Electrical Machines  Basic vocational knowledge (Institut für Berufliche Entwicklung, 144 p.) 
8. Transformer 
8.2. Operational behaviour of a transformer 

A transformer idles where mains voltage U_{1} remains applied to the primary side whilst no consumer is connected to the secondary side (Z_{a}) (Figures 125/126).
Primary circuit 
U_{1} applies 

I_{0} flows (idling current) 
Secondary circuit 
Z_{a} = ¥ 

I_{2} = 0 

U_{2} = U_{20} 
Idling current
The applied voltage U drives the idling current I_{0}. This is needed to establish the magnetic field Iµ. This lags behind the voltage U_{1}.
Figure 127  Indicator image for
idling operation
1 Iron loss current I_{Fe}
The phase position of the idling current I_{0} to voltage U_{1} can be determined according to the circuitry of Figure 128.
Figure 128  Circuitry to determine
idling losses
1 Rated voltage
_{}
The value of idling current I_{0} is between 2 and 5 per cent of idling current in big transformers and up to 15 per cent in smaller transformers.
Noload curve
The idling curve I = f (U_{1}) in Figure 129 indicates that noload current I_{0} increases proportionally to the input voltage U_{1}. Noload current increases markedly over and beyond the input rated speed U_{1n}. It can, moreover, even attain values greater than the rated current.
Figure 129  Idling curve of a
transformer I_{0} = f (U_{1})
Transformers shall not be driven by voltages greater than the rated voltage.
Idling losses (iron losses)
The active power derived from the circuit in Figure 128 can only be transformed into heat in the input winding and iron core as no current flows into the secondary winding during idling. The active power P_{0}, which is converted into heat in the iron core, is made up of eddy current and hyteresis loss.
The following example shows that the iron losses almost always arise during idling.
Example:
The following idling values were measured in a transformer:
U_{1n} = 220 V; I_{0} = 0.5 A; P_{0} = 40 W; R_{1} = 3.
What percentage of winding losses are contained in idling power?
Solution:
P_{0} = P_{VFe} + P_{VW}
_{}P_{VW} = 0.75 W
_{}P_{VFe} = P_{0}  P_{VW} = 40 W  0.75 W = 39.25 W
Thus, the power loss determined during idling is an iron loss.
Iron losses are determined during noload operation and are independent of load.