THREE PHASE SYSTEMS - TUTORIAL No: 1 SOLUTIONS 1. Three loads, each of resistance 50  are connected in star to a 400 V, 3-phase supply. Determine (a) the phase voltage, (b) the phase current and (c) the line current. 400 V, 3-phase supply means that 400 V is the line voltage. (a) For a star connection, VL  3 VP Hence, phase voltage, VP  (b) Phase current, IP  VL 400 = 231 V  3 3 VP 231 = 4 A  R P 50 (c) For a star connection, IP  IL Hence, line current, I L = 4 A 2. If the loads in question 1 are connected in delta to the same supply determine (a) the phase voltage, (b) the phase current and (c) the line current. (a) For a delta connection, VL  VP Since VL = 400 V, then phase voltage, VP = 400 V (b) Phase current, IP  VP 400 =8A  RP 50 (c) For a delta connection, line current, IL  3 IP  3 8 = 13 A 3. Three identical capacitors are connected (a) in star, (b) in delta to a 400 V, 50 Hz, 3-phase supply. If the line current is 12 A determine in each case the capacitance of each of the capacitors. (a) For a star connection, IL  12A  IP VL  3 VP hence, VP  XC  VP 231   19  IP 12 VL 400  = 231 V 3 3 thus, 1  19 2 f C and capacitance, C = 1 2  50 19  = 165 F (b) For a delta connection, IL  3 IP hence IP  IL 12   6 A 3 3 VL  VP = 400 V XC  VP 400   57  IP 6 and capacitance, C = thus, 1  57 2 f C 1 2  50  57  = 55 F 4. A 400 V, 3-phase, 4 wire, star-connected system supplies three resistive loads of 15 kW, 20 kW and 25 kW in the red, yellow and blue phases respectively. Determine the current flowing in each of the four conductors. For a star connected system, VL  3 VP from which, VP  Power, P = VI for a resistive load, hence I  IR  Thus, and IB  VL 400 = 230 V  3 3 P V PR 15000 = 64 A,  VR 230 IY  PY 20000 = 86 A  VY 230 PB 25000 = 108 A  VB 230 The phasor diagram of the three currents is shown in (i) below. Adding phasorially gives diagram (ii) below, where I N is the neutral current. (i) (ii) Total horizontal component = 64 cos 90 + 108 cos 210 + 86 cos 330 = - 18 X  9 and phase angle,   tan 1  L   tan 1    66 4  R  Hence, power factor of load = cos  = cos 66 = 0 (b) Power, P = 3IP 2 R P i. 1 103  3IP 2  4  from which, phase current, I P  1200 = 10 A 3(4) (c) In delta, line current, IL  3 (10) = 17 A (d) Power, P = from which, 3 VL IL cos  i. 1200 = supply voltage, VL  3 VL (17)(0) 1200 = 98 V 3 (17)(0) 8. The input voltage, current and power to a motor is measured as 415 V, 16 A and 6 kW respectively, Determine the power factor of the system. Power, P = 3 VL IL cos  from which, i. 6000 = 3 (415) (16) cos  cos  = power factor of system = 6000 = 0 3 (415)(16) 9. A 440 V, 3-phase a. motor has a power output of 11 kW and operates at a power facror of 0 lagging and with an efficiency of 84%. If the motor is delta connected determine (a) the power input, (b) the line current and (c) the phase current. (a) Efficiency = power output power input from which, (b) Power, P = i. power input = 0 = 11250 power input 11250 = 13393 W or 13 kW 0 3 VL IL cos  hence, line current, IL  (c) In delta, IL  3 IP from which, phase current, IP  P 13393  = 21 A 3 VL cos  3  440  0  IL 21  = 12 A 3 3 10. 8 kW is found by the two-wattmeter method to be the power input to a 3-phase motor. Determine the reading of each wattmeter if the power factor of the system is 0 Working in kilowatts, 8 = P1  P2 (1) If power factor = 0, then cos  = 0 and tan  = tan 31 = 0 Hence, tan  = 0 =  P P   P P  3 1 2   3 1 2   8   P1  P2  P1  P2  from which, and phase angle,  = cos1 0 = 31  0 8 3 = 2 (2) 2 P1  10 Adding equations (1) and (2) gives: P1  from which, 10 = 5 kW 2 and from equation (1), P2  8  5 = 2 kW Thus the readings of the two wattmeters are: 5 kW and 2 kW 11. Three similar coils, each having a resistance of 4  and an inductive reactance of 3  are connected (a) in star and (b) in delta across a 400 V, 3-phase supply. Calculate for each connection the readings on each of two wattmeters connected to measure the power by the twowattmeter method. (a) Star connection: VL  3 VP Hence, VP  IL  IP and VL 400 = 230 V  3 3 Phase impedance, ZP  4  3 = 5  Phase current, IP  VP 230 = 43 A  ZP 5 Total power, P = 3IP 2 R P  3  43   4  = 22 kW 2 If the wattmeter readings are P1 and P2 then: P1 + P2 = 22 X   3  Phase angle,  = tan 1  L   tan 1    40  4   R  and tan 40 86 =  P P   P P  3  1 2   3  1 2  from (1)  22   P1  P2  (1) Phase impedance, ZP  R P 2  XC 2  402  79  89  Phase current, IP  VP VL 415 = 4 A   ZP ZP 89 (since VP  VL in delta) (b) Line current, IL  3 IP  3 (4) = 8 A (c) From the impedance triangle, cos  = Hence, total power dissipated, P = (d) The kVA rating of the load, S = RP 40 = 0  ZP 89 3 VL IL cos   3  4158  0  = 2 kW 3 VL IL  3  4158  = 5 kVA 13. Three 24 F capacitors are connected in star across a 400 V, 50 Hz, 3-phase supply. What value of capacitance must be connected in delta in order to take the same line current? In star, VL  3 VP hence, VP  XC  VL 400 = 230 V  3 3 1 1 = 132   2 f C 2  50   24 106  Hence, IP  VP 230 = 1 A = line current for star connection.  X C 132

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3 Phase Systems Tutorial No 1 Solutions v1 PDF

Module: Electrical Systems (M2J906565)

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THREE PHASE SYSTEMS - TUTORIAL No: 1 SOLUTIONS

1. Three loads, each of resistance 50  are connected in star to a 400 V, 3-phase supply. Determine

(a) the phase voltage, (b) the phase current and (c) the line current.

400 V, 3-phase supply means that 400 V is the line voltage.

(a) For a star connection,

V 3 V

Hence, phase voltage,

V400

V33



= 231 V

(b) Phase current,

V231

IR 50



= 4.62 A



Hence, line current,

= 4.62 A

2. If the loads in question 1 are connected in delta to the same supply determine (a) the phase

voltage, (b) the phase current and (c) the line current.

(a) For a delta connection,

VV

Since

= 400 V, then phase voltage,

= 400 V

(b) Phase current,

V400

IR 50



= 8 A

 

I 3 I 3 8

= 13.86 A

3. Three identical capacitors are connected (a) in star, (b) in delta to a 400 V, 50 Hz, 3-phase

supply. If the line current is 12 A determine in each case the capacitance of each of the

capacitors.

(a) For a star connection,

I 12A I

V 3 V

hence,

V400

V33



= 231 V

V231

X 19.25

I 12

   

thus,

119.25

2 f C 



3 Phase Systems Tutorial No 1 Solutions v1 PDF

Electrical Systems (M2J906565)

Glasgow Caledonian University

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