Exact equivalent circuit of a transformer (with
lagging load) is shown below.
Let the lagging load has impedance angle Ï´2
.
Applying KVL in secondary circuit.
Draw I2R2
parallel to I2 , jI2X2
perpendicular and 900 ahead of I2
.
Add I2 and
jI2X2 phasors to phasor V2 to get E2 , as per equation (1)
above.
Let Ф = Фm
sinwt be the flux linking both
windings. Therefore,
By observing above E1 and E2 equations, we find that E1 and E2 lead flux Ф by 900 ,
therefore, draw Ф phasor 900
behind E2 . Draw E1 along E2 .
Current I0 is made of two components Ic and Im .
As the central part (in exact equivalent circuit
of a transformer) is an ideal transformer, therefore, by
relation
We find Ip to be in phase with I2 . Further resultant of I0
and IP gives I1 .
Apply KVL in primary circuit.
Draw I1R1 parallel to I1
, jI1X1 perpendicular to I1 and 900
ahead of it and add these to E1 to get V1 as per equation
(2).
This is the complete phasor diagram for a lagging power factor load. Exactly similarly, phasor diagram for leading power factor load
can be drawn by taking I2 ahead of V2 i.e Ï´2 in opposite direction.
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