forward biased acts as the perfect conductor and in reverse biased, it will act as the perfect insulator.
  • Therefore in an ideal case, the diode will offer zero resistance when forward biased.

 

Ques 59. If the resonant frequency of the circuit shown in figure I is 1 kHz, the resonant frequency of the circuit shown in figure II will be

Figure

  1. 4 kHz
  2. 2 KHz
  3. 0.05 KHz
  4. 0.25 KHz

Answer.3. 0.05 KHz

Explanation:

In figure second

Leq = 2L

Ceq = 2C

Solution 2

 

Ques 60. Two 100W, 200V lamps are connected in series across a 200V supply. The total power consumed by each lamp will be watts?

  1. 25 W
  2. 200 W
  3. 75 W
  4. 150 W

Answer.1. 25 W

Explanation:

We know that

P = V2/R

R = V2/P

R = (200*200)/100 = 400 ohms.

Since now the two bulbs are in series, the total resistance offered by them is :

R’ = 400 Ω + 400 Ω = 800 Ω

Now the current that flows through either of them is :

I = V/R

= 200/800 = 0.25A

Since the bulbs are in series, the same amount of current flows through each of them.

Thus the total power consumed by each lamp will be

P = I2R

= (0.25)2 × 400

=25 Watt 

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22 thoughts on “SSC JE electrical question paper explained 2016-2017”

  2. Arup Ghoshal

    respected sir,here is only 100 questions. Can you please send me the full set of 2017 solved papers with explanation. Please sir help me. This website is very helpful to me.

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