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.
Shortcircuit curves
Secondary current I_{2} increases if load resistance is decreased. Where Z_{a} = 0 the transformer has been shortcircuited.
Primary circuit 
U_{1} is applied 

I_{K} flows 
Secondary circuit 
Z_{a} = 0 

U_{2} = 0 
Shortcircuit voltage
The shortcircuited transformer can be replaced by resistor Z_{1} which corresponds to the transformer internal resistor.
Figure 130  Shortcircuited
transformer
1 Shortcircuit current I_{K}
Figure 131 depicts the commensurate duplicate circuit diagram.
Figure 131  Duplicate circuit
diagram for short circuit run
1 Ohmic winding resistance, 2 Scattered reactance (is made up of the scatter flow of the input and output coils), 3 Inner resistance of the transformer (impedance)
During a shortcircuit attempt (Figure 132) the input voltage given a shortcircuited output winding is increased until primary and secondary nominal currents flow. The voltage applied to the input side is then the shortcircuit voltage U_{K}.
Figure 132  Circuitry to determine
shortcircuit losses
1 Short circuit voltage
The shortcircuit voltage is the overall voltage decrease of a transformer during rated loading.
The relative shortcircuit voltage U_{K} in % is determined by the following equation:
_{}
The relative shortcircuit voltage is, on average, 2 to 10% of input rated voltage (U_{1n}) in mains transformers.
Shortcircuit losses (winding losses)
In the shortcircuit experiment (Figure 132) a power meter indicates shortcircuit losses as the primary and secondary rated currents generate winding losses. The iron core is only slightly magnetised by the applied shortcircuit voltage (U_{K} U_{1}).
The winding losses can be metered during the shortcircuit experiment. They are dependent on the load current (P_{VW} = I_{2} R).
In contrast to operational idling, during loading the secondary circuit is closed through an external resistance Z_{a} (Figure 126). Secondary current I_{2} flows. According to the energy conservation law the transformer must also take up commensurate primary power, thus a primary current I_{1} also flows.
Primary circuit 
U_{1} is applied 

I_{1} > I_{0} 
Secondary circuit 
Z_{a} < ¥ 

I_{2} > 0 

U_{2} ¹ U_{20} 
Voltage curve U_{2} = f (I_{2})
As the curve in Figure 133 shows, terminal voltage U_{2} decreases during loading.
Figure 133  Voltage behaviour
during loaded operation U_{2} = f (I)_{2}
1 U_{K} small, 2 U_{K} big
Figure 134 depicts the duplicate circuit diagram for the loaded transformer.
Figure 134  Duplicate circuit for
the loaded transformer with a transformation ratio = 1:1
The duplicate circuit diagram corresponds to a transformer with a transformation ratio _{}
_{}
_{}
As rated current flows the shortcircuit voltage U_{K} decreases at the internal transformer resistance Z_{i} as a result of which the terminal voltage U_{2} declines by the power decrease of the shortcircuit voltage U_{K}.
Transformers with considerable shortcircuit voltage U_{K} have powerful internal resistors, that is to say pronounced voltage changes as load alters.
U_{K} = 2...10% 
minimal voltage losses 

voltagerigid behaviour 
U_{K} = 20...50% 
considerable voltage losses 

voltageflexible behaviour 
Example:
A 220/42 V transformer has a shortcircuit voltage of 10%.
How great is the voltage change between idling and rated current loading?
Solution:
_{}
Output voltages at differing loads
Given differing loads with ohmic, inductive or capacitive external resistance gives rise to the dependence of output voltage on load current as shown in Figure 135.
Figure 135  Secondary terminal
voltage depending on the degree and nature of loading
1 Idling, 2 Rated load
Given capacitive load, the output voltage may even be greater than noload voltage.
The output voltage of a transformer depends on the
 degree of load current I_{2}
 the magnitude of relative shortcircuit voltage
 the nature of the load (ohmic, inductive or capacitive).
Efficiency represents the ratio of power output to power input.
_{}
Efficiency is determined by the separate loss indication.
P_{in} = P_{out} + P_{VFe} + P_{VW}
P_{out} 
rated power of the transformer given 

cos j = 1 (resistive load) 
P_{VW} 
metered winding losses in shortcircuit trial 
P_{VFe} 
metered iron losses in noload trial 
_{}
The following equation applies for determining efficiency for partially inductive or capacitive load:
_{}
Example:
Given:
S_{n} = 100 kVA
P_{VFe} = 570 W
P_{VW} = 2.1 kW
cos j = 0.85
Sought: h
Solution:
_{}