80 + Interview Questions in Polyphase Circuits in Network Theory Page 1 InterviewSolution

1.	A single wattmeter is connected to measure the reactive power of a three-phase, three-wire balanced load. The line current is 17A and the line voltage is 440V. Calculate the power factor of the load if the reading of the wattmeter is 4488 VAR.(a) 0.6(b) 0.8(c) 1(d) 1.2The question was posed to me in an online quiz.I'm obligated to ask this question of Power Measurement in Three-Phase Circuits in chapter Polyphase Circuits of Network Theory
Answer» RIGHT ANSWER is (B) 0.8 To EXPLAIN I would say: Wattmeter reading = VLILsinØ => 4488 = 440 x 17sinØ => sinØ = 0.6. POWER factor = cosØ = 0.8.

Discussion

2.	The total reactive power in the load while measuring the reactive power with wattmeter is?(a) √3VLILcosØ(b) √3VLILtanØ(c) √3VLILsinØ(d) √3 VLILsecØThe question was posed to me in final exam.Question is taken from Power Measurement in Three-Phase Circuits topic in portion Polyphase Circuits of Network Theory
Answer» RIGHT choice is (C) √3VLILsinØ To EXPLAIN: To obtain the reactive POWER, wattmeter reading is to be multiplied by √3. Total reactive power = √3VLILsinØ.

Discussion

3.	The wattmeter reading while measuring the reactive power with wattmeter is?(a) VLILsecØ(b) VLILsinØ(c) VLILtanØ(d) VLILcosØThe question was posed to me in an online interview.Query is from Power Measurement in Three-Phase Circuits in chapter Polyphase Circuits of Network Theory
Answer» The CORRECT option is (b) VLILsinØ The explanation is: The WATTMETER reading while measuring the REACTIVE power with wattmeter is wattmeter reading = VLILsinØ VAR.

Discussion

4.	The readings of the two watt meters used to measure power in a capacitive load are -3000W and 8000W respectively. Find the power factor.(a) 0.25(b) 0.5(c) 0.75(d) 1I had been asked this question in semester exam.The above asked question is from Power Measurement in Three-Phase Circuits topic in section Polyphase Circuits of Network Theory
Answer» Correct answer is (a) 0.25 The explanation: As the load is CAPACITIVE, the wattmeter connected in the leading phase GIVES LESS value. WR=-3000. WY=8000. tanØ = √3 (8000-(-3000))/5000=3.81 => Ø = 75.29⁰ => cosØ = 0.25.

Discussion

5.	The readings of the two watt meters used to measure power in a capacitive load are -3000W and 8000W respectively. Calculate the input power. Assume RYB sequence.(a) 5(b) 50(c) 500(d) 5000I have been asked this question in a national level competition.My doubt is from Power Measurement in Three-Phase Circuits in division Polyphase Circuits of Network Theory
Answer» Right OPTION is (d) 5000 Explanation: Toatal POWER is the sum of the power in R and power in Y. So total power = WR+WY = -3000+8000 = 5000W

Discussion

6.	The input power to a three-phase load is 10kW at 0.8 Pf. Two watt meters are connected to measure the power. Find the reading of lower reading wattmeter.(a) 1.835(b) 2.835(c) 3.835(d) 4.835This question was addressed to me in class test.Question is taken from Power Measurement in Three-Phase Circuits in chapter Polyphase Circuits of Network Theory
Answer» Right option is (B) 2.835 To elaborate: WR + WY = 10KW. Ø = cos^-10.8=36.8^o => tanØ = 0.75 = √3 (WR-WY)/(WR+WY)=(WR-WY)/10. WR-WY=4.33kW. WR+WY=10kW. WY=2.835kW.

Discussion

7.	The input power to a three-phase load is 10kW at 0.8 Pf. Two watt meters are connected to measure the power. Find the reading of higher reading wattmeter.(a) 7.165(b) 6.165(c) 6.165(d) 4.165I got this question in semester exam.Enquiry is from Power Measurement in Three-Phase Circuits in chapter Polyphase Circuits of Network Theory
Answer» RIGHT choice is (a) 7.165 Explanation: WR + WY = 10KW. Ø = cos^-10.8=36.8^o => tanØ = 0.75 = √3 (WR-WY)/(WR+WY)=(WR-WY)/10. WR-WY=4.33kW. WR+WY=10kW. WR=7.165kW.

