Practical Transformer on No-Load

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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 I₀ 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 (I₀) does not lag the applied voltage V₁ by 90° but lags it by an angle φ₀ which is less than 90°.

Therefore,

$$\mathrm{N_{o} \: - \: \text{ load input power, } \: P_{0} \: = \: V_{1} \: I_{0} \: \cos \: \varphi_{0}}$$

From the phasor diagram, it can be seen that the no-load primary current (I₀) 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 V₁ 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,

$$\mathrm{I_{W} \: = \: I_{0} \: \cos \: \varphi_{0}}$$

Magnetising Component

The component I_m is lagging behind the applied voltage V₁ 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,

$$\mathrm{I_{m} \: = \: I_{0} \: \sin \: \varphi_{0}}$$

Therefore, the no-load current I₀ is the phasor sum of I_W and I_m i.e.

$$\mathrm{I_{0} \: = \: \sqrt{I_{W}^{2} \: + \: I_{m}^{2}}}$$

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

$$\mathrm{\cos \: \varphi_{0} \: = \: \frac{I_{W}}{I_{0}}}$$

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.

$$\mathrm{N_{o} \: - \: \text{ load input power, } \: P_{0} \: = \: \text{ Iron losses}}$$

Also, at no-load, there is no current flowing in the secondary winding. Hence, E₂ = V₂.

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 (I₀), 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 R₀ - X_m in parallel with the primary winding.

The resistance R₀ 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 R₀ 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 (R₀) is given by,

$$\mathrm{R_{0} \: = \: \frac{V_{1}}{I_{W}}}$$

The magnetising reactance (X_m) is given by,

$$\mathrm{X_{m} \: = \: \frac{V_{1}}{I_{m}}}$$

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 (cosφ₀) 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,

$$\mathrm{\text{primary winding cu - loss } \: = \: I_{0}^{2}R_{1} \: = \: 5^{2} \: \times \: 0.08 \: = \: 2\:W}$$

The Core Losses −

$$\mathrm{\text{Core losses } \: = \: P_{0} \: - \: \text{ primary cu loss}}$$

$$\mathrm{\Rightarrow \: \text{ Core losses } \: = \: 200 \: - \: 2 \: = \: 198 \:W}$$

No-load Power Factor −

$$\mathrm{\cos \: \varphi_{0} \: = \: \frac{P_{0}}{V_{1}I_{0}} \: = \: \frac{200}{240 \: \times \: 5} \: = \: 0.167 \: (lagging)}$$

Active Component of No-load Current −

$$\mathrm{I_{W} \: = \: I_{0} \: \cos \: \varphi_{0} \: = \: 5 \: \times \: 0.167 \: = \: 0.835 A}$$

Magnetising Current −

$$\mathrm{I_{m} \: = \: \sqrt{I_{0}^{2} \: - \: I_{W}^{2}} \: = \: \sqrt{(5)^{2} \: - \: (0.835)^{2}} \: = \:4.93\:A}$$