3.Three Ph. Circuits p - Copy

MODULEOF

THREE PHASE CIRCUITS

Reference Books:

1. B.L. Thareja, Volume-I,Basic Electrical Engineering.2. R.G. Karandikar, Basic Electricity and Electronics - II

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Zagdu Singh Charitable Trust’s (Regd.)

Thakur College of Engineering & Technology

(Approved by AICTE, Govt. of Maharashtra & Affiliated to University of Mumbai)ISO 9001 : 2000 Certified

3THREE PHASE CIRCUITS

MOTIVATION:

Present chapter deals with three phase systems and the behavior of the three phase system for 3-ph circuit with the concept of phaser diagram. The Chapter also highlights the comparision of single phase and three phase supply system.

SYLLABUS:

OBJECTIVE:

1. Generation of three phase voltages and current.2. Balanced star and delta connection.3. Relationship between phase and line currents and voltages.4. Measurement of power by wattmeter method.5. Numericals related to topic.

KEY NOTATIONS :

(Refer fig. 3.2,3.3)

eR = Instantaneous value of generated voltage in winding RR1

eY = Instantaneous value of generated voltage in winding YY1

eB = Instantaneous value of generated voltage in winding BB1

Em = Maximum voltage of induced voltage in each Winding.

IL = line current

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Iph = phase current

VL = RMS value of line voltage

Vph = RMS value of phase voltage

Zph = impedance/phase

P = Total power

Q = Total reactive power

S = Total apparent power

KEY DEFINITIONS:

Single Phase System – It is a system which generates single alternating voltage and current.

It uses only one winding.

Phase sequence or phase order – The sequence in which the voltage in the three phases reaches maximum positive value. Phase sequence of three phase system is R-Y-B. The direction of rotation of the three phase machines depends on the phase sequence.

Line voltage – It is the voltage at any instant between two lines. Between R and Y line we have VRY.

Phase voltage – It is the voltage of individual phases with respect to neutral. We have VR, VY, VB.

Line current – It is the Phase current in any phase. We have IR, IY, IB

Phase current – It is the current in individual phases. We have iR, iY, iB

Balanced system – a three phase system is said to be balanced if

(a) The voltages and current in the three phase are equal in magnitude and differ in phase from one another by 1200.

(b) The all loads connected across the three phase have same magnitude and power factor (identical).

(c) Star connection (wye connection) – It is obtained by connecting similar terminals (start or finish) of the three windings together. This common point is called as neutral point. The remaining three ends of the windings are brought out for the external connection.

Delta connection (mesh connection) – It is obtained by connecting one end of the winding to the starting end of the other winding such that it produces a closed loop.

LEARNING RESOURCES:

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1. B.L. Thareja, Volume-I,Basic Electrical Engineering.2. R.G. Karandikar, Basic Electricity and Electronics – II

IMPORTANT RELATIONS:

1. Voltage, current and power relations in balanced star connected load:

VL = √3Vph

IL = Iph

Reactive power Q = 3Vph Iph sin ǿ

2. Voltage, cyrrent and power relation in balanced delta connected load:

VL = Vph

IL = √3 Iph

P = 3Vph Iph cos ǿ

S = 3Vph Iph

= √3 VL IL

Q = √3 VL IL sin ǿ

S = √3 VL IL

No.of hours allotted: 8 Hours

Week allotted: 1.5 Lecture scheduleSr. no. Topic Hours required

1. A.C.fundamental 2hrs

2. Waveforms, vector diagram, phasor diagram

4hrs

3. A.C circuit 4hrs

4. Resonance circuit 2hrs

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5.

6. Application area 2hrs

INTRODUCTION TO TOPIC:

Generation of three phase voltages: A system which generates a single alternating voltage and current is termed as single phase system. It utilizes only one winding. A polyphase system utilizes more than one winding. It will produce as many induced voltages as the number of winding. A three phase system consists of three separate but identical windings that are displaced by 120 electrical degrees from each other. When these three windings are rotated in an anticlockwise direction with constant angular velocity in a uniform magnetic field, the emfs are induced in each winding which have same magnitude and frequency but displaced 1200 from one another

Fig:3.1

The instantaneous values of generated voltage in winding RR1, YY1 and BB1 are given by eR = Em sin өeY = Em sin (ө - 1200)eB = EB sin (ө - 2400) (All are in time domain)Where Em is the maximum value of induced voltage in each winding. The waveform of these three voltages is shown in Fig

Fig 3.2.

