In this blog we will discuss **Ampere-turns relation.** After going through this topic we will get an insight of how the mutual flux sets-up in the core of a transformer and also we will see **why the mutual flux ɸ _{M }remains almost constant?** We will also learn

**why transformer is a**

**constant frequency device?**So, here we go.

At all ordinary loads, the e.m.f. E_{1} induced in the primary by the mutual flux ɸ_{M}, is nearly equal to the primary terminal voltage V_{1} in magnitude. The e.m.f. E_{1} differs from primary terminal voltage V_{1} only by a small magnitude due to the small impedance drop (I_{1}Z_{1}) in the primary winding. Hence, E_{1} and V_{1} are nearly equal. Also, since V_{1} is constant, the induced e.m.f. must also be nearly constant.

Applying KVL in primary winding

V_{1} – I_{1}Z_{1} – E_{1} = 0

∵ I_{1}Z_{1 }is a small impedance drop.

Now,
since E_{1} is nearly constant the mutual flux ɸ_{M} also must
be nearly constant at all normal loads and therefore the m.m.f. (N_{1}I_{1})
producing mutual flux ɸ_{M} as well as the iron losses must be nearly
constant. Thus the exciting current I_{o} must be constant at all
normal loads on the transformer. Also I_{o} is small in magnitude,
generally being 2 to 6% of the rated primary current.

The exciting current is of small magnitude and generally differs considerably in phase from the total primary current. Therefore, it is normally neglected in comparison with the total primary current.

Now,
if secondary is loaded then current I_{2} starts flowing in the
secondary and due to this current I_{2} demagnetizing flux ɸ_{2}
is produced by the secondary ampere-turns (N_{2}I_{2}) and this
flux ɸ_{2 }opposes the mutual flux ɸ_{M} according to the **Lenz’s
law. **Due to the opposition offered by demagnetizing flux ɸ_{2 }the
main flux ɸ_{M} starts decreasing which results in decrease of the
induced e.m.f. E_{1}.

But as we know that e.m.f. E_{1} is almost constant and to keep this constant, the mutual flux ɸ_{M} must be restored to the original value. To achieve this, the primary winding draws an additional current **I _{1}’** from the a.c. mains, such that

**the primary ampere turns (N**

_{1}I_{1}) is equal to the secondary ampere-turns (N_{2}I_{2}).Therefore, under loading conditions,

N_{1}I_{1} = N_{2}I_{2}

∵\( \frac { I_{ 1 } }{ I_{ 2 } } =\frac { N_{ 2 } }{ N_{ 1 } } \)

The above equations are known as **“Ampere turns Relation”.**

If the transformer losses be neglected and unity power factor be assumed, then

\(\frac { E_{ 1 } }{ E_{ 2 } } =\frac { V_{ 1 } }{ V_{ 2 } } =\frac { N_{ 1 } }{ N_{ 2 } } =\frac { I_{ 2 } }{ I_{ 1 } } \)i.e. \(V_{ 1 }I_{ 1 }=V_{ 2 }I_{ 2 } \)

**Explanation**:-

Referring
to the figure given below, **V _{1}** is the supply voltage source
connected to the primary winding of the transformer.

**I**current flows through primary winding and due to magnetizing ampere-turns

_{1}**(N**mutual flux

_{1}I_{1})**ɸ**sets up in the core of the transformer which in turns induce e.m.f.

_{M}**E**in primary winding.

_{1}Now,
let the secondary terminal be loaded and the load current be **I _{2}.**
Due to secondary ampere-turns

**(N**a flux

_{2}I_{2})**ɸ**is produced and according to Lenz’s law, this flux

_{2}**ɸ**oppose the main flux

_{2}**ɸ**This results in decrease of the main flux

_{M}.**ɸ**and hence primary induced e.m.f.

_{M}**E**also decreases.

_{1}As
we know that primary terminal voltage **V _{1}** and induced e.m.f.

**E**is almost equal. But due to decrease in

_{1}**E**a potential difference between

_{1}**V**and

_{1}**E**occurs which results in drawing of additional current

_{1}**I**through primary winding and this current

_{1}’**I**produces an additional flux

_{1}’**ɸ**by primary ampere-turns

_{1}**(N**in same direction of the main flux. This additional flux

_{1}I_{1}’)**ɸ**is equal and opposite to the flux

_{1}**ɸ**Hence flux

_{2}.**ɸ**compensate the flux

_{1 }**ɸ**and keep the main flux

_{2}**ɸ**constant.

_{M}The additional current **I _{1}’ **in primary winding is given by

## Why the mutual flux (ɸ_{M}) of a transformer remains constant ?

As
shown in the figure three fluxes **(****ɸ _{M },ɸ_{1}
and ɸ_{2}**

**)**are circulating in the core of transformer in which

**ɸ**and

_{M }**ɸ**are circulating in the same direction and

_{1 }**ɸ**is circulating opposite direction. As flux

_{2 }**ɸ**and

_{1}**ɸ**are equal and opposite they cancel out each other and

_{2 }**that’s why the main or mutual flux**

**ɸ**

_{M }**of the transformer remains almost constant.**

**Why Transformer is a constant frequency device ?**

As
the supply frequency to the transformer causes production of alternating flux **ɸ _{M} **in the core. So,
the frequency of supply and flux

**ɸ**is same. This flux

_{M }**ɸ**links both primary as well as secondary winding which induce e.m.f.

_{M}**E**and

_{1}**E**in the primary and secondary windings respectively.

_{2}As
e.m.f. E_{1} and E_{2 }are induced by the same flux **ɸ _{M }**therefore the frequency
of primary and secondary induced e.m.f. (E

_{1}and E

_{2}) is same as the frequency of flux

**ɸ**Hence the frequency of flux

_{M}.**ɸ**and mutually induced e.m.f. in the secondary winding

_{M}**(E**remains same as that of primary winding supply.

_{2})The other reason for same frequency on both sides
is that since *transformer is a static
device so there is no relative motion between primary and secondary windings
therefore the frequency of the induced voltage in secondary is same as the
frequency of the applied voltage to the primary.*

Hence, we can say that **transformer is a constant frequency device because whatever is the supply frequency at input of the transformer, same is the value of frequency at the output of the transformer. **

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Also, read

Transformer and its working Principle

Ideal transformer with phasor diagram