Electrical Machine - IES Academy

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Transformer
Chapter-1
turns and it is seen to oppose V 1 . Since primary winding resistance is negligible, e 1 , at
every instant must be equal and opposite to V 1 . Hence, V 1 = – E 1.
dφ
The emf induced in the secondary e2 =− N2 =− N2ωφ
maxcosωt
dt
Rms valve of emf E2 is
E2max 2 fN2 max
E π φ
2
= = = 4.44fφ
maxN2
volts
2 2
E1 N1
Hence, = = 4.44f
φ m
i.e. emf per
E2 N2
turn in primary = emf per turn in the
secondary.
The phasor diagram of an ideal transformer under no load conditions is as below:
Emfs E 1 and E 2 lag the mutual flux that induces them by 90° The applied voltage V 1
leads the flux by 90°.
4. If secondary circuit is completed, it will produce a secondary current (load current). This
load current produces a demagnetizing mmf (Lenz’s Law). The direction of secondary
current I 2 should be such that the secondary mmf F 2 (= I 2 N 2 ) is opposite to mutual flux
in the core. For F 2 to be directed against , the current I 2 must leave the terminal n, pass
through the load and enter the terminal m. The secondary winding behaves like a voltage
source, therefore, terminal n must be treated as positive and terminal m as negative. If
secondary winding is wound in a manner opposite to that shown in the Figure, terminal
m would be positive with respect to terminal n. This shows that polarity markings of
the windings in transformers depend upon the manner in which the windings
are wound around the legs with respect to each other.
In order to maintain constant mutual flux,
the primary current will increase and produce
additional magnetizing mmf such that the net
mmf and flux in the core is same as in the
case of no-load. This increased primary
current I1' is called load component of primary
current.
I1' N1 = I 2 N 2
The phasor diagram of an ideal transformer is
shown figure.
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