Phasor diagram of these three induced voltage.

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Fig3.3

STAR OR WYE CONNECTION: In the method, similar terminals (start or finish) of the three winding are joined together as shown in Fig. 3.5. The common point is called star or neutral point.

Fig.3.4

iR = Im sin өiY = Im sin (ө - 1200)iB = Im sin (ө - 2400)iR + iy + iB = Im sin ө +Im sin (ө - 1200) + Im sin (ө - 2400) = 0(All are in time domain)

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Fig.3.5DELTA OR MESH CONNECTION: In this method, dissimilar terminals of the three winding are joined together i.e. ‘finish’ terminal of one winding is connected to ‘start’ terminal of other winding and so on as shown in Fig.3.8 This system is also called three phase, three wire system.

Fig.3.6

For a balanced system sum of the three phase voltages round the closed mesh is zero. The three emfs are equal in magnitude but differ in phase from one another by 1200.eR = Em sin өeY = Em sin (ө - 1200)eB = Em sin (ө - 2400)eR + eY + eB = Em sin ө + Em sin (ө – 1200) + Em sin (ө - 2400) = 0(All are in time domain)

APPLICATIONS:

Application of three phase systems are-1. Three phase systems are generally used for Industrial purpose.2. It is generally used where large power application required e.g. Sub-stations, in transmission

systems where high efficiency required, etc.3. In High voltage transmission system it is mostly used.

KEY DERIVATIONS:

1. VOLTAGE, CURRENT AND POWER RELATIONS IN BALANCED STAR CONNECTED LOAD -

Relation between line voltage and phase voltage: Since the system is balanced, the three phase voltages VRN, VYN and VBN are equal in magnitude and differ in phase from one another by 1200.

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Fig. 3.7

Let VRN = VYN = VBN = Vph

Where Vph indicates rms value of phase voltage = Vph ∠ 00

= Vph 0

= Vph ∠-2400

Let VRY = VYB = VBR = VL

Where VL indicates rms value of line voltage. Applying Kircoff ‘s voltage law,

= + = -

= Vph ∠ 00 - Vph ∠ -1200 = Vph + j0 + 0.5 Vph + j0.866 Vph

= 1.5 Vph + j0.866 Vph = √3 Vph ∠ 300

Similarly, = + = √3 Vph ∠300

= + = √3 Vph ∠300

Thus in a star connected three phase system VL = √3 Vph and line voltages lead respective phase voltages by 300 .

Phasor Diagram: (Lagging power factor)

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Fig. 3.8

Relation between line current and phase current:From the fig 3.9, it is clear that line current is equal to the phase current. IL = Iph Power:The total power in three phase system is the sum of powers in the three phases. For a balanced load, the power consumed in each load phase is the same.

Total power P = 3 power in each phase = 3 Vph Iph cos ǿ

In a star connected three phase system,

Vph = VL /

Iph = IL

P = 3 VL/ 3 IL cos ǿ

= VL IL cos ǿ

Where ǿ is the phase difference between phase voltage and corresponding phase current. Similarly, total reactive power Q = 3Vph Iph sin ǿ

= VL IL sin ǿ

Total apparent power S = 3 Vph Iph = 3 VL IL

2. VOLTAGE, CURRENT AND POWER RELATIONS IN BALANCED DELATA CONNECTED LOAD-

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Fig. 3.9

Relation between line voltage and phase voltage :From Fig 3.11, It is clear that line voltage is equal to phase voltage. VL = Vph

Relation between line current and phase current :Since the system is balanced, the three phase current IRY,IYB and IBR are equal in magnitude but differ in phase from one another by 1200. Let IRY = IYB = Iph Where Iph indicates rms value of phase current.

