Quiz: Electrical Engineering
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute
Q1. In a vacuum cleaner, we generally use ………. motor.
(b) cumulatively compounded
(d) differentially compounded
Q2. A 220 V series motor takes a current of 35 A and runs at 955 r.p.m. Armature resistance = 0.25 Ω and series field resistance = 0.3Ω If iron and friction losses amount to 600 W, what is the armature torque?
(a) 140.52 Nm
(b) 118.5 Nm
(c) 70.26 Nm
(d) 35.13 Nm
Q3. The armature torque of a d.c. motor running at 850 r.p.m. is 97.3 Nm and shaft torque is 91.3 Nm. The iron and friction losses are
(a) 502.90 W
(b) 400 .00W
(c) 534.03 W
(d) 332.46 W
Q4. A d.c. motor runs at 1825 r.p.m. at full-load and 1875 r.p.m. at no-load. The speed regulation is ………
Q5. A transformer transfers electrical energy form primary to secondary usually with a change in ……….
Q6. A 230/2300 V transformer takes no-load current of 7.5 A and absorbs 195 W. If the resistance of the primary is 0.07 Ω, what is the core loss?
(a) 128.5 W
(b) 191.07 W
(c) 288.4 W
(d) 199.3 W
Q7. A generating station has a connected load of 49 MW and a maximum demand of 21 MW; the units generated being 61.5 × 106 kWh per annum. The demand factor is
Q8. For quick operation from zero to full load condition which of the following power plant is most suited…?
(a) nuclear power plant
(b) hydro-electric plant
(c) steam power plant
(d) gas turbine plant
Q9. The cost of an overhead line will increase if
(a) conductor spacing is decreased
(b) Conductor spacing is increased
(c) ground clearance is decreased
(d) Cheaper conductor is used
Q10. Corona is affected by
(a) condition of atmosphere
(b) size and spacing of conductors
(c) line voltage
(d) all of the above
Sol. we generally use series motor.
Sol. Motor resistance, Rm=0.25+0.3=0.55 Ω;Eb=V 〖-I〗a Rm=220-35×0.55= 200.75 V
Armature torque, Ta=9.55×(Eb Ia)/N=9.55×(200.75×35)/955=70.2625 Nm
Sol. Lost torque = Ta-Tsh = 97.3 – 91.3 = 6 Nm
Also, Lost torque = 9.55 × (Iron and friction loss)/N
∴ Iron and friction loss = (6 × 850)/9.55= 534.03 W
Sol. Speed regulation = (N0 -NFL)/NFL ×100=(1875-1825)/1825×100 = 2.74%
Sol. No. load loss, W0=195 W ; Primary Cu loss = I0^2 R1=(7.5)^2×0.07=3.93 W
Iron loss = W0-Primary Cu loss = 195 – 3.93 = 191.07 W
Demand Factor=(Max.demand)/(connected load)=21/49=0.428
Sol. Hydro-electric plant does not require a long starting time. In fact, such a plant can be put into service instantly.
Sol. Conductor spacing is increased
Sol. Corona is affected by condition of atmosphere, size and spacing of conductors & line voltage.