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Power electronic control in electrical systems 389

(ii) From supply diagram, I L ˆ I ph in all three lines/phases. From the voltage
p

diagram, V RY ˆ 2cos 30 V RN ˆ 3V RN ; other lines/phases likewise.
From load diagram, I R ˆ I 1 I 3 and if balanced, I L ˆ 2 cos 30
p
I ph ˆ 3I ph .
j0
(iii) 6000 ‡ j0 ˆ V RY I ˆ 415e I
RY RY so I RY ˆ 14:458A ˆ I 1
1
(4500/0:8)e j cos (0:8) ˆ 415e j120 I so I YB ˆ 13:554e j156:87 A ˆ 12:464
YB
j5:324A ˆ I 2
1
(2700/0:5)e j cos (0:5) ˆ 415e j120 I so I BR ˆ 13:012A ˆ I 3
BR
I R ˆ I 1 I 3 ˆ 27:470A
I Y ˆ I 2 I 1 ˆ 26:922 j5:324A ˆ 27:444e j168:813 A
I B ˆ I 3 I 2 ˆ 0:548 j5:324A ˆ 5:352e j95:877 A
Average line current/Average phase current ˆ (27:470 ‡ 27:444 ‡ 5:352)/
(14:458 ‡ 13:554 ‡ 13:012) ˆ 1:469.
17. (i) Draw a circuit diagram showing the connection of two wattmeters to measure
the power in a three-wire supply to a three-phase load.
(ii) Using a suitable phasor diagram for the two-wattmeter connection, prove that
the power-factor angle f of a balanced load can be determined from the equation

p  P 1 P 2
tan f ˆ 3
P 1 ‡ P 2
where P 1 and P 2 are the readings on the individual wattmeters.
(iii) A three-phase AC motor draws balanced currents from a three-phase supply.
Its power factor is 0.85 lagging. The output power of the motor is 9.7 kW and
the efficiency is 92%. Determine the individual readings P 1 and P 2 of two
wattmeters measuring the input power to the motor in the two-wattmeter
connection.

(i)

Fig. 9.11