Quiz: Electrical Engineering
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute
Q1. A galvanometer with a full-scale current of 1 mA has a resistance of 1000 Ω. The multiplying power of a 100 Ω shunt with this galvanometer is
Q2. It has been estimated that a minimum run-off of approximately 94 m3/sec will be available at a hydraulic project with a head of 39 m. The efficiency of the plant is 80%. The firm capacity of the plant is
(a) 25500 kW
(b) 28770 kW
(c) 42424 kW
(d) 17882 kW
Q3. In a 3-phase system, the line losses are W at a p.f. of 1. If the power factor becomes 0.6 lagging, the losses will become
(a) 2 W
(b) 0.8 W
(c) 1.25 W
(d) 2.77 W
Q4. The annual expenditure on the central organization and salaries of high officials for a power plant is included in
(a) annual fixed cost
(b) annual semi-fixed cost
(c) annual running cost
(d) both (i) and (ii)
Q5. The annual working cost of a power plant is represented by the formula Rs. (a + b kW + c kWh) where the various terms have their usual meaning. The value of a is maximum for
(a) Hydro power plant
(b) Steam power plant
(c) Diesel power plant
(d) Nuclear power plant
Q6. The relation between annual cost C of energy wasted in an overhead transmission line and area of cross-section a of the conductor is
Q7. In India, secondary distribution is carried out at
(a) 400/230 V
(b) 600/300 V
(c) 700/350 V
(d) none of the above
Q8. In a 3-wire balanced d.c. system, the voltage between outer conductor and neutral is V. The voltage between the outers is
Q9. If the fault current is 2000 A, the relay setting 50% and C.T. ratio 400/5, the plug setting multiplier will be
Q10. Four alternators, each rated at 5 MVA, 11 kV, with 20% reactance are working in parallel. The short-circuit level at the bus-bars is
(a) 6.25 MVA
(b) 20 MVA
(c) 25 MVA
(d) 100 MVA
Sol. With shunt S = 100 Ω , suppose the maximum current that can be measured is I mA.
Now, S = 100 Ω ; G = 1000 Ω ; Ig = 1 mA ; I = ?
Now, (I – Ig) S = Ig G or (I – 1)100 = 1 × 1000 ∴ I = 11 mA
∴ Multiplying power of shunt =I/I_g =(11 mA)/(1 mA)=11
Sol. Mass of water available = 94 × 1000 = 94000 kg/sec ; Head, H = 39 m
∴ Work done/sec = W × H = 94000 × 9.81 × 39 = 35963 × 103 W
= 35963 kW
This is the gross plant capacity.
Firm capacity = Plant efficiency × Gross plant capacity
= 0.80 × 35,963 = 28770 kW
Sol. We know that line losses ∝ 1/cos2 ϕ .
For the first case, W ∝ 1/cos2 ϕ1 ; For the second case, W’ ∝ 1/cos2 ϕ2
∴ W’/W=cos^2〖ϕ_1 〗/cos^2〖ϕ_2 〗 =((1)²)/(0.6)²=2.77 ∴ W’ = 2.77 W
Note that line losses are greatly increased at low power factor.
Sol. The answer is annual fixed cost because these expenses have to be met whether the plant has high or low maximum demand or it generates less or more energy (k Wh).
Sol. The annual fixed cost (a) of nuclear power plant is very high because sophisticate technology is required for the manufacture of nuclear power plant.
Sol. 400/230 V
Sol. The voltage between the outers is 2V.
Sol. Pick-up value = Rated secondary current of C.T. × Current setting
= 5 × 0.5 = 2.5 A
Fault current in relay coil = 2000 × 5/400 = 25 A
∴Plug setting multiplier = 25/2.5 = 10
Sol. Equivalent reactance = 20/4 = 5%
Short-circuit MVA at bus-bars =5×100/5=100 MVA