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# Practical Transformer on No-Load

When the secondary winding of a practical transformer is open circuited, the transformer is said to be on no-load (see the figure). Under this condition, the primary winding will draw a small no-load current I_{0} from the source, which supplies the iron losses and a very small amount of copper loss in the core and primary winding respectively. Thus, the primary no-load current (I_{0}) does not lag the applied voltage V_{1} by 90° but lags it by an angle φ_{0} which is less than 90°.

Therefore,

$$\mathrm{N_{o} − load\:input\: power,\:π_{0} = π_{1}πΌ_{0} cos\varphi_{0}}$$

From the phasor diagram, it can be seen that the no-load primary current (I_{0}) can be resolved into two rectangular components viz. *active component* and *magnetising component*.

## Active Component

The component I_{W} is in phase with the applied voltage V_{1} and is known as *active component* or *iron loss component*. This component of no-load current is responsible for supplying iron loss and a very small primary winding copper loss in the transformer. The I_{W} is given by,

$$\mathrm{πΌ_{π} = πΌ_{0} cos\varphi_{0}}$$

## Magnetising Component

The component I_{m} is lagging behind the applied voltage V_{1} by an angle of 90° and is known as *magnetising component* of the no-load current. The magnetising component is responsible for producing the mutual flux φm in the core of the transformer and is given by,

$$\mathrm{πΌ_{π} = πΌ_{0} sin \varphi_{0}}$$

Therefore, the no-load current I_{0} is the phasor sum of I_{W} and I_{m} i.e.

$$\mathrm{πΌ_{0} = \sqrt{πΌ_{π}^{2} + πΌ_{π}^{2}}}$$

Also, the no-load power factor is given by,

$$\mathrm{cos \varphi_{0} =\frac{πΌ_{π}}{πΌ_{0}}}$$

It should be noted that, in a practical transformer the copper loss in primary winding at no-load is very small and may be neglected. Therefore, the input power at no-load in a practical transformer is equal to the iron losses in the core of the transformer i.e.

$$\mathrm{N_{o} − load\:input\:power,\: π_{0} = Iron \:losses}$$

Also, at no-load, there is no current flowing in the secondary winding. Hence, E_{2} = V_{2}.

## No-Load Equivalent Circuit of Practical Transformer

When the transformer is on no-load, no current flows in the secondary winding. Although, the primary winding draws a small no-load current (I_{0}), which supplies the magnetising current (I_{m}) to produce flux in the core and the current I_{W} to supply the core losses. Therefore, the noload primary current is divided into two components and hence it can be represented by two parallel braches which composed of a parallel circuit R_{0} – X_{m} in parallel with the primary winding.

The resistance R_{0} is called *core-loss resistance* which represents the iron losses (i.e., hysteresis and eddy current losses), thus the current I_{W} flows through the R_{0} branch. The inductive reactance X_{m} is known as magnetising reactance which represents a loss-less coil that produces the magnetic flux in the core, thus the magnetising current I_{m} passes through it.

The core loss resistance (R_{0}) is given by,

$$\mathrm{π _{0} =\frac{π_{1}}{πΌ_{π}}}$$

The magnetising reactance (X_{m}) is given by,

$$\mathrm{π_{π} =\frac{π_{1}}{πΌ_{π}}}$$

It should be noted that, in a practical transformer the current I_{W} is very small as compared to current I_{m}. Therefore, the no-load power factor (cosφ_{0}) of a practical transformer is very small.

## Numerical Example

A 240/2200 V transformer takes a no load current of 5 A and absorbs 200 W. If the resistance of the primary winding is 0.08 Ω. Determine the following −

The core loss

No-load power factor

Active component of no-load current

Magnetising current

## Solution

The power absorbed by the transformer at no-load supplies the iron losses and primary winding copper loss. Thus,

$$\mathrm{primary\:winding\:cu − loss = πΌ_{0}^{2}π _{1} = 5^2 × 0.08 = 2\:W}$$

The core losses

$$\mathrm{Core\:losses = π_{0} − primary\:cu\:loss}$$

$$\mathrm{⇒ Core\:losses = 200 − 2 = 198 \:W}$$

No-load power factor

$$\mathrm{cos \varphi_{0} =\frac{π_{0}}{π_{1}πΌ_{0}}=\frac{200}{240 × 5}= 0.167 (πππππππ)}$$

Active component of no-load current

$$\mathrm{πΌ_{π} = πΌ_{0} cos\varphi_{0} = 5 × 0.167 = 0.835 A}$$

Magnetising current

$$\mathrm{I_{m}=\sqrt{I_0^2-I_W^2}=\sqrt{(5)^{2}-(0.835)^{2}}=4.93\:A}$$

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