Discussion

8.	The wattmeter method is used to measure power in a three-phase load. The wattmeter readings are 400W and -35W. Find the reactive power.(a) 751.44(b) 752.44(c) 753.44(d) 754.44This question was addressed to me by my college professor while I was bunking the class.This intriguing question originated from Power Measurement in Three-Phase Circuits topic in portion Polyphase Circuits of Network Theory
Answer» Correct option is (c) 753.44 For EXPLANATION I WOULD say: Reactive power = √3VLILsinØ. We know that WR – WY = 400-(-35)) = 435 = VLILsinØ. Reactive power = √3 X 435 = 753.44 VAR.

Discussion

9.	The wattmeter method is used to measure power in a three-phase load. The wattmeter readings are 400W and -35W. Find the power factor.(a) 0.43(b) 0.53(c) 0.63(d) 0.73This question was addressed to me at a job interview.I'd like to ask this question from Power Measurement in Three-Phase Circuits topic in division Polyphase Circuits of Network Theory
Answer» Correct OPTION is (a) 0.43 To EXPLAIN: We know tanØ = √3((WR – WY)/(WR + WY)) => tanØ = √3 (400-(-35))/(400+(-35))=2.064 => Ø = 64.15⁰. POWER FACTOR = 0.43.

Discussion

10.	The wattmeter method is used to measure power in a three-phase load. The wattmeter readings are 400W and -35W. Calculate the total active power.(a) 360(b) 365(c) 370(d) 375I have been asked this question in quiz.Query is from Power Measurement in Three-Phase Circuits in portion Polyphase Circuits of Network Theory
Answer» Right choice is (b) 365 To EXPLAIN I would say: Wattmeters are GENERALLY used to measure POWER in the circuits. TOTAL active power = W1 + W2 = 400 + (-35) =365W.

Discussion

11.	The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the B phase.(a) 0(b) 1(c) 3(d) 2I got this question in examination.The question is from Three-Phase Unbalanced Circuits topic in division Polyphase Circuits of Network Theory
Answer» Correct answer is (a) 0 Easy explanation: The TERM power is defined as the PRODUCT of square of current and the IMPEDANCE. So the power in the B PHASE = 40^2 X 0 = 0W.

Discussion

12.	The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the Y phase.(a) 1000(b) 2000(c) 3000(d) 4000The question was asked by my school principal while I was bunking the class.My enquiry is from Three-Phase Unbalanced Circuits topic in division Polyphase Circuits of Network Theory
Answer» RIGHT option is (b) 2000 The explanation is: The term power is DEFINED as the product of square of CURRENT and the impedance. So the power in the Y phase = 10^2 x 20 = 2000W.

Discussion

13.	The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the power in the R phase.(a) 6628(b) 6728(c) 6828(d) 6928I had been asked this question in examination.My question is from Three-Phase Unbalanced Circuits in chapter Polyphase Circuits of Network Theory
Answer» The correct answer is (d) 6928 The best I can explain: The TERM POWER is DEFINED as the product of square of CURRENT and the impedance. So the power in the R phase = 20^2 X 17.32 = 6928W.

Discussion

14.	The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I3.(a) (24.646+j20) A(b) (-24.646+j20) A(c) (-24.646-j20) A(d) (24.646-j20) AI got this question by my college director while I was bunking the class.My doubt stems from Three-Phase Unbalanced Circuits topic in portion Polyphase Circuits of Network Theory
Answer» The CORRECT ANSWER is (c) (-24.646-j20) A The best explanation: The LINE current I3 is the DIFFERENCE of IB and IY.So the line current I3 is I3 = IB – IY = (-24.646-j20) A.