= Iph ∠0 0

= Iph ∠-120 0

= Iph ∠-2400

Let IR = IY = IB = IL Where IL indicates rms value of line current. Applying Kirchhoff’s current law,

+ =

= - = Iph ∠0 0 – Iph ∠-240 0

= Iph + j0 + 0.5 Iph – j0.866 Iph

= Iph ∠-30 0

Similarly ,

= - = Iph ∠-30 0

= - = Iph ∠-30 0

Thus in a delta connected three phase system, IL = Iph and line current are 30 0 behind the

respective phase current.Phasor Diagram: (Lagging power factor)

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Fig.3.10

Power:Total power P = 3 Vph Iph cos ǿIn a delta connected three phase system, Vph = VL

Iph = IL/

P = 3 VL IL / 3 cos ǿ

= VL IL cos ǿ

Total reactive power Q = 3Vph Iph sin ǿ

= VL IL sin ǿ

Total apparent power S = 3 Vph Iph

= VL IL

BALANCED Y / AND /Y CONVERSION:

Any balanced star connected system can be completely converted into equivalent delta connected system and vice versa. For a balanced star connected load, Line voltage = VL

Line current = IL

Impedance/ phase = Zy

Vph = VL /

Iph = IL

ZY = Vph/Iph = VL/ IL

For an equivalent delta connected system, the line voltage and current must have the same values as in the star connected system i.e. Line voltage = VL

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Line current = IL

Impedance/phase = Z

Vph = VL

Iph = IL/

Z = Vph / Iph = VL / IL / VL / IL = 3ZY

ZY = Z / 3

3. RELATION BETWEEN POWER IN DELTA AND STAR SYSTEM

Let a balanced load be connected in star having impedance per phase as Zph.For star connected load,

Vph = VL /

Iph = Vph / Zph = VL / Zph

IL = Iph = VL / Zph

Now PY = VL IL cos ǿ

= VL VL / Zph cos ǿ = / Zph cos ǿ

For delta connected load, Vph = VL

Iph = Vph / Zph = VL / Zph

IL = Iph

= VL/Zph

Now P = VL IL cos ǿ

= VL VL / Zph cos ǿ

= 3 Zph cos ǿ

=3 PY

Thus power consumed by a balanced star connected load is one third of that in the case of delta connected load.

SOLVED NUMERICALS:

1. Three equal impedances each 8 + j10 ohms are connected in star. This is further connected to a 440 V, 50 Hz, three phase supply. Calculate the active and reactive power and line and phase currents. (Dec-10, 10marks)

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Data: Zph = 8 + j10 Ω

VL = 440 V

f = 50 Hz For star connected load,Vph = VL/√3 = 440/√3 = 254.03V

Zph = 8+j10 = 12.81 <51.340 Ω

Zph = 12.81 Ω ǿ = 51.340

Iph = Vph/Zph = 250.03/12.81 = 19.83 A

IL = Iph = 19.83 A

P = √3 VL IL cos ǿ = √ 3 440 19.83 cos (51.340)

= 9.44 kW

Q = √3 VL IL sin ǿ = √3 440 19.83 sin (51.340)

= 11.81kVAR. (Ans)

2. 415 V. 50 Hz, three phase voltage is applied to three star connected identical impedances. Each impedance consists of a resistance of 15 Ώ, capacitance of 177 µF and inductance of 0.1 Henry in series. Find (i) phase current (ii) line current (iii) power factor (iv) active power (v) reactive power (vi) total VA.Draw a neat phasor diagram. If the same impedances are connected in delta, find (i) line current (ii) power consumed.

Data: VL = 415 Ω F = 50 Hz

R = 15 Ω C = 177 µF

L = 0.1 HFor star connected load,

Vph = VL/√3 = 415/√3 = 239.6 V

XL = 2πfL =2π 50 0.1 = 31.42 Ω

Xc = 1/2 fC = 1/2π 50 177 10.6 = 17.98 Ω

Zph = R + jXL – jXC = 15 + j31.42 – j17.98 = 15 + j13.44

= 20.14 <41.860 Ω

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Zph = 20.14

ǿ = 41.860

Power factor = cos ǿ = cos (41.860)

= 0.744 (lagging)

Iph = Vph/Zph = 239.6/20.14 = 11.9 A

IL = Iph = 11.9 A

P = √3VL IL cos ǿ

= √3 415 11.9 0.744

= 6.36 kW

Q = √3 VL IL sin ǿ

= √3 415 11.9 sin (41.860)

= 5.71 kVAR

S = √3VL IL

= √3 415 11.9

= 8.55 kVA

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If same impedances are connected in delta,

