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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  Test: Poly Phase AC Circuit- 2 - Electrical Engineering (EE) MCQ

Test: Poly Phase AC Circuit- 2 - Electrical Engineering (EE) MCQ


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15 Questions MCQ Test - Test: Poly Phase AC Circuit- 2

Test: Poly Phase AC Circuit- 2 for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Test: Poly Phase AC Circuit- 2 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Test: Poly Phase AC Circuit- 2 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Poly Phase AC Circuit- 2 below.
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Test: Poly Phase AC Circuit- 2 - Question 1
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Three equal resistances connected in star take a line current of 10 A when fed from 400 V, 50 Hz source. If the load resistances are reconnected in delta, the line current would be

Detailed Solution for Test: Poly Phase AC Circuit- 2 - Question 1


or, 

Test: Poly Phase AC Circuit- 2 - Question 2
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The minimum number of wattmeters required to measure the real power in an n-phase system with unbalanced load is

Detailed Solution for Test: Poly Phase AC Circuit- 2 - Question 2

Minimum number of wattmeters required
 


  • In an n-phase system with unbalanced load, the minimum number of wattmeters required to measure the real power is n - 1.

  • Each wattmeter measures the real power in one of the phases, and since the load is unbalanced, each phase may have a different power consumption.

  • Therefore, in order to accurately measure the real power in all n phases, we need n - 1 wattmeters.


    •  
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Test: Poly Phase AC Circuit- 2 - Question 3
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The input to the 3-phase, 50 Hz circuit shown in figure below is 100 V. For a phase sequence of ABC, the wattmeters would read

Detailed Solution for Test: Poly Phase AC Circuit- 2 - Question 3

Wattmeter readings are:
W1 = VLIL cos (30 + φ)
and W2 = \/LIL cos (30 - φ)
Here, 

and φ = 30°

Now, 


and 

Test: Poly Phase AC Circuit- 2 - Question 4
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Three-phase power can be measured by two-wattmeter method in case of
Test: Poly Phase AC Circuit- 2 - Question 5
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In the case of power measured by two-wattmeter method in a balanced 3-phase system with a pure capacitive load
Detailed Solution for Test: Poly Phase AC Circuit- 2 - Question 5

W1 = VL IL cos (30° + φ)
and W2 = VL IL cos(30° - φ)
When the load is purely capacitance, then φ = - 90°.
So, W1 = VLI L cos(30°- 90°)

and W2 = VLIL cos (30° + 90°)

Hence, |W1| = |W2| but W1 = - W2

Test: Poly Phase AC Circuit- 2 - Question 6
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Measurement of power factor of a 3-phase system by two-wattmeter method can be obtained in case of
Test: Poly Phase AC Circuit- 2 - Question 7
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Two wattmeters used to measure power of a 3-phase balanced load reads W1, and W2. The reactive Dower drawn bv the load is
Detailed Solution for Test: Poly Phase AC Circuit- 2 - Question 7

We know that:
(W1 - W2) = VLILsin φ = Q1 - φ
So, Q3 - φ = √3(W1 - W2)

Test: Poly Phase AC Circuit- 2 - Question 8
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Two wattmeters used to measure power of a 3-phase balanced load reads W1 and W2. The' reactive Dower drawn bv the load is
Detailed Solution for Test: Poly Phase AC Circuit- 2 - Question 8

We know that:
(W1 - W2) = VL IL sin φ = Q1-φ
so, Q3 - φ = √3(W1 - W2)

Test: Poly Phase AC Circuit- 2 - Question 9
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A three-phase 220 V supply is applied to a balanced Δ-connected three phase load. The phase current being Iab =. 10∠-30°A as shown in the given figure, The total power received by the Δ-load is
Detailed Solution for Test: Poly Phase AC Circuit- 2 - Question 9

Iab = 10∠-30°A
So, Ibc = 10∠-150°A
and Ica = 10∠-270° A
Phase current,
Ia = Iab - Ica = (10∠-30° - 10∠-270°) A 
= 10[cos 30° - jsin 30° - cos 270° + jsin 270°]
= 10[0.866 - j0.5 - 0 - j1]
= 8.66 - j15 = 17.32 ∠-60° A
∴ Ia = 17.32 ∠-60°A
∴ Total power received
= √3 VLIL cos φL
= 3VPIP cos φph
= 3 x 220 x 10 x cos 30°
= 5715W

Test: Poly Phase AC Circuit- 2 - Question 10
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The magnitude of line current in the circuit shown below is
Detailed Solution for Test: Poly Phase AC Circuit- 2 - Question 10

Impedance per phase
= 1 + j1 + 3 + j5
= (4 + j6) Ω

Test: Poly Phase AC Circuit- 2 - Question 11
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A three-phase, 400 V ac system is connected across a delta connected load having load resistance per phase of 200 Ω. If one of the phases of the delta connected load experiences an open circuit, then power consumed by the load will be
Detailed Solution for Test: Poly Phase AC Circuit- 2 - Question 11

Initially, in closed - Δ, power consumed by the load is

= 2400W
Let, P2 be the power consumed in open delta.
then, 

Test: Poly Phase AC Circuit- 2 - Question 12
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Assertion (A): Generated voltage or current wave have no odd harmonics.
Reason (R): The field system and armature coils of the generators are all symmetrical and so generates mostly symmetrical voltage wave.
Test: Poly Phase AC Circuit- 2 - Question 13
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Assertion (A): RMS value of line voltage is more than √3 times of the rms value of phase voltage.
Reason (R): In a star-connected system, the line voltage contains no triple frequency harmonic content.
Test: Poly Phase AC Circuit- 2 - Question 14
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Assertion (A): In a balanced 3-phase star connected system, line voltages are 30° ahead of the respective phase voltages.
Reason (R): The line voltages are √3 times of the respective phase voltages,
Test: Poly Phase AC Circuit- 2 - Question 15
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Match the List - I (Power factor) with List - II (Wattmeter readings) for the measurement of power in a three-phase system using two-wattmeter method and select the correct answer using the codes given below the lists:
List-I
A. 0.5 lag
B. Unity
C. Zero
D. 0.866

List-II
1. W1 = W2 
2. W1 = 2W2
3. W1 > 0,  W2 = 0
4. W1 > 0, W2 < 0

Codes:
Detailed Solution for Test: Poly Phase AC Circuit- 2 - Question 15

W1 = VLIL cos (30 + φ)
and W2 = VLIL cos (30 - φ)
By finding φ from cos φ, we can find the values of W1 and W2.

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