Discussion

15.	The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I2.(a) (-27.32+j10) A(b) (27.32+j10) A(c) (-27.32-j10) A(d) (27.32-j10) AThe question was posed to me in exam.I'm obligated to ask this question of Three-Phase Unbalanced Circuits in section Polyphase Circuits of Network Theory
Answer» Correct option is (a) (-27.32+j10) A For explanation: The line CURRENT I2 is the DIFFERENCE of IY and IR. So the line current I2 is I2 = IY – IR = (-27.32+j10) A.

Discussion

16.	The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the line current I1.(a) (-51.96-j10) A(b) (-51.96+j10) A(c) (51.96+j10) A(d) (51.96+j10) AThe question was asked by my college professor while I was bunking the class.This intriguing question comes from Three-Phase Unbalanced Circuits topic in section Polyphase Circuits of Network Theory
Answer» CORRECT option is (c) (51.96+j10) A Best explanation: The LINE CURRENT I1 is the difference of IR and IB. So the line current I1 is I1 = IR – IB = (51.96+j10) A.

Discussion

17.	The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the phase current IB.(a) (34.64-j20) A(b) (34.64+j20) A(c) (-34.64+j20) A(d) (-34.64-j20) AThe question was posed to me in an online interview.My enquiry is from Three-Phase Unbalanced Circuits topic in portion Polyphase Circuits of Network Theory
Answer» Correct ANSWER is (d) (-34.64-j20) A Explanation: The VOLTAGE VBR is VBR = 400∠-240⁰V. The impedance Z3 is Z3 = 10∠-90⁰Ω => IB = (400∠240^o)/(10∠-90^o)=(-34.64-j20)A.

Discussion

18.	The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Find the phase current IY.(a) (10-j0) A(b) (10+j0) A(c) (-10+j0) A(d) (-10-j0) AThe question was posed to me in semester exam.My query is from Three-Phase Unbalanced Circuits in chapter Polyphase Circuits of Network Theory
Answer» The correct answer is (C) (-10+j0) A The BEST I can explain: The VOLTAGE VYB is VYB = 400∠-120⁰V. The impedance Z2 is Z2 = 40∠60⁰Ω => IY = (400∠-120^o)/(40∠60^o)=(-10+j0)A.

Discussion

19.	If the system is a three-wire system, the currents flowing towards the load in the three lines must add to ___ at any given instant.(a) 1(b) 2(c) 3(d) zeroI got this question during an interview.This key question is from Three-Phase Unbalanced Circuits in chapter Polyphase Circuits of Network Theory
Answer» The correct option is (d) ZERO To EXPLAIN I would SAY: If the system is a three-wire system, the currents flowing TOWARDS the load in the three lines must add to zero at any GIVEN instant.

Discussion

20.	The three impedances Z1 = 20∠30⁰Ω, Z2 = 40∠60⁰Ω, Z3 = 10∠-90⁰Ω are delta-connected to a 400V, 3 – Ø system. Determine the phase current IR.(a) (17.32-j10) A(b) (-17.32-j10) A(c) (17.32+j10) A(d) (-17.32+j10) AThis question was addressed to me during an online exam.This intriguing question comes from Three-Phase Unbalanced Circuits in division Polyphase Circuits of Network Theory
Answer» The CORRECT answer is (a) (17.32-j10) A Explanation: Taking VRY = V∠0⁰ as a reference phasor, and ASSUMING RYB PHASE SEQUENCE, we have VRY = 400∠0⁰V Z1 = 20∠30⁰Ω = (17.32+j10)Ω IR = (400∠0^o)/(20∠30^o) = (17.32-j10) A.

Discussion

21.	Determine the power (kW) drawn by the load.(a) 21(b) 22(c) 23(d) 24The question was asked during an interview for a job.My question is from Three-Phase Balanced Circuits topic in section Polyphase Circuits of Network Theory
Answer» CORRECT option is (d) 24 Easiest explanation: Power is DEFINED as the PRODUCT of voltage and CURRENT. So the power drawn by the load isP = 3VPhIPhcosØ = 24kW.

