Ques.91. How much energy is stored in a 100 mH inductance when a current of 1A is flowing through it? (SSC-2015)
5.0 J
0.05
0.5 J
0.005 J
Answer.4. 0.005 J
Explanation:-
The energy stored in the magnetic field of an inductor can be expressed as
E = 1/2 L I2
where
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)
1 Ω
2 Ω
3 Ω
4 Ω
Answer.1. 1Ω
Explanation:-
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)
8 V
16 V
2 V
0.8 V
Answer.1. 8V
Explanation:-
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)
6 V
5.66 V
6.66 V
5 V
Answer.1. 6V
Explanation:-
Ques 95. In a pure inductive circuit if the supply frequency is reduced to half the current will? (SSC-2015)
Be four times as high
Be doubled
Be reduced to half
Be reduced to one fourth
Answer.3. Be reduced to half
Explanation:-
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)
Below 50 %
Above 50%
50 %
Maximum
Answer.3. 50 %
Explanation:-
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)
25 W
5 W
0.25 W
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)
Closed Path
Group of interconnected Elements
Open terminal of the elements
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)
Candela
Lumen
Lux
Steradian
Answer.2. Lumen
Explanation:-
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.
