Harmonics in Transformers

The no load current or the magnetizing current in case of transformers is not sinusoidal in nature due to presence of non-linearities. Thus the phase magnetizing currents contain third and higher order harmonics in order to produce sinusoidal flux. This is observed in case of bank of three single phase transformers and single unit of three phase shell type transformers as the phases are magnetically separate.
If the phase voltage is sinusoidal then the magnetizing current in each phase can be expressed using fourier series as,
IRY = I1m sinωt + I3m sin(3ωt + Φ3) + I5m sin(5ωt + Φ5)                  .............(I)
Even harmonics are absent and only odd harmonics are present.
Similarly,   IYB = I1m sin(ωt - 120o) + I3m sin{3(ωt - 120o) + Φ3} + I5m sin{5(ωt - 120o) + Φ5} + ..
= I1m sin(ωt - 120o) + I3m sin(3ωt - 360o + Φ3) + I5m sin {5(ωt - 120o) + Φ5}+ ....
...               IYB = I1m sin(ωt - 120o) + I3m sin (3ωt + Φ3)  + I5m sin{5(ωt- 120o) +  Φ5}+ ...      ......(II)
IBR = I1m sin (ωt - 240o) + I3m sin {3(ωt - 240o) + Φ3} + I5m sin{5(ωt - 240o) + Φ5} + ..
...               IBR = I1m sin (ωt - 240o) + I3m sin(3ωt + Φ3) + I5m sin {5(ωt - 240o) + Φ5}  + ...     .....(III)
It can be seen from equation (I), (II) and (III) that the third harmonic current in the three currents are in phase or they are cophasal while other higher order harmonic currents are displaced from each other with some angular displacement.
If the transformer is connected in delta fashion then the line currents can be computed from phase currents as,
ĪRY = ĪYB - ĪBR
= {I1m sinωt + I3m sin(3ωt + Φ3) + ..} - { I1m sin(ωt - 240o) + I3m sin(3ωt + Φ3) + ...}
= I1m {sinωt - sin(ωt - 240o)} + I5m {sin(5ωt + Φ5) - sin{5(ωt - 240o) + Φ5}}+ ...
...               IRY = √3I1m sin(ωt - 30o) - √3I5m sin (5ωt - 30o + Φ5) + ...                        ......(IV)
From equation (IV), It can be seen that the third harmonic is absent in the line current which was present in the corresponding phase currents. Since the third harmonic currents are cophasal, they cancel each other But these third harmonic currents flow in the closed path formed by delta connection.
Key Point  : A delta connection thus provides sinusoidal flux and voltage with absence of third harmonic and all its multiples (triples) in currents. Hence the supply line is free from all triplen harmonics, hence delta connection is used for majority of transformers.
In those cases where it is not possible to have delta connection for primary and secondary winding then tertiary winding with delta connection is provided. This winding carries the circulating third harmonic current so as to provide sinusoidal flux in the core.
In case of star connected transformers we have,
IRN = I1m sinωt + I3m sin(3ωt + Φ3) + ..
IYN = I1m sin(ωt - 120o) +  I3m sin{(3ωt - 120o) + Φ3)} + ...
IBN = I1m sin(ωt - 240o) + I3m sin{3(ωt - 240o) + Φ3} + ...
If IN is current in neutral wire then as per KCL, we have,
IN = IRN + IYN + IBN
If higher harmonics are neglected then,
IN = IRN + IYN + IBN = 3I3m sin(3ωt + Φ3)
Under balance conditions, the current in neutral wire is a third harmonic current having magnitude three times that of current in each phase. This current produces inductive interference with neighbouring communication circuits.
For three phase three wire system, the neutral current is zero.
...                3I3m sin(3ωt + Φ3) = 0
I3m = 0
Key Point : Thus flow of third harmonic current is prohibited in case of 3 phase 3 wire system while in 3 phase 4 wire system third harmonic current flows through neutral wire.
The three phase voltages in star connection are given as,
VRN = V1m sin( ωt + Φ1' ) + V3m sin( 3ωt + Φ3' ) +  ...
VYN = V1m sin( ωt - 120o+ Φ1' ) + V3m sin(3ωt + Φ3') + ....
VBN = sin(ωt - 240o + Φ1') + V3m sin( 3ωt + Φ3' ) +  ...
The third harmonic voltages in the three phases are found to be cophasal i.e. having the same phase.
The third harmonic voltage is absent in supply lines as it cancels out. All triplen harmonics are thus absent.

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