SSC JE Basic Electrical Questions (2009 – 2018) Solved (Part-2)

SSC Topic wise Solved paper / March 15, 2022

Ques.170. A current of 5 mA flows in a resistance less choke from a 200 V alternating source. The energy consumed in the choke is (SSC-2014, E-shift)

  1. 0 J
  2. 4.4 J
  3. 500 J
  4. 1000 J

Answer.1. 0J

Explanation:-

As the choke is resistance less therefore

R = 0

Energy consumed = I2R = I20 = 0

 

Ques.171.  Find RAB for the circuit shown in the figure (SSC-2014, E-shift)

Numerical33

  1. 18Ω
  2. 30Ω
  3. 45Ω
  4. 68Ω

Answer.1. 18Ω

Explanation:-

When balanced, the Wheatstone bridge can be analyzed simply as two series strings in parallel, therefore, the current through the 50Ω resistor is zero. Hence the circuit become

solution33 1

RAB = 30Ω || 45Ω

= 30 x 45/ 30 + 45

= 18 Ω

 

Ques.172.  Calculate the total susceptance of the circuit as shown in the figure (SSC-2014, E-shift)

Numerical34

  1. 6.67 mho
  2. 1.87 mho
  3. 0.16 mho
  4. 0.08 mho

Answer.4. 0.08 mho

Explanation:-

The susceptance of the given circuit will be B1 + B2

where B1 =  XL/(R2L + X2L)

=  4/(32 + 42)

= 0.16 mho

B2 = Xc/(R2c + X2c)

= -10/[52 + (-102)]

= -0.08 mho

∴ susceptance = 0.16 – 0.08 = 0.08 mho

 

 

Ques.173.  In an R-L series circuit, the phase difference between the applied voltage and circuit current will increase if (SSC-2014, E-shift)

  1. XL is increased
  2.  R is increased
  3. XL is decreased
  4. Supply frequency is decreased

Answer.1. XL is increased

Explanation

The voltage of inductor directly proportional to the reactance of inductor XL when Voltage across the inductor is greater than the voltage across Resistor then under this condition Xl is greater than R  and the circuit becomes inductive in nature. So the phase angle increases.

Ques.174. A series circuit has R = 4 Ω, XL = 12 Ω and XC = 9 Ω and is supplied with 200 V, 50 Hz Calculate the power (SSC-2014, E-shift)

  1. 6400 W
  2. 8000 W
  3. 14,400 W
  4. 19,200 W

Answer.1. 6400 W

Explanation:-

The Impedance of series RLC circuit is given as

solution37

Ques.175. Two sinusoidal currents are given by the equations i1 = 50 sin (ωt + π/4) and i2 = 25 sin (ωt – π/6). The phase difference between them in ______ degrees. (SSC-2014, E-shift)

  1. 15
  2. 30
  3. 45
  4. 75

Answer.4. 75

Explanation:-

Phase difference Φ = Φ1− Φ2

= π/4 −(−π/6)

= 5π/12 = 75°

 

Ques.176. The reactance of 1-farad capacitance when connected to a DC circuit is (SSC-2014, E-shift)

  1. Infinite
  2. 1 ohm
  3. 0.5 Ohm
  4. Zero Ohm

Answer.1. Infinite

Explanation:-

Capacitive reactance XC = 1/2πfc

And for DC the frequency is equal to zero hence

XC = 1/2π0 = ∞

Ques.177. A supply voltage of 230 V, 50 Hz is fed to a residential building. Write down its equation for instantaneous value (SSC-2014, E-shift)

  1. 163 sin 314.16t
  2. 230 sin 314.16t
  3. 325 sin 314.16t
  4. 361 sin 314.16t

Answer.3. 325 sin 314.16t

Explanation:-

The AC voltage is supplied to our home, therefore, the equation of alternating voltage E is

E = Vm sinωt

and Vm =√2Vrms

= √2 x 230 = 325.27 V

= ω = 2πf = 2 x π x 50 = 314.16

= E = 325.27 sin314.16t

 

Ques.178. If the voltage is increased by ‘n’ times, the size of the conductor would (SSC-2014, E-shift)

  1.  Increase by ‘n’ times
  2.  Reduce by ‘1/n’ times
  3.  Increase by ‘n2‘ times
  4.  Reduce by ‘1/n2 ‘ times

Answer.4.  Reduce by ‘1/n2 ‘ times

Explanation:-

The volume of the conductor is Inversely proportional to the square of Voltage ( Vol ∝1/V2).

so if the voltage is increased by n times than volume is reduced to

Volume = 1/n2

Ques.179. The north pole of a magnet is moved away from a metallic ring. The induced current in the ring flows (SSC-2014, E-shift)

  1. Clockwise
  2. Anticlockwise
  3. First anticlockwise and then clockwise
  4. First clockwise and then anticlockwise

Answer.1. Clockwise

Explanation:-

When the north pole of the magnet is moved away from the coil the direction of the induced current is such as to make the nearer pole of the coil as a south pole which makes clockwise induced current to flow in the coil or the ring.

solution 20

Ques.180. Energy stored in the inductor is given by (SSC-2014, E-shift)

  1. (LI)2/√2
  2. L2I/2
  3. 1/√LI
  4. LI2/2

Answer.4. LI2/2

Explanation:-

The energy stored in the magnetic field of an inductor can be expressed as

W = 1/2 L I2

where

W = energy stored (joules, J)

L = inductance (henrys, H)

I = current (amps, A)

For SSC JE Basic Electrical Questions (Part-1) Solved (2009 – 2018) Click Here

For SSC JE Electrical Transformer Questions Solved (2009 – 2018) Click Here

For SSC JE 2018 SET-1  Electrical paper with complete solution Click Here

For SSC JE 2018 SET-2  Electrical paper with complete solution Click Here

For SSC JE 2018 SET-3  Electrical paper with complete solution Click Here

For SSC JE 2018 SET-4  Electrical paper with complete solution Click Here

For SSC JE 2018 SET-5  Electrical paper with complete solution Click Here

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