QUIZ-RSMSSB-JE ELECTRICAL, RAJASTHAN AE/JE ELECTRICAL MCQS |_00.1
Engineering Jobs   »   Quiz: Electrical Engineering 21 Oct 2020

Quiz: Electrical Engineering 21 Oct 2020

Quiz: Electrical Engineering
Exam: RSMSSB-JEn
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/3 mark
Time: 10 Minute

Q1. In a three-phase system, the relation V_L=V_ph is applicable to a ________.
(a) star-connected without neutral point
(b) star-connected load
(c) single-phase system also
(d) delta-connected load

Q2. In a star connected three-phase balanced system. If the line current is 125A then the phase current is?
(a) 216.5 A
(b) 72.16 A
(c) 250 A
(d) 125 A

Q3. A 8-pole lap-wound generator has 400 conductors; the e.m.f. induced per conductor
being 5 V. The generated voltage of the generator is ………
(a) 60 V
(b) 1500 V
(c) 360 V
(d) 250 V

Q4. The curve B in Fig. below is the external characteristic of a ……… generator.
QUIZ-RSMSSB-JE ELECTRICAL, RAJASTHAN AE/JE ELECTRICAL MCQS |_40.1
(a) shunt
(b) series
(c) flat-compounded
(d) under-compounded

Q5. Out of the following, ……… wave is the peakiest.
(a) sinusoidal
(b) square
(c) rectangular
(d) triangular

Q6. When a 15-V square wave is connected across a 50-V a.c. voltmeter, it will read ………
(a) 15 V
(b) 15×√2 V
(c) 15/√2 V
(d) none of the above

Q7. An alternating voltage υ=200 √2 sin 100 t is connected to a 1 μ F capacitor
through an a.c. ammeter. The reading of the ammeter shall be
(a) 40 mA
(b) 80 mA
(c) 100 mA
(d) 20 mA

Q8. A choke is preferred to a resistance for limiting current in an a.c. circuit because
(a) choke is cheap
(b) there is no wastage of energy
(c) current becomes wattless
(d) current strength increases

Q9. The value of current ‘I’ is flowing in the 1Ω resistor in the circuit shown in the figure
below will be
QUIZ-RSMSSB-JE ELECTRICAL, RAJASTHAN AE/JE ELECTRICAL MCQS |_50.1
(a) 10 A
(b) 6 A
(c) 5 A
(d) zero

Q10. The no load speed of DC series motor is
(a) very small
(b) medium
(c) very high
(d) small

SOLUTIONS
S1. Ans.(d)
Sol. for delta connection –
Line voltage (VL )= Phase voltage (Vph)
Line current (IL) = √3× Phase current (Iph )

S2. Ans.(d)
Sol. for star (Y) – connection: –
Line current (IL) = Phase current (Iph )
Line voltage (VL )=√3×Phase voltage (Vph )
Or, phase voltage = (Line voltage)/√3
⇒ ▭(Vph=VL/√3) – For star connection
S3. Ans.(d)
Sol. No of conductors per parallel path
=400/8=50
∴ Eg=50×5=250 V

S4. Ans.(c)
Sol. Flat-compounded. (voltage regulation=zero)

S5. Ans.(d)
Sol. That wave is peakiest which has the greatest form factor (triangular wave).
The Form Factor for the various sinusoidal waveforms are as follows:
For a sine wave, it is π/2√2 = 1.11072073
For a half-wave rectified sine wave, it is π/2 = 1.5707963
For a full-wave rectified sine wave, it is π/2√2 = 1.11072073
For a square wave, it is equal to 1
For triangle waveform, it is 2/√3 = 1.15470054
For sawtooth waveform, it is 2/√3 = 1.15470054

S6. Ans.(a)
Sol. A voltmeter records the r.m.s. value and the r.m.s. value of 15-V square wave is 15 V.

S7. Ans.(d)
Sol. V(r.m.s.)=Vm/√2=200 √2 /√2=200V; ω=100 rad/s; X_C=1/ωC=1/(100×1×10^(-6) )=10^4 Ω
∴ I(r.m.s.)=V(r.m.s.)/X_C =200/10^4 =20×10^(-3) A=20 mA

S8. Ans.(b)
Sol. In an AC circuit, the coil of high inductance and negligible resistance used to control current, is called the choke coil. The power factor of such a coil is given by
cos⁡φ=R/√(R^2+ω^2 L^2 )≈ R/ωL (as R <<<ωL).
cos⁡φ is very small. Thus, the power absorbed by the coil is very small. The only loss of energy is due to hysteresis in the iron core, which is much less than the loss of energy in the resistance that can also reduce the current if placed instead of the choke coil.

S9. Ans.(c)
Sol.
Method-1: From ohm’s law, if we know the voltage across a resistor, then we can determine the current through resistor R as I=V/R.
Here, V = 5 V (constant)
So, I=5/1 = 5A
QUIZ-RSMSSB-JE ELECTRICAL, RAJASTHAN AE/JE ELECTRICAL MCQS |_60.1
Note: The voltage across any current source is purely arbitrarily. The voltage across it depends purely upon the voltage source connected in parallel across it. Hence, in present case voltage across the current source = 5 V.
Method2:
Applying superposition theorem:
Current due to 5 V source: current source open circuit & I=5/1=5 A
Current due to 5 A source: voltage source short circuit and current flows through short-circuit. So, current through 1-ohm resistor=0 A
∴I_TOTAL=5+0=5 A

S10. Ans.(c)
Sol. The load-speed curve of DC series motor is shown below.
QUIZ-RSMSSB-JE ELECTRICAL, RAJASTHAN AE/JE ELECTRICAL MCQS |_70.1
It is clear that the speed of the DC series motor at no load is very high. So, it is never stared at no load.

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