Question

2.60 430) Z. (a + jb) Vos Zb (b + c) D ze liedjes Z. (d-je) Three loads of impedance Za (1.4+j6.1) (phase A), Z (20 + 134.64) (phase B), Zc (0-j10) (phase C) are connected in star and supplied from a 400V symmetrical 3-phase, 4 wire system. NB. there is an star to neutral impedance of ZSN (296.0 +j0.0) due to terminal corrosion in the junction box. Q: Calculate the magnitude of neutral shift voltage Vsn Hint use Millman's theorem).

Please show clear working out, so I understand better

Answer 1

According to Millman's theorem,

"Any number of parallel voltage source can be reduced to one equivalent one". Here the circuit can reduced as shown in the Figure 1.

Vs= (van + b + Ven). Zeq zon.

z by Von Zc S I = y You = 0.16_-77.07 **Van + Von + Ven 46.00254-60° Za 4c =0.1<90° (Here, Cartesian Coordinate converted I to Polae Coordinate) vanta + Von Y6 + Vente 1 Yan=vor vi 40020 ant tot ze Yeqa 64 477.04 + 10260°+402-90 9.6289.7 1=(14.32 + 3 62.37) + (5 + j 8.66) + (0-140) = 19.32+ 31.03. This is the cartesian Coordonate of Pan Ya+Vbny + Venlo VSA , & magnitude of neutral shift voltage = 3.80 L-31,62° = 3.80 v L-31.620