VL = Vph = 415 V

Zph = 20.14 Ω

Iph = Vph/Zph = 415/10.14 = 2061 A

IL = √3 Iph = √3 20.61 = 35.69 A

P = √3 VL IL cos ǿ

= √3 415 35.69 0.744

= 19.09 kW (Ans)

STAR AND DELTA CONNECTION:1. A circuit has XL=20Ω at 50Hz its resistance being 15Ω.For an applied voltage of 200V at 50Hz,Calculate:

1. P.F. 2. Current 3. Value of shunting capacitance to bring the resultant current into phase with the applied voltage4.resultant curren tin case 3. 10Marks (Dec-10)

2. Three idnetical coils, each having resistance of 15 Ώ and inductance of 0.03 H, are connected in delta across a three phase ,50 Hz ,230 V supply. Calculate the phase current, line current, and total power absorbed. draw phasor diagram.

10Marks (June 2008 )

3. Three non inductive resistances each of 100 ohm connected in star to 3 phase 440V supply. Three ideal equal choke are also connected in delta to the same supply, the reactance of one of the coil being equal to 100 ohm. Calculate i) line current ii) power factor of the system. 10Marks (Dec2007 ) [ , p.f=0.317 ]

4. A balanced three phase load connected in delta draws a power of 10 kW at 440 V at a power factor of 0.6 lead. Find the lead, find the values of the circuit elements and the reactive voltamperes drawn. (May 2004)

8 Marks [2.873 , 4.975 , 18.079 KVAR]

5.Three equal impedances each 8 + j10 ohms are connected in star. This is further connected to a 440 V, 50 Hz, 3 supply. Calculate the active and reactive power and line and phase currents. (Dec-10,Dec. 88)

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10Marks[9.458 KW, 11.806 KVAR, 19.85 A, 19.85 A]

6.A balanced delta connected load of impedance (8 – j6) ohm per phase is connected to a three phase 230 V, 50 Hz supply. Calculate : (i) line current (ii) power factor (iii) reactive power. Draw phasor diagram. (May 94)

10Marks [39.83 A, 0.8 leading, 9.52 KVAR]

7.Three coils, each having resistance of 8 inductance of 0.02 H are connected in (i) star (ii) delta to a 3, 230 V, 50 Hz ac supply. Calculate in both cases, the various line and phase quantities and the total power absorbed. (May 98)

[ 13.05 A, 132.79V, 10.17Ω, 4.088 kW, 39.33 A, 68.12A, 10.17Ω, 37.124kW]

8.Three similar coils A, B, and C are available. Each coil has 9 _ resistance and 12 _ reactance. They are connected in delta to a 3, 440 V, 50 Hz supply. Calculate phase current, line current, power factor, total kVA, active power and reactive power. If these coils are connected in star across same supply, calculate all the above quantities. 10Marks [May 91]

[29.33 A, 50.8 A, 0.6 lagging, 38.72 KVA, 23.23 KW, 30.97 KVAR, 16.94 A, 16.94 A, 0.6 lagging,12.91 KVA, 7.74 KW, 10.33 KVAR]

9.415 V, 50 Hz, three phase voltage is applied to three star connected identical impedances. Each impedance consists of a resistance of 15 _, capacitance of 177 µF and inductance of 0.1 henry in series. Find (i) phase current (ii) line current (iii) power factor (iv) active power (v) reactive power (vi) total VA. Draw a neat phasor diagram. If the same impedances are connected in delta, find (i) line current (ii) power consumed.

10Marks (May 88)

[11.9 A, 11.9 A, 0.744 lagging, 6.36 KW, 5.71 KVAR, 8.55 KVA, 35.69 A, 19.09 KW]

10.Each phase of a delta connected load consists of a 50 mH inductor in series with a parallel combination of 50 Ω resistor and 50 µF capacitor. The load is connected to three phase, 550 V, 800 rad/sec AC supply. Find (i) phase current (ii) line current (iii) power drawn (iv) power factor (v) reactive power (vi) kVA rating of the load. 10Marks (May 95,Dec-10)

[24.59 A, 42.604 A, 18.151 KW, 0.447 lagging, 36.3 KVAR, 40.59 KVA]

11.Calculate the active and reactive components of the current in each phase of a star connected 5000 V, 3, alternator supplying 3000 KW at a pf of 0.8. 8 Marks (May 98)

