SSC JE Basic Electrical Questions (2009 – 2018) Solved (Part-3)

SSC JE Basic Electrical Questions (2009 – 2018) Solved (Part-3)

Ques 1. The voltage wave v = Vm sin⁡(ωt − 15°) volt is applied across an ac circuit. If the current leads the voltage by 10° and the maximum value of current is Im, then the equation of current is

  1. I = Im sin⁡(ωt + 5°) amps
  2. I = Im sin⁡(ωt – 25°) amps
  3. I = Im sin⁡(ωt + 25°) amps
  4. I = Im sin⁡(ωt – 5°) amps

Answer.4. I = Im sin⁡(ωt – 5°) amps

Explanation:

From the above question, it is clear that the given circuit is capacitive in nature, therefore, the current is leading the voltage by 10°. Hence,

I = Im sin⁡(ωt − 15° + 10°)

=  I = Im sin⁡(ωt – 5°) amps

 

Ques 2. The average value of current (Iav) of a sinusoidal wave of peak value (Im) is

  1. Iav = Im/2
  2. Iav = Im x Π/2
  3. Iav = Im x 2/Π
  4. Iav = Im/√2

Answer 3. Iav = = Im x 2/Π

Explanation:

Average Value Of Alternating Current

The average value is the DC value that produces the same charge as it is produced by an AC source in the given circuit for the given time.
average-value-of-alternating current

Thus the average value of an alternating quantity = 0.637 x maximum value of that alternating quantity.

NOTE: The average value for a sinusoidal wave cannot be calculated over a complete cycle as it is zero so it is calculated over half cycle from 0 to 180 degrees.

 

 

Ques 3. The emf induced in a coil is given by f = -N dΦ/dt, where ‘e’ is the induced emf, N is the number of turns and dΦ’ is the instantaneous flux linkage with the coil in time ‘dt’. The negative sign of the above expression is due to

  1. Hans Christian Oersted
  2. Andre-Marie Ampere
  3. Michael Faraday
  4. Emil Lenz

Answer.4. Emil Lenz

Explanation:

  • According to Lenz’s law, the induced EMF sets up current is such a direction so as to oppose the very cause of producing it. Mathematically this expression is expressed in the negative sign
  • So, In Faraday’s law, the negative sign shows that the polarity of induced emf is such that it opposes any change in the magnetic flux of coil.

 

Ques.4. The temperature coefficient of resistance of copper at 20°c is

  1. 0.0045/°C
  2. 0.0017/°C
  3. 0.0393 /°C
  4. 0.0038/°C

Answer.3. 0.0393 /°C

Explanation:

Temperature Coefficient of Copper

The  formula for temperature effects on resistance is

R = Rref [1 + α(T – Tref)]

Where

R = Conductor resistance at temperature T

Rref = Conductor resistance at reference temperature “Tref” usually 20°C

α = Temperature coefficient of the resistance of the conducting material

T = Conductor temperature in degree

Tref = Reference temperature for α is specified

The temperature coefficient for some common materials are listed below (@ 20ºC):

Resistivity and Temperature Coefficient at 20 C

 

 

Ques.5.  How many watt-seconds are supplied by a motor developing 2 hp (British) for 5 hours?

  1. 2.68452×10 7 watt-seconds
  2. 4.476×10 5 watt-seconds
  3. 2.646×10 7 watt-seconds
  4. 6.3943×10 6 watt-seconds

Answer.1. 2.68452×10 7 watt-seconds

Explanation:

1 HP=745.7 Watt of Power

2 HP = 1491.4 Watt of Power

So, 2 HP  motor working for 5 hours = 1491.4 x 5= 7457 Watt-hour

To convert Watt-Hour into Watt-sec multiply it by 3600

7457 x 3600 = 26845200

= 2.68452×10 7 watt-seconds

 

Ques.6.  Phasor diagram of load voltage (V), current in pressure coil (I P ) and current in the current coil (I C ) is shown in the figure when an electrodynamic wattmeter is used to measure power. The reading of the wattmeter will be proportional to

Phasor diagram of load voltage (V), current in pressure coil

  1. cos⁡(β + ψ)
  2. cos⁡ψ
  3. cosβcosψ
  4. cosβcos⁡(β + ψ)

 Answer 3. cosβcosψ

Explanation: 

P = VI cos β. Ic cos ψ

Where β is the angle by which Ip is lagging behind applied voltage V

ψ is the power factor of the load

Therefore the reading of wattmeter will be proportional to

P  ∝ cos β cos ψ

 

Ques.7. Two parallel conductors carrying current in opposite directions will exert on each other

  1. An attractive force
  2. A repulsive force
  3. An axial force
  4. No force

Answer.2. A repulsive force

Explanation:

If the current flowing in the same direction then the conductor attracts each other and if the current flowing in the opposite direction then the conductor generates a repulsive force.

 

Ques.8. The unit of reluctance of magnetic circuit is

  1. AT/m
  2. Weber/m
  3. AT/Weber
  4. Weber/AT

Answer.3. AT/Weber

Explanation:

Reluctance is defined as

ℜ = F/Φ

F = magnetomotive force (MMF) in ampere-turns

Φ =  the magnetic flux in webers.

Therefore the unit of Reluctance is ampere-turns per Weber ( AT/Weber) (a unit that is equivalent to turns per Henry).

 

Ques.9.  A balanced 3-phase, 3-wire supply feeds balanced star connected resistors. If one of the resistors is disconnected, then the percentage reduction in load will be

  1. 33.33 %
  2. 50 %
  3. 66.67 %
  4. 75 %

Answer.2. 50

Explanation:

Underbalanced condition P = √3

A balanced 3-phase, 3-wire supply feeds balanced star connected resistors. If one of the resistors is disconnected, then the percentage reduction in load will be

If one of the resistors is disconnected it becomes the single-phase circuit with power

A balanced 3-phase, 3-wire supply feeds balanced star connected resistors. If one of the resistors is disconnected, then the percentage reduction in load will be

 

Ques.10. To minimize the error due to lead and contact resistances, low resistances used in electrical measurement work are provided with

  1. Guard Rings
  2. Four Terminals
  3. Thick Insulation
  4. Metal shields

Answer.2. Four Terminals 

Explanation:

Four-terminal sensing

Four-point-measurement

  • Four-terminal sensing is also known as Kelvin sensing, which measures very low resistances using four-terminal sensing. Each two-wire connection can be called a Kelvin connection.
  • When a Kelvin connection is used, the current is supplied via a pair of source connections (current leads). These generate a voltage drop across the impedance to be measured according to Ohm’s law V=IR.
  • A pair of sense connections (voltage leads) are made immediately adjacent to the target impedance so that they do not include the voltage drop in the force leads or contacts.
  • Since almost no current flows to the measuring instrument, the voltage drop in the sense leads is negligible hence accurate measurement can be obtained.
  • It is usual to arrange the sense wires as the inside pair, while the force wires are the outside pair.

Leave a Comment

Do NOT follow this link or you will be banned from the site!