Quiz: Electrical Engineering 27 Aug 2020

Quiz: Electrical Engineering
Exam: NLC-GET
Topic: Miscellaneous

Each question carries 1 mark.
Negative marking: 1/4 mark
Time: 10 Minute

Q1. For the equation, s^3-4s^2+s+6=0 the number of roots in the left half of s-plane will be
(a)Zero
(b)One
(c)Two
(d)Three

Q2. The de-ionization of the medium in current zero method of an arc extinction is NOT achieved by
(a)Splitting the arc
(b)Lengthening of the gap between the contacts
(c)Increasing the pressure of the vicinity of the arc
(d)Cooling the arc

Q3. In a synchronous motor, during hunting when the rotor speed exceeds the synchronous speed then damper bar develop
(a)Induction generator torque
(b)Harmonic
(c)DC motor torque
(d)Synchronous motor torque

Q4. Power consumed by a balanced 3-phase; 3-wire load is measured by two wattmeter method. The first wattmeter reads twice that of the second. Then what will be the load impedance angle in radian?
(a) (π/6)
(b) (π/3)
(c) (π/2)
(d) (π/4)

Q5. Single or One Wattmeter method can only be used for
(a) Balanced three-phase load
(b) Imbalanced two-phase load
(c) Balanced one-phase load
(d) Imbalanced one-phase load

Q6. A series RLC circuit consisting of R = 10 Ω, X_L=20 Ω and X_C=20 Ω, is connected across an AC supply of 100 V (rms). The magnitude and pulse angle (with respect to supply Voltage) of the voltage across the induction coil are respectively.
(a) 100 V; 90°
(b) 100 V; -90°
(c) 200 V; -90°
(d) 200 V; 90°

Q7. The impedance of a circuit is given by Z = 3 + j4. its conductance will be:
(a) 3/4
(b) 3/25
(c) 1/3
(d) 3/7

Q8. At any power factor of the load the efficiency of transformer will be maximum when:
(a) copper loss is equal to core loss
(b) copper loss is equal to eddy current loss
(c) copper loss is less then core loss
(d) copper loss is greater than core loss

Q9. In a synchronous machine, all of the following losses are independent of the load EXCEPT:
(a) iron loss
(b) bearing friction
(c) windage loss
(d) copper loss

Q10. A synchronous machine with low value of short circuit ratio has:
(a) higher stability limit
(b) good voltage regulation
(c) good speed regulation
(d) lower stability limit

SOLUTIONS
S1. Ans.(b)
Sol. s^3-4s^2+s+6=0
by R-H criterion:
s^3 1 1
s^2 -4 6
s^1 (-4-6)/(-4)=+2.5
s^0 6
Here there is two sign changes( s^3 to s^2 and s^2 to s^1).
So, number of roots in RHS=2
∴Number of roots in LHS=3-2=1.

S2. Ans.(a)
Sol. There are two methods of extinguishing the arc in CB viz.
(a)High Resistance Method
(b)Low resistance Method or current zero method
Low resistance Method or current zero method: This method is employed for arc extinction in AC circuits only. In this method, the rapid increase of dielectric strength of the medium near current zero can be achieved by:
(a)Lengthening of the gap
(b)High pressure in the vicinity of the arc
(c)Blast effect
(d)Cooling
NOTE: Splitting the arc method is related with arc extinction by High resistance method.

S3. Ans.(a)
Sol. When hunting occurs, the difference in the speed of stator and rotor poles develops an induces emf in the damper winding, which acts in such a way to suppress the rotor oscillation. The induced emf generated develop induction torque in the synchronous motor.

S4. Ans.(a)
Sol. W1=2W2
W(1 ):W2=2:1
Let W1=2
W2=1
Then,
tanθ= √3 ((W1-W2)/(W1+W2 ))
=√3 ((2-1)/(2+1))
=√3×1/3=1/√3
⇒θ=tan^(-1) (1/√3)=30°=(π\/6)radian

S5. Ans.(a)
Sol.
I. one wattmeter → for 3-ϕ balanced load
II. Two wattmeter → 3-ϕ balanced, unbalanced star, Delta connected load
III. Three wattmeter →3-ϕ, 4 wire system

S6. Ans.(d)
Sol.

Here XL=XC=200 Ω.
So, it is case of series Resonance.
So, Z =R = 10 Ω.
IR =V/R=100/10 =10 A
IR = IL = IC (series combination)
So, VL=I× XL
= 10 × 20
= 200 V

├ ■(VL=200V@θ=90° leading)⌉

S7. Ans.(b)
Sol. given z = 3 + 4j
conductance (G) =?
Admittance (γ) = 1/Z
= 1/(3+j 4)
= 1/(3+j4) × ((3-j))/((3-j4))
= (3-j4)/25
= 3/25-4/25 j

▭(γ = 1/z = G + j B)
↓ ↓
conductance susceptance
on comparing
G = 3/25

S8. Ans.(a)
Sol. At any power factor of the load, the efficiency of transformer will be maximum when copper loss is equal to core loss.

S9. Ans.(d)
Sol. In a synchronous machine, copper loss is load dependent.

S10. Ans.(d)
Sol. Xd in p.u = 1/SCR
SCR = 1/ Xd in per unit
Thus, short circuit ratio is equal to the reciprocal of per unit value of direct axis synchronous reactance. The more the value of Xd, the lesser will be the short circuit ratio.
For the small value of the short circuit ratio (SCR), the synchronizing power is small. As the synchronizing power keeps the machine in synchronism, a lower value of the SCR has a low stability limit. In other words, a machine with a low SCR is less stable when operating in parallel with the other generators.

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