# Ideal power transformer

## Session 10

Contents

10.1 E.M.F. Equation of a transformer:

10.2 Wave form representation of flux induced E.M.F. and applied voltage:

10.3 Ideal transformer on no load:

Aim

To discuss and obtain the properties of an Ideal power transformer.

Specific objectives:

At the end of this lesson you should be able to –

- Derive expressions for the induced e.m.f. on both windings of an ideal power transformer.
- Define the properties of an ideal power transformer.
- Draw the phasor diagram of an ideal power transformer with load and without load.

Introduction:

In your last lesson, you have identified the power transformers and their necessities. With the information that you have got about transformers, if I define a transformer in the following manner, you would agree with me.

“A transformer is a static electromagnetic device designed for the transformation of one alternating current system into another one of the same frequency but with other characteristics voltage and current in particular.”

You also have studied the principle of operation of an ideal power transformer and obtained one of the basic properties.

(i.e.) For the arrangement shown in Fig 10.1. You will have,

V_{1} / V_{2
=} E_{1} / E_{2} = N_{1}/N_{2}, Where N_{1}
and N_{2} are the number of turns in P and S coils respectively.

Now we will discuss it further and obtain the other properties also.

## 10.1 E.M.F. Equation of a transformer:

Look at Fig.10.1. The coil P is supplied by an alternating source. Hence an alternating flux will be set up in the core.

Let us say that
this flux is donated by f = f_{m} sin wt. As you know f_{m}_{ }is the peak flux
and w is the angular frequency._{}

As we found earlier, this flux will link the coil P and coil S and hence an e.m.f. will be induced. Consider the coil P; by Farrady’s law and Lenz’s Law. Induced e.m.f. per turn (say e,)

e = -df/dt

= -d (f_{m} sin wt) / dt

= -f_{m }w cos wt

= -f_{m }w sin (p/2 - wt)

= f_{m }w sin (wt-p/2)

If f_{m}.w = e_{m}, then

e_{1
}= e_{m} sin (wt-p/2)

If coil P has N_{1} turns, then the total
e.m.f. induced = N_{1}e_{1}

=
N_{1}.e_{m} sin (wt-p/2)

As you know the r.m.s. value of this induced e.m.f. can be obtained by dividing the peak value by.

Let E_{1}
be the r.m.s. value of the induced e.m.f. in coil P.

but w = 2pf

(i.e.) E_{1}=
4.44 f_{m} f N_{1}
volts

Question:

Could you find an expression for the induced e.m.f. in coil S?

Answer:

You might get the expression easily if you remember your work under section 9.1 of session.

If the whole of the flux produced by coil P passes through coil S, (i.e) if there is no flux leakage, the e.m.f. induced per turn will be equal to that of the coil P.

(i.e.) Induced e.m.f at coil S per turn (say e_{2})

= -df/dt

=
-d (f_{m}
sin wt) / dt

e_{2} =
f_{m .}w sin (wt-p/2)

_{ }= e_{m} sin (wt-p/2)

If N_{2} is the number of turns in coil S,
then the total e.m.f. induced will be N_{2}.e_{2}.

(i.e.) Total e.m.f. induced in coil S = N_{2}e_{m}
sin (wt-p/2)

Let the r.m.s. value of the e.m.f is E_{2}.

= 4.44 f_{m} f N_{2}

For coil P,

Induced
e.m.f. = E_{1} = 4.44 f_{m} f N_{1}

For coil S,

Induced
e.m.f. = E_{2} = 4.44 f_{m} f N_{2}

Therefore E_{1} / E_{2} = N_{1}
/ N_{2}

If the winding resistance of coil P and S are negligible, then the voltage appearing across the terminals of coil S is the same as its induced e.m.f.

(i.e.) E_{2}
= V2

Similarly if the
resistance of coil P is also negligible, the supply voltage V_{1} is
equal in magnitude to the induced e.m.f. in coil P and they will oppose each
other.

(i.e) V_{1}
= E1_{}

Therefore

E_{1} / E_{2} = V_{1} / V_{2
=} N_{1 }/ N_{2}

N_{1 }/ N_{2} = a, where ‘**a’**
is known as the transformation ratio.

(i.e.) when the
voltage V_{1} is applied across the coil P, the coil S gives the
supply, at a potential V_{2} to the load.

The coil P is
known as the __primary winding__ and coil S is known as __secondary winding__.
This action is known as transformer action.

## 10.2 Wave form representation of flux induced E.M.F. and applied voltage:

You have studies
in the previous section that when an alternating flux, f = f_{m}
sin wt is set up in the core, the induced
e.m.f. per turn will be e_{m} sin(wt-p/2).

Question

VV

Could you draw the forms of flux-induced e.m.f. and the applied voltage for an ideal transformer, on the same diagram?Answer:

The induced e.m.f. should lag the flux by
90^{0} as indicated by the expressions.

By Lenz’s law,
the applied voltage should be opposite to the induced e.m.f. Therefore it
should lead the flux by 90^{0}

Question:

Draw the phasor representations of flux, induced e.m.f. and the applied voltage.

Answer:

As you found in the last section, the induced e.m.f. lags the flux in the core.

Take the flux as the reference phasor.

(i.e) __f__ = fÐ0^{0}

Since the
induced e.m.f. lags by 90^{0}

__E___{1} = E_{1 }Ð-90^{0}

and __E___{2}
= E_{2 }Ð-90^{0}

By lenz’s law the applied voltage is opposed by the induced e.m.f.

Therefore __V___{1}
= V_{1}Ð90^{0}

When the secondary winding resistance is negligible,

__E___{2} = __V___{2} = V_{2}Ð-90^{0}

## 10.3 Ideal transformer on no load:

Let I_{0} be the primary current when the
secondary is open circuited.

N_{1} – number of turns in the primary

N_{2} – number of turns in the secondary

If you could remember your magnetic circuit theory,
you will be able to get an expression
for the no-load current I_{0}. It may be written that

f = N_{1}I_{0 }/ S_{1}

Where S_{1 } is the reluctance of the magnetic circuit.

(i.e) I_{0} = S_{1}f / N_{1}

If the instantaneous value of I_{0} is i_{0}
then

I_{0} = (S_{1}/N_{1}) f_{m} sinwt

This will be in-phase with the flux.

Therefore the phasor diagram will be as follows:

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## Comments 2 comments

GOOD...........

In a laoded tranformer the I(s) is the seconday current the oppsite current I(s)' flows in primary to canel the effect, what does this mean??