Ampere turns relation

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. E₁ induced in the primary by the mutual flux ɸ_M, is nearly equal to the primary terminal voltage V₁ in magnitude. The e.m.f. E₁ differs from primary terminal voltage V₁ only by a small magnitude due to the small impedance drop (I₁Z₁) in the primary winding. Hence, E₁ and V₁ are nearly equal. Also, since V₁ is constant, the induced e.m.f. must also be nearly constant.

Applying KVL in primary winding

           V₁ – I₁Z₁ – E₁ = 0

∵ I₁Z₁ is a small impedance drop.

\({ V }_{ 1 }\)\(\approx\)\({ I }_{ 1 }\)

Now, since E₁ is nearly constant the mutual flux ɸ_M also must be nearly constant at all normal loads and therefore the m.m.f. (N₁I₁) 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 I₂ starts flowing in the secondary and due to this current I₂ demagnetizing flux ɸ₂ is produced by the secondary ampere-turns (N₂I₂) and this flux ɸ₂ opposes the mutual flux ɸ_M according to the Lenz’s law. Due to the opposition offered by demagnetizing flux ɸ₂ the main flux ɸ_M starts decreasing which results in decrease of the induced e.m.f. E₁.

But as we know that e.m.f. E₁ 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 I₁’ from the a.c. mains, such that the primary ampere turns (N₁I₁) is equal to the secondary ampere-turns (N₂I₂).

Therefore, under loading conditions,

N₁I₁ = N₂I₂

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

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, V₁ is the supply voltage source connected to the primary winding of the transformer. I₁ current flows through primary winding and due to magnetizing ampere-turns (N₁I₁) mutual flux ɸ_M sets up in the core of the transformer which in turns induce e.m.f. E₁ in primary winding.

Now, let the secondary terminal be loaded and the load current be I₂. Due to secondary ampere-turns (N₂I₂) a flux ɸ₂ is produced and according to Lenz’s law, this flux ɸ₂ oppose the main flux ɸ_M. This results in decrease of the main flux ɸ_M and hence primary induced e.m.f. E₁ also decreases.

As we know that primary terminal voltage V₁ and induced e.m.f. E₁ is almost equal. But due to decrease in E₁ a potential difference between V₁ and E₁ occurs which results in drawing of additional current I₁’ through primary winding and this current I₁’ produces an additional flux ɸ₁by primary ampere-turns (N₁I₁’) in same direction of the main flux. This additional flux ɸ₁ is equal and opposite to the flux ɸ₂. Hence flux ɸ₁ compensate the flux ɸ₂ and keep the main flux ɸ_M constant.

The additional current I₁’ in primary winding is given by

\(I_{ 1 }^{ ‘ }=(V_{ 1 }-E_{ 1 })/Z_{ 1 }\)

Why the mutual flux (ɸ_M) of a transformer remains constant ?

As shown in the figure three fluxes (ɸ_M ,ɸ₁ and ɸ₂) are circulating in the core of transformer in which ɸ_M and ɸ₁ are circulating in the same direction and ɸ₂ is circulating opposite direction. As flux ɸ₁ and ɸ₂ are equal and opposite they cancel out each other and 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_M is same. This flux ɸ_M links both primary as well as secondary winding which induce e.m.f. E₁ and E₂ in the primary and secondary windings respectively.

As e.m.f. E₁ and E₂ are induced by the same flux ɸ_M therefore the frequency of primary and secondary induced e.m.f. (E₁ and E₂) is same as the frequency of flux ɸ_M. Hence the frequency of flux ɸ_M and mutually induced e.m.f. in the secondary winding (E₂) remains same as that of primary winding supply.

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

Emf equation of transformer

Ideal transformer with phasor diagram