SSC JE electrical 2019 question paper with solution SET-1

Ques.11. What would the total loss of the 2kVA transformer corresponding to maximum efficiency be, provided the transformer has the iron of 150 W and full-load copper loss of 250W?

  1. 500 W
  2. 400 W
  3. 100 W
  4. 300 W

Answer.4. 300 W


When variable loss becomes equal to the constant loss, efficiency is maximum.

Losses = Pi + Pc

Since copper loss is a variable loss therefore

Losses = Pi + Pi = 2pi

Thus at a maximum efficiency of this transformer total loss

= 150 x 2 = 300 W


Ques.12. An alternator has 9 slots/pole. What is the value of the pitch factor if each coil spans 8 slot pitches.

  1. Cos 45°
  2. Cos 30°
  3. Cos 10°
  4. Cos 60°

Answer.3. Cos 10°


Coil Span = 8

Pole Pitch = 9 slots/pole

Slot angle (α) = 180° / Pole Pitch = 180°/9 = 20°

Pitch Factor = cos(α/2) = cos (20/2)

Pitch Factor = cos 10°


Ques.13. Which of the following is the function of the moderator in a nuclear reactor?

  1. To slow down the fast-moving electron
  2. Controlling chain reaction
  3. Stopping chain reaction
  4. Starting chain reaction

Answer.1. To slow down the fast moving electron


The main function of the moderators is to reduce the energy of neutrons evolved during fission. By slowing down the high energy neutron, the possibility of escape of neutrons is reduced while the possibility of absorption of neutrons by fuel to cause further fission is increased. This also reduces the amount of fuel required for the chain reaction. Moderator reduces the high speed (equal to 1.5 x 107 m/s) of the fast neutron to the corresponding speed (equal to 2200 m/s) of molecules in gas at Nuclear Thermal Plant.

The commonly used moderators are graphite, beryllium, and heavy water. Some other functions of moderators include the prevention of corrosion of fuel element, retain the radioactivity, and to provide structural support.


Ques.14. A single-phase circuit’s power factor can be measured with a/an

  1. Energy meter alone
  2. Combination of a voltmeter, ammeter, and wattmeter
  3. Wattmeter alone
  4. Combination of a voltmeter and ammeter

Answer.2. Combination of a voltmeter, ammeter, and wattmeter


The power factor in a single-phase circuit (or balanced three-phase circuit) can be measured with the wattmeter—ammeter—voltmeter method, where the power in watts is divided by the product of measured voltage and current.

Power in an ac circuit is given as:

P= V.I cosφ


P = Power input to the load circuit

V = Voltage across the load

I = Current through the load

Cosφ = Load power factor

The factor, Cosφ is called the power factor of the circuit and its value depends upon the angle φ by which the current I leads or lags the voltage V. Power factor of a circuit depends upon the nature and value of the circuit elements viz. resistance. inductance and capacitance. The power factor varies between 0 and 1 since φ can vary between 0° and 90°.

Power Measurment

Power in a single-phase ac circuit can be measured either directly by a wattmeter or by measuring V, I and cosφ separately and multiplying their values. The wattmeters and power factor meters are dynamometer type instruments. They have two coils viz. current coil and potential coil. In most of such instruments, there are more than one current and voltage coils of different ratings.

single Phase Wattmeter

Digital instruments can be made that either directly measure the time lag between voltage and current waveforms and so calculate the power factor, or by measuring both true and apparent power in the circuit and calculating the quotient. The first method is only accurate if the voltage and current are sinusoidal; loads (e.g., rectifiers) distort the waveforms from the sinusoidal shape.


Ques.15. The cheap and temporary system of the internal wiring is?

  1. Cleat wiring
  2. Casing-capping
  3. CTS or TRS wiring
  4. Conduit wiring

Answer.1. Cleat wiring


Cleat wiring is normally used for temporary wiring purposes. As a permanent system of wiring, it is not preferred in domestic premises. However, it is quite suitable for taking a temporary connection, such as for function, marriages, etc. The cleat wiring system is very cheap and can be laid very easily. The disadvantage of this wiring system is that its life is short as there is no continuous base for the wires.

In this type of wiring, vulcanized Indian rubber (VIR) or polyvinyl chloride (PVC) insulated wires are used as conductors.

cleat wiring

Advantages of Cleat Wiring:

  • It is a simple and cheap wiring system
  • Most suitable for temporary use i.e. under construction building or army camping
  • As the cables and wires of the cleat wiring system are in the open environment, Therefore fault in cables can be seen and repair easily.
  • Cleat wiring system installation is easy and simple.
  • Customization can be easily done in this wiring system e.g. alteration and addition.
  • Inspection is easy and simple.

