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

Ques.91. How much energy is stored in a 100 mH inductance when a current of 1A is flowing through it? (SSC-2015)

  1. 5.0 J
  2. 0.05
  3. 0.5 J
  4. 0.005 J

Answer.4. 0.005 J


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

E = 1/2 L I2


E = energy stored (joules, J)

L = inductance (henrys, H)

I = current (amps, A)

1/2(100 x 10-3 x 1)

50 x 10-3 J

0.05 J


Ques.92. For the circuit shown below, find the resistance between points P & Q. (SSC-2015)

Numerical 1

  1. 1 Ω
  2. 2 Ω
  3. 3 Ω
  4. 4 Ω



solution 1

Ques.93. The rate of change of current in a 4 H inductor is 2 Amps/sec. Find the voltage across the inductor (SSC-2015)

  1. 8 V
  2. 16 V
  3. 2 V
  4. 0.8 V

Answer.1. 8V


Here L = 4H and di/dt = 2 A

Hence voltage Across inductor

= V = L(di/dt)

= 4 x 2 = 8 V


Ques 94. Find the node voltage VA (SSC-2015)


  1. 6 V
  2. 5.66 V
  3. 6.66 V
  4. 5 V

Answer.1. 6V



Ques 95. In a pure inductive circuit if the supply frequency is reduced to half the current will? (SSC-2015)

  1. Be four times as high
  2. Be doubled
  3. Be reduced to half
  4. Be reduced to one fourth

Answer.3. Be reduced to half


The amplitude of the current in a pure inductive circuit is given by

Im = Vm/ωl

And ω = 2πf

Im = Vm/2πfl

Im ∝ 1/f

Hence, when the frequency is halved, the amplitude of the current is doubled in a pure inductive circuit.


Ques 96. When a source is delivering maximum power to the load the efficiency will be? (SSC-2015)

  1. Below 50 %
  2. Above 50%
  3. 50 %
  4. Maximum

Answer.3. 50 %


solution 3

The maximum power is transferred when the load resistance RL is equal to internal resistance Rth or the equivalent resistance Rth.

RL = Rth


Under this condition, the same amount of power is dissipated in the internal resistance Rth and hence the efficiency is 50%.

Ques 97. The internal resistance of a voltage source is 10 ohm and has 10 volts and its terminals. Find the maximum power that can be transferred to the load (SSC-2015)

  1. 25 W
  2. 5 W
  3. 0.25 W
  4. 2.5 W

Answer.4. 2.5 W

For maximum power to be transferred from a source to a load resistance, the value of load resistance should be equal to the internal resistance of the source.

RL = R = 10Ω


current through the circuit = 10/(10+10) = 0.5 A

power transferred = I2R = 0.25 x 10 = 2.5 W



Ques.98. A node in the circuit is defined as the (SSC-2015)

  1. Closed Path
  2. Group of interconnected Elements
  3. Open terminal of the elements
  4. Junction of two or more elements

Answer.4. Junction of two or more elements

In electrical engineering, the node refers to any point or junction on a circuit where two or more circuit elements meet.

Ques.99. The unit of luminous flux is (SSC-2015)

  1. Candela
  2. Lumen
  3. Lux
  4. Steradian

Answer.2. Lumen


The SI unit of luminous flux is the lumen (lm). One lumen is defined as the luminous flux of light produced by a light source that emits one candela of luminous intensity over a solid angle of one steradian.

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