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Quiz4 9-4-2011 Power distribution and utilization A 3-phase, 4-wire distributor supplies a balanced voltage of 400/230 V to a load consisting of 30.4 at p.f. 0-866 lagging for R-phase, 30 A at pf. 0-866 leading for Y phase and 30 A at unity pf. for B phase. The resistance of each line conductor is 0-2 Q. The area of X-section of neutral is half of any line conductor. Calculate the supply end voltage for R phase. The phase sequence is RYB. Solution. The circuit diagram is shown in Fig Since neutral is hal the cross-section, its resistance is 0-4 Q. Considering the load end and taking 1’, as the reference vector, the phase voltages can be written as : Fz = 230 20° volts ; WH =230 4—120° volts : Vy =230 2 120° volts 304 0-866 lead The vector diagram of the cirenit is shown in Tig. 14.20. The linc current J, lags behind V7, by an angle cos? 0-866 = 30°. The current J, leads by 30° and the current /, is in phase with ’. Referring to the vector diagram of Fig.14.20. the line currents can be expressed as : Ip = 30 Z4—30° amperes T, = 30 Z—90° amperes Tz; — 30 Z 120° amperes Current in nentral wire. Ty = Tet+h+Ty Fig, 14.20 = 30 730° +30 7—90° +30 / 120° — 30(0-866 f0-5) 30()+30( 0-5 +7 0-866) = 10-98—j 19-02 Let the supply voltage of phase R to neutral be ER: Then, Be = Ve + Drop in & phase + Drop in neutral (230 +7 0) + 0-2 x 30 4 — 30° + (10-98 —7 19-02) x 0-4 230 + 6 (0-866 — 7 0-5) + 0-4 (10-98 — 7 19-02) = 239-588 —j 10-608 = 239-8 £ —2-54° volts docsity.com 
