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WBSETCL-JE’21 EE: Daily Practices Quiz 03-Jan-2022

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WBSETCL-JE’21 EE: Daily Practices Quiz 03-Jan-2022

Each question carries 1 mark.
Negative marking: NO negative marking
Total Questions: 06
Time: 06 min

Q1. Maxwell bridge is used for the measurement of ……………having the Q-factor in the
range ……..
(a) Inductors, 1 < Q < 10
(b) capacitors, 1 < Q < 100
(c) Inductors, 100 < Q < 1000
(d) Capacitors and inductors, 1 < Q < 100

Q2. Find the coefficient of coupling of two coils having L1=2 H,L2=8 H and M=4 H.
(a) 0.5
(b) 1.0
(c) 0.25
(d) 0.75

Q3. Which device is used in sub-stations to improve the power factor?
(a) Synchronous reactor
(b) Series inductor
(c) Synchronous condenser
(d) Series capacitor

Q4. Ferranti effect in overhead line is experienced when?
(a) Sending end voltage is more than receiving end voltage
(b) Sending end voltage is less than receiving end voltage
(c) Sending end voltage is equal to receiving end voltage
(d) Voltage at all points in line is equal.

Q5. In a three-phase system, the relation VL=Vph is applicable to a ________.
(a) star-connected without neutral point
(b) star-connected load
(c) single-phase system also
(d) delta-connected load

Q6. A 8-pole lap-wound generator has 400 conductors; the e.m.f. induced per conductor
being 5 V. The generated voltage of the generator is ………
(a) 60 V
(b) 1500 V
(c) 360 V
(d) 250 V


S1. Ans.(a)
Sol. Bridges used for measurement of inductance:
Anderson bridge (Q<1).
Maxwell bridge (1<Q<10).
Hay bridge (Q>10).

S2. Ans.(b)
Sol. K=M/√(L1 L2 )=4/√(2×8)=4/4=1.0

S3. Ans.(c)
Sol. For power factor improvement, synchronous condenser is used.
Synchronous condenser is basically a synchronous motor operating at no-load and leading power factor.

S4. Ans.(b)
Sol. Ferranti effect in overhead transmission lines is experienced at light load condition in this phenomenon, V_R>V_S ie, receiving end voltage is greater than sending end voltage. To overcome this problem, shunt reactors are used at receiving end of lines.

S5. Ans.(d)
Sol. For delta connection –
Line voltage (VL )= Phase voltage (Vph)
Line current (IL) = √3× Phase current (Iph )

S6. Ans.(d)
Sol. No of conductors per parallel path=400/8=50
∴ Eg=50×5=250 V


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