Symmetrical Components, Per Unit Representation

1. 1. Three resistances of 10, 15 and 20 ohms are connected in star across a three-phase supply
of 200V per phase as shown in the figure. The supply neutral is earthed while the load
neutral is isolated. Find the currents in each load branch and the voltage of load neutral
above earth. Use the method of symmetrical components.
VAN = 200
VBN = 200a2
VCN = 200a
Let us represent it this way, it will be easier.
Apply KVL to the loop AacC
VAN – 10 Ia +20Ic – VCN = 0
And also apply KVL to the loop AabB
VAN – 10Ia +15Ib – VBN = 0
Now write the currents in symmetrical components and also we know that the zero
sequence components will be zero as the load neutral current is not grounded.And if you
we apply KCL at load neutral we get Ia0 = 0.
Hence we get,
Ia1(20a - 10) + Ia2(20a2 - 10) = 200a – 200
And
Ia1(15a2 – 10 ) + Ia2(15a – 10 ) = 200a2 – 200
Solving the above equation in terms of ‘a’ and in the end substitute ‘a’ value.
Ia1 = 13.8461 - 0.0000i
Ia2 = 2.3077 + 1.3324i
Hence Ia = Ia1+Ia2 = 16.1539 + 1.3323i
Ib = Ib1 + Ib2 = - 9.2302 - 10.6584i
Ic = Ic1 +Ic2 = - 6.9230 + 9.3262i
To get Vn(voltage of the neutral with respect to ground) – start from ground and come to
neutral of load

2. Vn = 200 - 10*Ia
Vn = 38.4614 -13.3234i
Magnitude = 40.7037
2. The voltages at the terminals of a balanced load consisting of three 20 ohm Y-connected
resistors are 200(0◦), 10(255.5◦) and 200(151◦) V. Find the line currents from the
symmetrical components of the line voltages if the neutral of the load is isolated. What
relation exists between the symmetrical components of the line and………..
Solution :
Convert star to delta and you get the value of all resistances which are balanced as 60 ohms
Iab = 200/60 = 10/3 (00)
Ibc = 100(255.50)/60 = 5/3(255.50)
Ica = 200(1510)/60 = 10/3(1510)
Using symmetrical components resolution formula involving A-1
From this we get Iab1 = 2.7(15.10)
Iab2 = 1.03(- 44.530)
Ia1 = Iab1 – Ica1 = 4.67(-150)
Ia2 = Iab2 – Ica2 = 1.784(-14.50)
Ia = Ia1 + Ia2(zero seq. Comp. Currents will not appear in the lines of delta connected or star
connected load with neutral isolated)
Ia = 6.453(-14.870)
Ib1 = a^2Ia1 = 4.67(-1350)
Ib2 = aIb2 = 1.784(105.50)
Ib = Ib1+Ib2 = 4.105(-156.90)
Ic1 = a*Ia1 = 4.67(1050)
Ic2 = a^2 Ia2 = 1.784(-134.50)
Ic = Ic1+ Ic2 = 4.066(127.40)
There the power dissipated in 3 phase 20 ohms resistors is

3. P = Ia^2(20)+ Ib^2(20) + Ic^2(20)
= (6.453^2 + 4.1^2+ 4.066^2)(20)
P = 1500W