The current ratio on a transformer
In previous section we saw that the total no-load current in the primary winding of the
transformer can be given by,
iex  im  ih  e
where, ih+e is responsible for hysteresis and eddy current loss (core loss), and im is responsible
for the mutual flux M and the primary leakage flux LP. This M again depends on the applied
primary voltage, vP(t), minus the induced voltage in primary winding due to the primary
leakage flux, LP.
LP
M
iex


NP
vS
NS


Fig(a) Transformer with secondary open circuited.
LP

MS
MS
LS
M
iex+ip2
iS

vP
NP
NS
vS
ZL


Fig(b) Transformer with a load connected at secondary.
Now suppose that a load is connected across the secondary winding. A secondary winding
current will start to flow depending on:
i) the induced voltage in the secondary winding due to the mutual flux M, that is
produced by the primary magnetization current, im, and
ii) the load impedance at the secondary winding.
According to the Lenz’s law, the polarity of the induced voltage in the secondary winding
will be such that the flux produced by the secondary current will oppose the original mutual
flux M. The flux produced by the secondary current has two components:
i) the main component in the core that opposes the original mutual flux,M produced
by the primary current, and
ii) the leakage flux in the secondary winding.
Now, if we ignore the secondary leakage flux, the contribution to the mutual flux by
secondary current can be give as,
MS 
where, NSiS = mmf of secondary winding;
N S iS

(1)
 = reluctance of the total magnetic path of the
transformer core.
However, to maintain the same total mutual flux in the transformer core and hence the
induced voltage in the primary winding, that is approximately equal to the applied voltage,
vP(t), another current, iP2 will flow in the primary winding, such that the flux produced by this
current will be exactly equal and opposite to MS. So we can write,
MS 
N Pi p 2

(2)
where, NPiP2 = mmf in the primary winding due to iP2 only. The phase of iP2 will be same as
that of secondary current iS.
From equations (1) and (2) we have,
ip2 
N S iS
NP
(3)
So, in loaded condition the total primary current of a transformer can be given as,
iP  iex  iP 2
(4)
In a practical transformer, when load current is high (near to full load condition) the
component iP2 becomes very high compared to the excitation current iex, i.e., iP2 >> iex. In that
condition we can write,
iP  iP 2
(5)
Now, from equations (3) and (5) we can write,
iP N S 1


iS N P a
where, a = turns ratio of the transformer.
(6)