Discussion

22.	A three-phase balanced delta connected load of (4+j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current IB.(a) 44.74∠303.4⁰A(b) 44.74∠-303.4⁰A(c) 45.74∠303.4⁰A(d) 45.74∠-303.4⁰AThis question was addressed to me in examination.Origin of the question is Three-Phase Balanced Circuits in portion Polyphase Circuits of Network Theory
Answer» Right choice is (B) 44.74∠-303.4⁰A The explanation is: Taking the LINE voltage VRY = V∠0⁰ as a reference VRY = 400∠0⁰V, VYB = 400∠-120⁰V and VBR = 400∠-240⁰V. Impedance per phase = (4+j8) Ω = 8.94∠63.4⁰Ω. Phase current IB = (400∠240^o)/(8.94∠63.4^o) = 44.74∠-303.4⁰A.

Discussion

23.	A three-phase balanced delta connected load of (4+j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current IY.(a) 44.74∠183.4⁰A(b) 45.74∠183.4⁰A(c) 44.74∠183.4⁰A(d) 45.74∠-183.4⁰AThis question was addressed to me during an online exam.I'd like to ask this question from Three-Phase Balanced Circuits topic in section Polyphase Circuits of Network Theory
Answer» The correct choice is (c) 44.74∠183.4⁰A Easy EXPLANATION: Taking the LINE voltage VRY = V∠0⁰ as a REFERENCE VRY = 400∠0⁰V, VYB = 400∠-120⁰V and VBR = 400∠-240⁰V. Impedance per phase = (4+j8)Ω = 8.94∠63.4⁰Ω. Phase current IY = (400∠120^o)/(8.94∠63.4^o)= 44.74∠-183.4⁰A.

Discussion

24.	A three-phase balanced delta connected load of (4+j8) Ω is connected across a 400V, 3 – Ø balanced supply. Determine the phase current IR. Assume the phase sequence to be RYB.(a) 44.74∠-63.4⁰A(b) 44.74∠63.4⁰A(c) 45.74∠-63.4⁰A(d) 45.74∠63.4⁰AThe question was posed to me in exam.I want to ask this question from Three-Phase Balanced Circuits in chapter Polyphase Circuits of Network Theory
Answer» The correct OPTION is (a) 44.74∠-63.4⁰A The explanation is: Taking the line VOLTAGE VRY = V∠0⁰ as a reference VRY = 400∠0⁰V, VYB = 400∠-120⁰V and VBR = 400∠-240⁰V. Impedance per PHASE = (4+j8) Ω = 8.94∠63.4⁰Ω. Phase current IR = (400∠0^o)/(8.94∠63.4^o )= 44.74∠-63.4⁰A.

Discussion

25.	If the load impedance is Z∠Ø, the expression obtained for current (IB) is?(a) (V/Z)∠-240+Ø(b) (V/Z)∠-240-Ø(c) (V/Z)∠240-Ø(d) (V/Z)∠240+ØI have been asked this question in an international level competition.Enquiry is from Three-Phase Balanced Circuits in chapter Polyphase Circuits of Network Theory
Answer» Correct ANSWER is (b) (V/Z)∠-240-Ø Explanation: As the LOAD impedance is Z∠Ø, the current flows in the THREE load impedances and the current flowing in the B impedance is IB = VBR∠240⁰/Z∠Ø = (V/Z)∠-240-Ø.

Discussion

26.	If the load impedance is Z∠Ø, the expression obtained for current (IY) is?(a) (V/Z)∠-120+Ø(b) (V/Z)∠120-Ø(c) (V/Z)∠120+Ø(d) (V/Z)∠-120-ØI have been asked this question during an online exam.My question is from Three-Phase Balanced Circuits in portion Polyphase Circuits of Network Theory
Answer» Correct choice is (d) (V/Z)∠-120-Ø Explanation: As the LOAD IMPEDANCE is Z∠Ø, the current FLOWS in the three load impedances and the current flowing in the Y impedance is IY = VYB∠120⁰/Z∠Ø = (V/Z)∠-120-Ø.