Disadvantages of Cleat Wiring:

  • Appearance is not so good.
  • Cleat wiring can’t be used for permanent installation because Sag may occur after some time of the usage.
  • It is not a lasting wire system because of the weather effect, the risk of fire, and wears & tear.
  • It can be only used on 250/440 Volts at low temperatures.
  • There is always a risk of fire and electric shock.
  • it can’t be used in important and sensitive locations and places.
  • It is not a lasting, reliable, and sustainable wiring system.
  • In this wiring system, the cables and wiring are in the open air, therefore, oil, Steam, humidity, smoke, rain, chemical and acidic effect may damage the cables and wires.


Ques.16. If the copper loss of a transformer at half full load is 400 W, then the copper loss corresponding to full load is

  1. 400 W
  2. 800 W
  3. 1200 W
  4. 1600 W

Answer.4. 1600 W


Copper loss is proportional to the square of load current. At half load, load current becomes half as voltage remains the same, so the copper loss will become (1/2)2 i.e 1/4 times of full load copper loss.

At full load copper Loss = I2R

At half load copper Loss = (I/2)2 × R = I2/4 × R

400 = I2/4 × R

I2R = 4 × 400

I2R = Full load copper Loss = 1600 W


Ques.17. Find H =  _____ A/m at the center of a circular coil of diameter 1 m and carrying a current of 2A

  1. 1 A/m.
  2. 2 A/m.
  3. 4 A/m.
  4. 8 A/m.

Answer.B. 2 A/m


Magnetic Field Strength (H)  gives the quantitative measure of strongness or weakness of the magnetic field.

H = B/μo


B = Magnetic Flux Density

μo = Vacuum Permeability

Magnetic Field strength at the center of circular loop carrying current I is given by

B = μoI/2r

B/μo = I/2r

H = I/2r

Where r = Radius

Now Given Parameters

Diameter = 1m

Current = 2A

∴ Magnetic field Intensity H =  (2 / 2 × 1/2) = 2 A/m


Ques.18. In the circuit shown, if R = 0, then the phase angle between v(t) and i(t) is

SSC 2019 Ques.20

  1. 30°
  2. 60°
  3. 90°

Answer.4. 90°


In the given circuit if R = 0 then the circuit becomes purely inductive. So the phase angle between v(t) and i(t) is 90°.

In the circuit shown, if R = 0, then the phase angle between v(t) and i(t) is

Ques.19. The resistive component of the rotor with resistance R2 offered to the backward rotating flux wave in a single-phase motor runs with slip s is given as

  1. R2 ⁄ (1 + S)
  2. R2 ⁄ (2 − S)
  3. R2 ⁄ (1 − S)
  4. R2 ⁄ S

Answer.2. R2 ⁄ (2 − S)


The flux rotating in the clockwise direction (i.e. the direction of spinning) is called the forward rotating flux (φf) and that in the other direction is called the backward rotating flux (φb). The forward rotating flux has synchronous speed Ns (= 120 f/P) and the synchronous speed of rotating backward flux (anticlockwise) is –Ns.

Slip in forward Direction:-

Sf =  (Ns – N)/Ns = s

Slip in backward Direction:-

Sb =  (–Ns – N)/–Ns = s

Sb =  (2Ns + N+ N)/Ns

Sb = (2 – s)

At standstill, N = 0 so that Sf = Sb = 1. For forward rotating flux, the slip is s (less than 1) and for backward rotating flux, the slip is 2 – s (greater than 1). We know that in a 3-phase induction motor, the torque developed is directly proportional to effective rotor resistance.

The effective rotor resistance for single-phase motor is R2/s in the forward direction and R2/2 – s in the backward direction.


Ques.20. When the only current source is active in the circuit, find the current through the 10Ω resistor.

When the only current source is active in the circuit, find the current through the 10Ω resistor.

  1. 1.66 A
  2. 0 A
  3. 1.33 A
  4. 0.66 A

Answer.4. 0.66 A


Applying the  Thevenin’s Theorem in the given circuit and replacing the  voltage source in the circuit with its internal resistance as shown in the figure.

When the only current source is active in the circuit, find the current through the 10Ω resistor.

As per current divider rule

[latex]\begin{array}{l}{I_{10\Omega }} = I\dfrac{{{R_2}}}{{{R_1} + {R_2}}}\\\\{I_{10\Omega }} = 2\dfrac{5}{{5 + 10}}\\\\{I_{10\Omega }} = 0.66A\end{array}$

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