Quiz: Electrical Engineering
Exam: DFCCIL
Topic: Miscellaneous
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute
Q1. A transformer has a core loss of 64 W and copper loss of 144 W, when it is carrying 20 % overload current. The load at which this transformer will operate at the maximum efficiency……
(a) 120 %
(b) 80 %
(c) 44 %
(d) 66 %
Q2. A 40 KVA transformer has a core loss of 450 W and total loss of 800 W. find the copper loss for maximum efficiency.
(a) 800 W
(b) 350 W
(c) 450 W
(d) None of the above
Q3. In transmission line generally, the wire used are
(a) ACSR wires
(b) Iron wires
(c) Copper wires
(d) Aluminium wires
Q4. Under no-load conditions the current in a transmission line is due to
(a) Capacitance of the lines
(b) Corona effect
(c) Spinning reserve
(d) Back low from earth
Q5. A FET is essentially a
(a) Solar device
(b) Current driven device
(c) Power driven source
(d) Voltage driven device
Q6. FETs are
(a) Unipolar device
(b) Bipolar device
(c) Either unipolar or bipolar
(d) None of the above
Q7. Idc in a full wave rectifier is
(a) Im/π
(b) 2Im/π
(c) 3Im/π
(d) 4Im/π
Q8. A 10 mH inductor carries a sinusoidal current of 1 A r.m.s. at a frequency of 50 Hz. The average power dissipated by the inductor is
(a) 0 W
(b) 0.25 W
(c) 0.5 W
(d) 1 W
Q9. The difference between the two half-power frequencies is called ……….
(a) resonant frequency
(b) quality factor
(c) mid-frequency
(d) bandwidth
Q10. In a 2-pole lap winding dc machine, the resistance of one conductor is 2Ω and total number of conductors is 100. Find the total resistance
(a) 50 Ω
(b) 100 Ω
(c) 200 Ω
(d) 10 Ω
SOLUTIONS
S1. Ans.(b)
Sol. At 20% overload current (i.e.,I=1.2I_fl) copper loss=144 W
As copper loss(Pcu ) α I^2
∴Copper loss at full-load, Pcu(fl) =144/(1.2)^2 =100 W
Core loss or constant loss=64 W
Load at which efficiency is maximum,
∴ηmax=rated KVA×√(Pi/Pcu(fl) )=rated KVA×√(64/100)=0.8×rated KVA
∴At 80% load, efficiency is maximum.
S2. Ans.(c)
Sol. For maximum efficiency, copper loss or variable loss=core or iron loss
i.e., Pcu=Pi=450 W
S3. Ans.(a)
Sol. ACSR (Aluminium Conductor Steel Reinforced) Conductors
Center of the conductors comprised galvanized steel and outer layers of the galvanized steel comprise Aluminum conductors. These types of conductors are much use for long spans of transmission lines because these lines got high tensile strength.
S4. Ans.(a)
Sol. During the no-load condition, the current flowing is only charging current due to line capacitance.
S5. Ans.(d)
Sol. Field effect transistor (FET):
It is unipolar device.
It is voltage-controlled device.
S6. Ans.(a)
Sol. Field effect transistor (FET):
It is unipolar device.
It is voltage-controlled device.
S7. Ans.(b)
Sol.
For FWR: Idc=(2Im)/π
For HWR: Idc=Im/π
S8. Ans.(a)
Sol. The average power dissipated by an inductor is zero.
S9. Ans.(d)
Sol. Half power frequencies are the frequencies (ω1,ω2) at which power dissipated is one half of the power dissipated at resonate frequency(wo).
The bandwidth (BW) is defined as the frequency band between half power frequencies;
i.e.,BW=ω2-ω1
S10. Ans.(a)
Sol. method-1:
X = Resistance of one conductor =2 Ω
Z = Total number of conductors=100
A = No of parallel paths=2(For lap winding A=P)
∴RT=(2×100)/2^2 =50 Ω
Method-2:
Total resistance depends upon no of parallel path. In lap winding parallel path is no of poles and here it is two. Half of conductor are in series i.e., 50 in series and rest of 50 in series and they are parallel together. 50 no 2Ω in series = 100Ω. When two such paths are parallel, their equivalent will be 50 Ω.