Discussion

27.	If the load impedance is Z∠Ø, the current (IR) is?(a) (V/Z)∠-Ø(b) (V/Z)∠Ø(c) (V/Z)∠90-Ø(d) (V/Z)∠-90+ØThe question was asked in class test.This interesting question is from Three-Phase Balanced Circuits in portion Polyphase Circuits of Network Theory
Answer» The correct OPTION is (a) (V/Z)∠-Ø Best EXPLANATION: As the load IMPEDANCE is Z∠Ø, the current flows in the three load impedances and the current flowing in the R impedance is IR = VBR∠0⁰/Z∠Ø = (V/Z)∠-Ø.

Discussion

28.	In a balanced three-phase system-delta load, if we assume the line voltage is VRY = V∠0⁰ as a reference phasor. Then the source voltage VBR is?(a) V∠120⁰(b) V∠240⁰(c) V∠-240⁰(d) V∠-120⁰The question was asked during an interview.Enquiry is from Three-Phase Balanced Circuits in portion Polyphase Circuits of Network Theory
Answer» The correct choice is (c) V∠-240⁰ Explanation: As the LINE voltage VRY = V∠0⁰ is taken as a reference phasor. Then the SOURCE voltage VBR is V∠-240⁰.

Discussion

29.	In a balanced three-phase system-delta load, if we assume the line voltage is VRY = V∠0⁰ as a reference phasor. Then the source voltage VYB is?(a) V∠0⁰(b) V∠-120⁰(c) V∠120⁰(d) V∠240⁰I had been asked this question in semester exam.My enquiry is from Three-Phase Balanced Circuits in portion Polyphase Circuits of Network Theory
Answer» Correct answer is (B) V∠-120⁰ Explanation: As the line VOLTAGE VRY = V∠0⁰ is taken as a reference phasor. Then the SOURCE voltage VYB isV∠-120⁰.

Discussion

30.	In a delta-connected load, the relation between line voltage and the phase voltage is?(a) line voltage > phase voltage(b) line voltage < phase voltage(c) line voltage = phase voltage(d) line voltage >= phase voltageThis question was addressed to me during an online interview.This intriguing question originated from Three-Phase Balanced Circuits topic in chapter Polyphase Circuits of Network Theory
Answer» RIGHT choice is (c) LINE VOLTAGE = PHASE voltage To explain I would say: In a delta-connected LOAD, the relation between line voltage and the phase voltage is line voltage = phase voltage.

Discussion

31.	A balanced delta-connected load of (2+j3) Ω per phase is connected to a balanced three-phase 440V supply. The phase current is 10A. Find the apparent power.(a) 10955.67 VAR(b) 10.95567 VAR(c) 109.5567 VAR(d) 1.095567 VARThis question was addressed to me in my homework.My question comes from Voltage, Current and Power in a Delta Connected System topic in portion Polyphase Circuits of Network Theory
Answer» RIGHT choice is (a) 10955.67 VAR Easiest explanation: Sinφ = 0.83. Reactive power = √3 VLILsinφ. VL = 440V, IL = 17.32A. On substituting we GET reactive power = √3 x 440 x 17.32 = 10955.67 VAR.

Discussion

32.	A balanced delta-connected load of (2+j3) Ω per phase is connected to a balanced three-phase 440V supply. The phase current is 10A. Find the total active power.(a) 7.26W(b) 726W(c) 7260W(d) 72.6WThe question was posed to me in a national level competition.My doubt stems from Voltage, Current and Power in a Delta Connected System in section Polyphase Circuits of Network Theory
Answer» Right option is (c) 7260W To EXPLAIN I would say: ZPH = √(2^2+3^2) = 3.6∠56.3⁰Ω. cosφ = RPh/ZPh = 2/3.6 = 0.55. IL = √3× IPH = 17.32A. Active POWER = √3 VLILcosφ = √3×440×17.32×0.55= 7259.78W.

Discussion

33.	In a delta connected system, the expression of power (P) is?(a) VLILcosφ W(b) √3 VLILcosφ W(c) 3VLILcosφ W(d) 3√3VLILcosφ WThe question was posed to me at a job interview.My question is taken from Voltage, Current and Power in a Delta Connected System in portion Polyphase Circuits of Network Theory
Answer» RIGHT answer is (b) √3 VLILcosφ W The explanation is: The TOTAL power in the delta circuit is the sum of the powers in the three phases. In a delta CONNECTED SYSTEM, the expression of power (P) is P = √3VLILcosφ W.

