Lecture 4

Transformer on Load

Fig. 32. 17(a) Transformer without load

It has already seen that at no load condition only current I0 flows in the circuit as shown in Fig. 32.17(a). This current sets up the flux .

When the secondary is loaded, the secondary current I2 is set up.

The magnitude and phase of I2 with respect to V2 determined by the characteristics of the load.

• The secondary current sets up its own mmf (=N2I2) and hence its on flux 2 which is in opposition to the main primary flux which is due to I0 as shown in Fig. 32.17(b).

• The secondary ampere-turns N2I2 are known as demagnetizing amp-turns.

• The opposing secondary flux 2 weakens the primary flux , hence primary back emf E1 tends to be reduced.

Fig. 32.17 (b) Transformer on load

For a moment V1 gains upper hand over E1 and hence causes more current to flow in primary.

Let the additional primary current be I2’ as shown in Fig. 32.17(c).

Fig. 32.17(c)

It is known as load component of primary current. This current is anti-phase with I2 .

The additional primary mmf N1I2’ sets up its own flux 2’, which is in

opposition to 2 (but is in the same direction as ) and is equal to it in magnitude.

Hence the two cancel each other out as shown in Fig. 32.17(d).

• So, we find that the magnetic effects of secondary current I2 are immediately neutralized by the additional primary current I2

’, which is brought into existence exactly at the same instant as I2.

• Hence, whatever the load conditions, the net flux passing through the core is approximately the same as at no-load.

Fig. 32.17(d)

Fig. 32.17

An important deduction is that due to the constancy of core flux at all loads; the core loss is also practically the same under all load conditions.

When transformer is on load, the primary winding has two currents in it; one is I0 and the other I2

’ is which is anti-phase with I2 and K times in magnitude (i.e. I2

’ = I2K).

The total primary current is the vector sum of I0 and I2

’.

Quick Quiz

1.

2. In a transformer (a) All turns are equally insulated (b) The end turns are more strongly insulated (c) The end turns are closely wound (d) The end turns are widely separated3. Select the correct statement: (b) emf per turn of both the windings are equal(b) emf per turn in HV winding is more than the emf per turn in LV

winding (c) emf per turn in HV winding is less than the emf per turn in LV

winding4. The flux involved in the emf equation of a transformer has (c) rms value (b) Average value (c) Total value (d) Maximum value

5. Transformer changes the value of (a) Power (b) Frequency (c) Voltage (d) Current

6. Draw the vector diagram of no load condition.

In Fig. 32.8 are shown the vector diagrams for a load transformer when load is non-inductive and when it is inductive.Voltage transformation ratio of unity is assumed so that primary vectors are equal to the secondary vectors.

With reference to Fig. 30.18(a), I2 is secondary current in phase with E2.

It causes primary current I2’ which is anti-phase with it and equal to it in magnitude.

Total primary current I1 is the vector sum of I0 and I2’ and lags behind V1 by an angle

1.

Here, I2 lags E2 (actually V2) by 2.

Current I2’ is again in anti-phase with I2

and equal to it in magnitude.

As before, I1 is the vector sum of I2’ and I0

and lags behind V1 by 1.

In Fig. 30.18(b) vectors are drawn for an inductive load.

In Fig. 32.8 are shown the vector diagrams for a load transformer when load is non-inductive and when it is inductive.Voltage transformation ratio of unity is assumed so that primary vectors are equal to the secondary vectors.

With reference to Fig. 30.18(a), I2 is secondary current in phase with E2.

It causes primary current I2’ which is anti-phase with it and equal to it in magnitude.

Total primary current I1 is the vector sum of I0 and I2’ and lags behind V1 by an angle

1.

Here, I2 lags E2 (actually V2) by 2.

Current I2’ is again in anti-phase with I2

and equal to it in magnitude.

As before, I1 is the vector sum of I2’ and I0

and lags behind V1 by 1.

In Fig. 30.18(b) vectors are drawn for an inductive load.