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 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,

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

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.

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

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,

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

• The core losses

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

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

• No-load power factor

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

• Active component of no-load current

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

• Magnetising current

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