Discussion

34.	The line currents are ________ behind respective phase currents in a delta connected system.(a) 120⁰(b) 90⁰(c) 60⁰(d) 30⁰I got this question during an interview for a job.This interesting question is from Voltage, Current and Power in a Delta Connected System in section Polyphase Circuits of Network Theory
Answer» Correct answer is (d) 30⁰ The EXPLANATION is: In a delta connected system, all the LINE currents are equal in magnitude but DISPLACED by 120⁰ from ONE another and the line currents are 30⁰ behind the respective phase currents.

Discussion

35.	In a delta-connected system, the currents IR = IB = IY =?(a) IPh(b) 2IPh(c) 3IPh(d) 4IPhI got this question by my college professor while I was bunking the class.This question is from Voltage, Current and Power in a Delta Connected System in chapter Polyphase Circuits of Network Theory
Answer» The correct option is (a) IPh To elaborate: In a delta-connected system, the currents IR = IB = IY = IPh. Since the system is balanced, the SUM of the THREE VOLTAGES ROUND the closed mesh is zero; consequently no current can flow around the mesh when the terminals are open.

Discussion

36.	The relation between IL and IPh is in a delta connected system is?(a) IL = IPh(b) IL = √3 IPh(c) IL = 3 IPh(d) IL = 3√3IPhI got this question in examination.The origin of the question is Voltage, Current and Power in a Delta Connected System topic in chapter Polyphase Circuits of Network Theory
Answer» Right answer is (b) IL = √3 IPh The EXPLANATION is: The relation between IL and IPh is in a delta connected system is IL = √3 IPh. The ARROWS placed ALONGSIDE the voltages of the three phases indicate that the terminals are positive during their positive HALF cycles.

Discussion

37.	In delta-connected system, the currents IR, IY, IB are equal in magnitude and they are displaced by _____ from one another.(a) 0⁰(b) 60⁰(c) 90⁰(d) 120⁰I had been asked this question in semester exam.Asked question is from Voltage, Current and Power in a Delta Connected System topic in portion Polyphase Circuits of Network Theory
Answer» Right answer is (d) 120⁰ The BEST I can explain: In delta-connected system, the currents IR, IY, IB are equal in magnitude and they are displaced by 120⁰ from one another. From the manner of interconnection of the three phases in the circuit, it may APPEAR that the three phase are short circuited AMONG themselves.

Discussion

38.	In a delta connected system, the voltage across the terminals R and Y is 400∠0⁰. Find the line voltage VBR.(a) 400∠240⁰(b) 400∠120⁰(c) 400∠-240⁰(d) 400∠-120⁰I got this question in an international level competition.This interesting question is from Voltage, Current and Power in a Delta Connected System in chapter Polyphase Circuits of Network Theory
Answer» Right choice is (c) 400∠-240⁰ Easiest explanation: We know, \|VBR\| = \|VPh\|, and is displaced by 120⁰, therefore the LINE voltage VBR is VBR = 400∠-240⁰V. DELTA connection is so CALLED because the THREE branches in the circuit can be arranged in the shape of delta.

Discussion

39.	In a delta connected system, the voltage across the terminals R and Y is 400∠0⁰. Find the line voltage VYB.(a) 400∠120⁰(b) 400∠-120⁰(c) 400∠240⁰(d) 400∠-240⁰This question was posed to me in unit test.This is a very interesting question from Voltage, Current and Power in a Delta Connected System topic in section Polyphase Circuits of Network Theory
Answer» The correct answer is (b) 400∠-120⁰ Explanation: As \|VYB\| = \|VPh\|, and is displaced by 120⁰, therefore the LINE voltage VYB is VYB = 400∠-120⁰V. A balanced three phase, three wire, delta connected system is referred to as mesh CONNECTION because it forms a CLOSED circuit.

