SSC JE 2015 Electrical question paper with solution

Ques 90.  A circuit breaker is rated as follows: 1500 A, 33 kV, 2000 MVA 3 sec, 3 phase oil circuit breaker, Determine the making current?

  1. 35 KA
  2. 1.5 KA
  3. 110 KA
  4. 89 KA
solution12

 

Ques 91. An electric heater draws 3.5 A from a 110 V source. The resistance of the heating elements is approximate?

  1. 31.43 ohms
  2. 3.1 ohms
  3. 385 ohms
  4. 38. 5 ohms

According to Ohm’s law

V=I*R

R=V/I

so R=110/3.5

=31.43 Ohm

 

Ques 92. The current ‘I’ in the electric circuit shown below is?

numerical5

  1. 3.7 A
  2. 1 A
  3. 2.7 A
  4. 1.7 A
solution 12

 

Ques 93. The acceptable value of grounding resistance for domestic application is?

  1. 2 ohm
  2. 0.5 ohm
  3. 1.5 ohm
  4. 1 ohm

Earth resistance should be as small as possible. It depends on the voltage level of the system to be grounded. As the voltage level increases the earth’s resistance is required to be nearer to zero as far as possible.  The NFPA and IEEE recommend a ground resistance value of 5 ohms or less.

For domestic application, the earth resistance should be between 0.2 – 0.5 ohms

 

Ques 94. An eight-pole wound rotor induction motor operating on a 60 Hz supply is driven at 1800 rpm by a prime mover in the opposite direction of the revolving magnetic field. The frequency of rotor current is

  1. 120 Hz
  2. 180 Hz
  3. 60 Hz
  4. 200 Hz

The speed of induction motor is

N = 120f/P

f = PN/120

f = (8 x 1800)/120

= 120

Frequency of rotor current is

f’ = f + 60 = 120 + 60

F’ = 180

Second Method

SOLUTION13

 

Ques 95. In a parallel RLC circuit if the lower cut-off frequency is 2400 Hz and the uppercut-off frequency is 2800 Hz, What is the bandwidth?

  1. 2800 Hz
  2. 2400 Hz
  3. 400 Hz
  4. 5200 Hz

Bandwidth = upper frequency – lower frequency

Bandwidth = f1 – f2

2800 – 2400

400 Hz

 

Ques 96. If 750 μA is flowing through 11 kΩ of resistance, what is the voltage drop across the resistor?

  1. 14.6 V
  2. 146 V
  3. 82.5 V
  4. 8.25 V
According to Ohm’s law, the voltage drop is the product of the current and the resistance

V=IR

V=(0.750)*(11)

V = 8.25

 

Ques 97. If two capacitances C1 and C2 are connected in parallel then the equivalent capacitance is given by

C1 C2

C1C2/(C1 + C2)

C1 + C2

C1 / C2

When two capacitance are connected in parallel the resultant capacitance is C1 + C2

 

Ques 98. The synchronous impedance method of finding voltage regulation of an alternator is called the pessimistic method because

  1. It is simplest to perform and compute
  2. Armature reaction is wholly magnetizing
  3. It gives a regulation value lower than its actual found by direct loading
  4. It gives the regulation value higher than its actual found by direct loading
The regulation calculated from the synchronous impedance method is higher than the actual value found by direct loading hence this method is called as the pessimistic method.

 

Ques99. If the coefficient of coupling between two coils is increased, mutual inductance between the coils

  1. Changes depend on current only
  2. Increased
  3. Decreased
  4. Remain unchanged

Mutual inductance between two coils are

M = K√L1L2

Where K is the coefficient of coupling and L1 & L2 are inductance between the coil

Therefore  from the above equation, we can say that the  mutual inductance is directly proportional to the coefficient of coupling i.e M ∝ K hence if the coefficient of coupling increases the mutual inductance will also increase

 

Ques 100. When a series resistance RL circuit is connected to a voltage source V at  t = 0, the current passing through the inductor L at t = 0 is

  1. Infinite
  2. Zero
  3. V/L
  4. V/R
  • When the switch is first closed, the current through the inductor is zero, because it cannot change instantaneously. This means that the inductor acts as an open circuit, so all the voltage is across the inductor.

 

For SSC JE 2017  Electrical paper with complete solution Click Here

For SSC JE 2013 Electrical paper with complete solution Click Here

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