[250 A, 150A]

12.A three phase, 208 volt generator supplies a total of 1800 W at a line current of 10 A when three identical impedances are arranged in Wye connection across the line terminals of the generator. Compute the resistive and reactive components of each phase impedance. 8 Marks (Dec. 89)

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[6 , 10.39 ]

13.Three similar choke coils are connected in star to a 3 supply. If the line current is 15 A, the total power consumed is 11 kW and the volt-ampere input is 15 kVA, find the line and phase voltages, the VAR input and the reactance and resistance of each coil. If these coils are now connected in delta to same supply, calculate phase and line currents, active and reactive power. 10Marks (Dec. 92)

(Dec. 91, May 97, Jan 2004)

[577.36 V, 333.35 V, 10.19 KVAR, 15.11 , 18.289 , 25.98 A, 45 A, 32.98 KW, 30.59 KVAR]

14.Three identical coils connected in delta to a 440 V, three phase supply take a total power of 50 kW and a line current of 90 A. Find phase current, power factor, total apparent power taken by the coils. (Dec. 92)

8 Marks [57.961 A, 0.7289 lagging, 68.589 KVA]

15.A 3 star connected source feeds 1500 kW at 0.85 power factor lag to a balanced mesh connected load. Calculate the current, its active and reactive components in each phase of the source and the load. The line voltage is 2.2 kV. 8 Marks (Dec. 95)

[ 463.11 A, 393.64 A, 243.95 A]

16.A balanced three phase load connected in delta draws a power of 10.44 kW at 200 V at a power factor of 0.5 lead. Find the values of the circuit elements and the reactive voltamperes drawn. (May 95)

10Marks [0.908 , 27.878 , 13.555 KVAR]

MEASUREMENT OF POWER:

Example: Three coils each with a resistance of 10 Ω and reactance of 10 Ω are connected in star across a three phase, 50 Hz, 400 V supply. Calculate (i) Line current and (ii) Readings on the two wattmeter’s connected to measure the power.

Data: R = 10 Ω XL = 10 Ω

VL = 400 V For star connected load, Vph = VL/√3 = 400/√3 = 230.94 V

Zph = R + jXL = 10 + j10 = 14.14 <450 Ω

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Zph = 14.14 Ω

ǿ = 450

Power factor = cos ǿ

= cos (450) = 0.707 (lagging)

Iph = Vph/Zph = 230.94/14.14 = 16.33 A

IL = Iph = 16.33 A

P = √3VLIL cos ǿ

= √3 400 16.33 0.707

= 7998.83 W W1 – W2 = 7998.83 WAlso

Tan ǿ = √3 W1–W2 / W1+W2

Tan 450 = √3 W1-W2 / 7998.83

W1-W2 = 4618.13 W

Solving equations (1) and (2),

W1 = 6308.48 W

W2 = 1690.35W

1. Three identical coils each having a resistance of 8 and inductance of 0.02 H are connected in (i) star and (ii) Delta across 3, 400 V, 50 Hz supply. Draw neat phasor diagram and calculate the readings of two wattmeters connected to measure power. Also calculate pf of the circuit. 10 Marks (May 2000)

[W1=8.99kW, W2=3.38kW, 0.7866 lag W1=26.979kW, W1=10.144kW, 0.7866 lag]

2. Thre identical coils each having a reactance of 20Ω and resistance of 10Ω are connected in (i) star (ii) delta across a 440 V, 3 phase line. Calculate for each method of connection the line current and readings on each of the two watt meters connected to measure the power. 10 Marks (Jan 2003)

[11.3609A, 4171.79 W. -299.073 W, 34.083 A, 12515.59W, -697.227W]

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3. Two wattmeters connected to measure the total input to a balanced 3 circuit indicate 2000 W and 500 W respectively; the latter reading being obtained by reversing the terminals of current coil. Find the pf of the circuit. 8 Marks (May 96)

[0.3272 lagging]

4. Find the power and power factor of the balanced circuit in which the wattmeter readings are 5 k and 0.5 kW, the latter being obtained after the reversal of current coil terminals of the wattmeter. (May 96)