Discussion

40.	A symmetrical star connected system has VRN = 230∠0⁰. The phase sequence is RYB. Find VBR.(a) 398.37∠210⁰(b) 398.37∠-210⁰(c) 398.37∠120⁰(d) 398.37∠-120⁰The question was posed to me during a job interview.My question is taken from Voltage, Current and Power in a Star Connected System in chapter Polyphase Circuits of Network Theory
Answer» The CORRECT answer is (b) 398.37∠-210⁰ The best I can EXPLAIN: All the line voltages are EQUAL in magnitude and are displaced by 120⁰. VBN = 230∠-240⁰V. VBR = √3×230∠(-240^o+30^o)V=398.37∠-210^oV.

Discussion

41.	In a delta connected system, the voltage across the terminals R and Y is 400∠0⁰. Calculate the line voltage VRY. Assume RRY phase sequence.(a) 400∠0⁰(b) 400∠120⁰(c) 400∠-120⁰(d) 400∠240⁰The question was posed to me in an interview.Asked question is from Voltage, Current and Power in a Delta Connected System topic in section Polyphase Circuits of Network Theory
Answer» Right ANSWER is (a) 400∠0⁰ For explanation I would say: In a balanced delta-connected system we know \|VRY\| = \|VPH\|, and it is displaced by 120⁰, therefore the LINE voltage VRY is VRY = 400∠0⁰V.

Discussion

42.	A symmetrical star connected system has VRN = 230∠0⁰. The phase sequence is RYB. Find VYB.(a) 398.37∠-30⁰(b) 398.37∠210⁰(c) 398.37∠90⁰(d) 398.37∠-90⁰I have been asked this question in an online interview.Origin of the question is Voltage, Current and Power in a Star Connected System topic in section Polyphase Circuits of Network Theory
Answer» RIGHT OPTION is (d) 398.37∠-90⁰ To elaborate: Corresponding line voltages are EQUAL to √3 times the phase voltages and are 30⁰ ahead of the RESPECTIVE phase voltages. VYN = 230∠-120⁰V. VYB = √3×230∠(-120^o+30^o)V=398.37∠-90⁰V.

Discussion

43.	A symmetrical star connected system has VRN = 230∠0⁰. The phase sequence is RYB. Find VRY.(a) 398.37∠30⁰(b) 398.37∠-30⁰(c) 398.37∠90⁰(d) 398.37∠-90⁰I had been asked this question in an interview for job.Query is from Voltage, Current and Power in a Star Connected System topic in chapter Polyphase Circuits of Network Theory
Answer» Right option is (a) 398.37∠30⁰ Easiest explanation: SINCE the system is a BALANCED system, all the PHASE VOLTAGES are equal in magnitude but displaced by 120⁰. VRN = 230∠0⁰V. VRY = √3×230∠(0^o+30^o)V=398.37∠30^oV.

Discussion

44.	The voltages, VBR ,Vph are related in star connected system is?(a) VBR = 3Vph(b) VBR = 3√3Vph(c) VBR = √3Vph(d) VBR = VphThe question was asked in examination.My enquiry is from Voltage, Current and Power in a Star Connected System topic in section Polyphase Circuits of Network Theory
Answer» Correct OPTION is (c) VBR = √3Vph Easiest explanation: The voltages, VBR, Vph in star CONNECTED system are related as VBR = √3Vph. The LINE voltage VYB is EQUAL to the phasor DIFFERENCE of VBN and VRN and is equal to √3Vph.

Discussion

45.	In a star connected system, the relation between VYB, Vph is?(a) VYB = Vph(b) VYB = 3√3Vph(c) VYB = 3Vph(d) VYB = √3VphThis question was posed to me in an interview for internship.My enquiry is from Voltage, Current and Power in a Star Connected System topic in section Polyphase Circuits of Network Theory
Answer» The correct answer is (d) VYB = √3Vph The explanation: In a star connected SYSTEM, the RELATION between VYB, Vph is VYB = √3Vph. The LINE VOLTAGE VYB is equal to the phasor DIFFERENCE of VYN and VBN and is equal to √3Vph.

