Quiz: Electrical Engineering
Exam: NLC-GET
Topic: Miscellaneous
Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute
Q1. A negative sequence relay is commonly used to protect
(a) An alternator
(b) A transformer
(c) A transmission-line
(d) A bus-bar
Q2. For enhancing the power transmission in along EHV transmission line, the most preferred method is to connect a
(a) Series inductive compensator in the line
(b) Shunt inductive compensator at the receiving end
(c) Series capacitive compensator in the line
(d) Shunt capacitive compensator at the sending end
Q3. An open loop system represented by the transfer function
(a) Stable and of minimum phase type
(b) Stable and of non-minimum phase type
(c) Unstable and of minimum phase type
(d) Unstable and of non-minimum phase type
Q4. A single-phase, 22 kVA, 2200 V/ 220 V, 50 Hz, distribution transformer is to be connected as an auto-transformer to get an output voltage of 2420 V. Its maximum kVA rating as an autotransformer is
(a) 22
(b) 24.2
(c) 242
(d) 2420
Q5. Determine the rating of a transformer to deliver a 100 W of dc power to a load under full-wave bridge rectifier.
(a) 144.5 VA
(b) 123 VA
(c) 100 VA
(d) 150 VA
Q6. In a p-n junction diode under reverse bias, the magnitude of electric field is maximum at………………………?
(a) the edge of the depletion region on the p-side
(b) the edge of the depletion region on the n-side
(c) the p-n junction
(d) the center of the depletion region on the n-side
Q7. 1 Volt = __________________________.
(a) 1 Joule/ 1 Coulomb
(b) 1 watt/ 1 Ohm
(c) 1 Joule/ 1 watt
(d) 1 watt/ 1 Coulomb
Q8. Ampere second could be the unit of:
(a) Power
(b) Conductance
(c) energy
(d) charge
Q9. The Y_bus matrix of a 100-bus interconnected system is 80% sparse. Then, the number of transmission lines in the system must be
(a) 1500
(b) 1900
(c) 3000
(d) 950
Q10. The starting current of a three-phase induction motor is 5 times the rated current, while the rated slip is 4%. The ratio of starting torque to full-load torque is…….
(a) 0.6
(b) 0.8
(c) 1.0
(d) 1.2
SOLUTIONS
S1. Ans.(a)
Sol. A negative sequence relay is commonly used to protect an alternator.
A relay which protects the electrical system from negative sequence component is called a negative sequence relay or unbalance phase relay. The negative sequence relay protects the generator(alternator) and motor from the unbalanced load which mainly occurs because of the phase-to-phase faults.
S2. Ans.(c)
Sol. For enhancing the power transmission in along EHV transmission line, the most preferred method is to connect a Series capacitive compensator in the line.
As we know, P α 1/X, where, X=XL-XC for series capacitive compensation
S3. Ans.(b)
Sol. Open loop system stability is depending only on pole locations ⇒ system is stable. There is one zero on right half of s-plane so system is non – minimum phase
S4. Ans.(c)
Sol. As voltage rating is additive. i.e. 2200+220=2420 V
And a_2wdg=2200/220=10
∴(KVA)auto=(a2wdg+1)×(KVA)2wdg=11×22=242 KVA
S5. Ans.(b)
Sol. Tranformer rating=Pdc/TUF
Where, TUF= transformer utilization factor
For, full wave bridge rectifier, TUF=0.812
For, centre-tapped, TUF=0.692
∴Transformer rating= 100/0.812=123 VA(case of bridge rectifier)
S6. Ans.(c)
Sol. Under the condition when a p-n junction is reverse biased, the applied reverse voltage is increased, the electric field across the p-n junction increases because in the depletion region across the p-n junction, there are equal two opposite charges stored. As a result of high electron field across the junction under reverse biased condition, the breakdown of charge carriers takes place. Therefore, current increases rapidly. This is because E=−dV/dr, when both p and n regions are heavily doped, a depletion layer is very thin. So, electron field is maximum at the junction.
S7. Ans.(a)
Sol. V = W/Q (joule/Coulomb)
Unit = Joule /coulomb or volt
S8. Ans.(d)
Sol. charge (q) = current × time
⇒ q = I t
↓ ↓
Ampere Sec
So, (1 C= 1Amp.Sec)
S9. Ans.(d)
Sol. Number of buses, N=100
Sparsity, s=0.8
Then, the number of transmission lines is given as: NT=(N^2 (1-s)-N)/2=(100^2 (1-0.8)-100)/2
⇒NT=(100×19)/2=950
S10. Ans.(c)
Sol. given that, Ist=5Ifl and sfl=4%=0.04
∴T_st=3/ωs Ist^2 R2=3/ωs (5Ifl )^2 R2
And T_fl=(3/ω_s I_fl^2 R_2)/s_fl
Therefore, Tst/Tfl =((5Ifl )^2×sfl)/(Ifl^2 )=(25×0.04)/1=1.0