8 Marks [ 4.5 kW, 0.427 lagging]

5. Three star connected impedances Z1 = ( 20+ j37.7 ) per phase are conncted in parallel with thre delta connected impedances Z2 = ( 30- j59.3 ) per phase. The line voltage is 398 V. Find line current, p.f, power and reactive VA taken by the combination. 10Marks ( Dec 2006 )

6. 3 - phase ,220 V ,50 Hz ,11.2 KW induction motor has full load efficiency 88% and draws a line current of 38 A under full load when connected to 3 hase 220 V supply. Find thwe reading on the two wattmeter connected in the circuit to measure the input to the motor. Determine also the pf at whichh the motor is operating. 8Marks ( June 2007 )

7. Two wattmeters connected to measure the power input to 3 phase circuit, indicate 2500W and 500W respectively, Find the power factor of the circuit

i) when both reading are positive

ii) When later reading is obtained after reversing the connection to the current coil of one instrument.

10Marks (Dec -07,Dec-10) [0.36 lag , 0.65lag]

8. A 3-ph RYB system has effective line voltage 173.2v.Wattmeter in lines R and Y read 301w and 1327w respectively. Find the impedence of the balance star connected load.

10 Marks (Dec-08) [Z==12.45<47.5Ω]

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9. Calculate the total power and reading of two wattmeters connected to measure power in t

hree phase balance load, if the reactive power is 15 KVAR, and load P.F. is 0.8 lagg.

10 Marks (Dec-08) [20KW]

SHORT QUESTIONS:

1. Explain two wattmeter method for power measurement in 3-ph delta connected balance load. Also draw the phasor diagram. 10 marks (Dec-10)

2. Show only two wattmeter are sufficient to measure 3-ph power. Also state advantages of two wattmeter.

10 marks (May-10)

3. Show that the algebraic sum of all potential difference in delta nerwork is zero.(Show graphically).

8 marks (Dec 2007)

4. Prove that for 3 phase balanced delta connected lode line current is times phase

current. Also define power triangle in 3 phase circuits. 10 marks (June 2008)

5.Deduce the relationship between phase and line quantities (voltage, current, power) in the circuit of a three phase delta connected system. Also draw neat diagram

10 marks. (Dec. 98, Dec. 2000, Dec. 2001, July 2002, May-10)

6.Deduce the relationship between phase and line quantities (voltage, current, power) in the circuit of a three phase star connected system. Also draw neat diagrams.

8 marks (Dec. 96, May 98, Dec. 98)

7. Explain merits of two wattmeter method for power measurement, giving circuit diagram and phasor.

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5marks (Jan 2004, June 2008)

8. Explain with phasor diagram of how two wattmeter can be used to measure power in 3phase system. Also

explain the variations in the wattmeter readings with load power factors. 5marks (Jan 2003)

OBJECTIVES QUESTIONS :

1. A delta circuit has each element of value R/2. The equivalent elements of star circuit will be

(a) 2R (b) 6R

(c) 3/2 R (d) R/6

Ans:(--)

2. In case of delta connected circuit, when one resistor is open, power will be

(a) zero (b) reduced to 1/3

(c) reduced by 1/3 (d) unaltered

Ans:(--)

3. For three phase star connected circuit,

(a) Line voltage = phase voltage (b) Line current= phase current

(c) Line current = 3 phase current (d) None of the above is true

Ans:(--)

4. Which of the following is true for a three phase delta connected circuit

(a) Line voltage = phase voltage (b) Line current= phase current

(c) Line current = 1/3 phase voltage (d) Line current = 3 phase current

Outcomes:

1. In some applications single phase ac supply is not suitable; so some load needs a polyphase ac supply. The word polyphase means many, thus a polyphase ac supply is the one which produces many phases simultaneously.

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2. A polyphase system has an advantage of more output, smaller size, more power is transmitted, small cross sectional area of conductors, better power factor, high transmission efficiency etc over the single phase system. Hence they are preferred practically.

3. Secondly the power is dependent on power factor cosΦ alongwith Voltage & Current. So in order to measure the 3 ph active power, we need to use an instrument which will sanse voltage, current as well as the power factor and give the reading directly in watts. The instrument which can do this is a Wattmeter.

Ans:(--)

5. In three phase star connections, line voltage is the same as

(a) Phase voltage (b) 1/3 phase voltage

(c) 3 phase voltage (d) 3 phase current

Ans:(--)

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