Discussion

46.	In a star connected system, the phasors VRN, VYN are ____ apart.(a) 15⁰(b) 30⁰(c) 45⁰(d) 60⁰The question was posed to me at a job interview.This intriguing question originated from Voltage, Current and Power in a Star Connected System in division Polyphase Circuits of Network Theory
Answer» CORRECT option is (d) 60⁰ The best explanation: In a star connected system, the phasors VRN, VYN are separated by θ = 60⁰. To subtract VYN from VRN, we REVERSE the phase VYN and FIND its phasor sum with VRN.

Discussion

47.	The relation between VRY, Vph in a star connected system is?(a) VRY = Vph(b) VRY = √3Vph(c) VRY = 3√3Vph(d) VRY = 3VphI had been asked this question in an interview for internship.The origin of the question is Voltage, Current and Power in a Star Connected System topic in section Polyphase Circuits of Network Theory
Answer» The correct option is (b) VRY = √3Vph To EXPLAIN I would say: The TWO phasors VYN and VBN are equal in length and are 60⁰apart. The relation between VRY, Vph in a star CONNECTED system is VRY = √3Vph.

Discussion

48.	The relation between the lengths of the phasors VRN and – VYN is?(a) \|VRN\| > – \|VYN\|(b) \|VRN\| < – \|VYN\|(c) \|VRN\| = – \|VYN\|(d) \|VRN\| >= – \|VYN\|This question was posed to me during an interview.The doubt is from Voltage, Current and Power in a Star Connected System in chapter Polyphase Circuits of Network Theory
Answer» The CORRECT option is (C) \|VRN\| = – \|VYN\| The explanation is: The voltage VRY is found by compounding VRN and VYN reversed. The relation between the lengths of the phasors VRN and – VYN is \|VRN\| = – \|VYN\|.

Discussion

49.	In three phase system, the line voltage VRY is equal to?(a) phasor sum of VRN and VNY(b) phasor difference of VRN and VNY(c) phasor sum of VRN and VNY(d) algebraic sum of VRN and VNYThe question was posed to me in quiz.This intriguing question originated from Voltage, Current and Power in a Star Connected System topic in division Polyphase Circuits of Network Theory
Answer» The CORRECT choice is (a) phasor SUM of VRN and VNY The EXPLANATION: In THREE phase system, the line voltage VRY is equal to the phasor sum of VRN and VNY which is also equal to the phasor difference of VRN and VYN.

Discussion

50.	If a star connected system has equal impedances Z1, then after converting into delta connected system having equal impedances Z2, then?(a) Z2 = Z1(b) Z2 = 2Z1(c) Z2 = 3Z1(d) Z2 = 4Z1I had been asked this question in a national level competition.Enquiry is from Star to Delta and Delta to Star Transformation in section Polyphase Circuits of Network Theory
Answer» Correct answer is (c) Z2 = 3Z1 Explanation: If a star connected SYSTEM has equal IMPEDANCES Z1, then after converting into DELTA connected system having equal impedances Z2, then Z2 = 3Z1.

Discussion

Explore topic-wise InterviewSolutions in .

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Determine the power (kW) drawn by the load.(a) 21(b) 22(c) 23(d) 24The question was asked during an interview for a job.My question is from Three-Phase Balanced Circuits topic in section Polyphase Circuits of Network Theory

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The relation between IL and IPh is in a delta connected system is?(a) IL = IPh(b) IL = √3 IPh(c) IL = 3 IPh(d) IL = 3√3IPhI got this question in examination.The origin of the question is Voltage, Current and Power in a Delta Connected System topic in chapter Polyphase Circuits of Network Theory

The voltages, VBR ,Vph are related in star connected system is?(a) VBR = 3Vph(b) VBR = 3√3Vph(c) VBR = √3Vph(d) VBR = VphThe question was asked in examination.My enquiry is from Voltage, Current and Power in a Star Connected System topic in section Polyphase Circuits of Network Theory

In a star connected system, the phasors VRN, VYN are ____ apart.(a) 15⁰(b) 30⁰(c) 45⁰(d) 60⁰The question was posed to me at a job interview.This intriguing question originated from Voltage, Current and Power in a Star Connected System in division Polyphase Circuits of Network Theory