Practical Transformer on Load

When a load Impedance is connected across the secondary winding of the practical transformer, then the transformer is said to be loaded and draws a load which flows through the secondary winding and the load.

We shall consider following two cases for analysing the practical transformer −

Case 1 – When the transformer is assumed to have no winding resistance and leakage flux

The figure shows a practical transformer with the assumption that the winding resistances and the leakage reactances are neglected. With this assumption,

$$\mathrm{𝑉_{1} = 𝐸_{1}\: and \:V_{2} = 𝐸_{2}}$$

Consider an inductive load is connected across the secondary winding which causes the secondary current I₂ to lag the secondary voltage V₂ by an angle ϕ₂. In this case, the total primary current I₁ must fulfil two requirements −

• First, it must supply the no-load current I₀ to produce the iron losses and the magnetic flux in core of the transformer.
• Secondly, it must supply a current I’₂ to neutralise the demagnetising effect of secondary current I₂.

The magnitude of current I’₂ is given by,

$$\mathrm{𝑁_{1}I'_{2}= 𝑁_{2}I_{2}}$$

$$\mathrm{⇒ I'_{2}=\frac{𝑁_{2}}{𝑁_{1}}I_{2} = 𝐾I_{2}}$$

Therefore, the total primary current drawn by a practical transformer under loaded condition is given the phasor sum of I’₂ and I₀ i.e.

$$\mathrm{𝑰_{𝟏} = 𝑰'_{𝟐}+ 𝑰_{𝟎}}$$

Where,

$$\mathrm{𝐼’_{2} = − 𝐾I_{2}}$$

The negative sign shows that the current I’₂ is 180° out of phase with current I₂.

The phasor diagram shows that both the emfs E₁ and E₂ lag behind the mutual flux (ϕ_m) by 90°. The current I’₂ denotes the portion of primary current which neutralise the demagnetising effect of the secondary current (I₂). Therefore, the current I_’2 must be in anti-phase with I₂. The phasor I₀ represents the no-load current of the transformer. The phasor sum of I’₂ and I₀ represents the total primary current (I₁). Hence,

$$\mathrm{primary\:power\: factor = cos \varphi_{1}}$$

$$\mathrm{Secondary\: power\: factor = cos \varphi_{2}}$$

Therefore,

$$\mathrm{Input\: power,𝑃_{1} = 𝑉_{1}𝐼_{1} cos \varphi_{1}}$$

$$\mathrm{Output \:power, 𝑃_{2} = V_{2}I_{2} cos \varphi_{2}}$$

Case 2 – When the transformer has winding resistance and leakage flux

The figure shows a practical transformer having winding resistances and leakage reactances. This is the actual case that exits in a practical transformer. Here, some of the applied voltage is dropped in the primary winding resistance R₁ and leakage reactance X₁, thus the primary EMF (E ₁) will be less than the applied voltage V₁.

similarly, there being a voltage drop in the secondary winding resistance R₂ and the leakage reactance X₂ so that the voltage across the secondary winding terminals V₂ will be less than the secondary EMF (E₂).

Now, consider an inductive load connected across the secondary winding of the transformer which causes the secondary current I₂ to lag behind the secondary voltage V₂ by an angle ϕ₂ and the total primary current (I₁) must fulfil two requirements −

• First, it must supply the no-load current I₀ to produce the iron losses and the magnetic flux in core of the transformer.
• Secondly, it must supply a current I’₂ to neutralise the demagnetising effect of secondary current I₂.

The magnitude of current I'₂ is given by,

$$\mathrm{𝑁_{1}I'_{2}= 𝑁_{2}I_{2}}$$

$$\mathrm{⇒ I'_{2} =\frac{𝑁_{2}}{𝑁_{1}}I_{2} = 𝐾I_{2}}$$

Therefore, the total primary current drawn by a practical transformer under loaded condition is given the phasor sum of I’2 and I₀ i.e.

$$\mathrm{𝑰_{𝟏} = 𝑰'_{𝟐} + 𝑰_{𝟎}}$$

Where,

$$\mathrm{𝐼’_{2} = − 𝐾I_{2}}$$

The negative sign shows that the current I’₂ is 180° out of phase with current I₂.

Now, by applying KVL in primary loop and secondary loop, we get the applied primary voltage (V₁) and secondary terminal voltage (V₂) as,

$$\mathrm{𝑽_{𝟏} = −𝑬_{𝟏} + 𝑰_{𝟏}(𝑹_{𝟏} + 𝒋𝑿_{𝟏})}$$

$$\mathrm{⇒ 𝑽_{𝟏} = −𝑬_{𝟏} + 𝑰_𝟏𝒁_𝟏}$$

And,

$$\mathrm{𝑽_{𝟐} = 𝑬_{𝟐} − 𝑰_{𝟐}(𝑹_{𝟐} + 𝒋𝑿_{𝟐})}$$

$$\mathrm{⇒ 𝑽_{𝟐} = 𝑬_{𝟐} − 𝑰_𝟐𝒁_𝟐}$$

The bold letters shows the phasor sum.

From the phasor diagram, it can be seen that both the EMFs E₁ and E₂ lag behind the mutual flux (ϕ_m) by 90°. The current I’₂ represents the primary current to counteract the demagnetising effect of secondary current (I₂) which in anti-phase with I₂. The current I₀ is the no-load current of the transformer. Thus, the total primary current (I₁) is obtained by the phasor sum of I’₂ and I₀.

Also, the primary voltage (V₁) is obtained by adding (phasor sum) the drop I₁R₁ and I₁X₁ to the counter EMF (– E₁). The secondary terminal voltage V₂ is obtained by subtracting (phasor difference) I₂R₂ and I₂X₂ from EMF (E₂).

The input and output power factors are given by,

$$\mathrm{Input\:power \:factor\: = cos\varphi_1}$$

$$\mathrm{Output\: power\: factor\: = cos\varphi_2}$$

Also, the input and output powers of the transformer is given by,

$$\mathrm{Input\: power,\:𝑃1 = 𝑉_{1}𝐼_{1} cos \varphi_1}$$

$$\mathrm{Output \:power, \:𝑃2 = V_{2}I_{2} cos \varphi_2}$$

Numerical Example

A 440/120 V single phase transformer takes no-load current of 6 A at 0.3 lagging power factor. If the secondary winding supplies a current of 100 A at a power factor 0.85 lagging. Determine the current taken by the primary winding.

Solution

Here, the primary current I₁ is given by phasor sum of I’₂ and I₀, thus,

$$\mathrm{cos\varphi_0 = 0.3;\:\: \therefore\:\: \varphi_0 = 72.54°}$$

$$\mathrm{cos\varphi_2 = 0.85;\:\: \therefore \:\:\varphi_2 = 31.79°}$$

Now, the transformation ratio is,

$$\mathrm{𝐾 =\frac{V_{2}}{𝑉_{1}}=\frac{120}{440} =\frac{3}{11}}$$

Therefore,

$$\mathrm{I'_{2}= 𝐾I_{2} = (\frac{3}{11}) × 100 = 27.27 A}$$

Refer the phasor diagram, the angle between I’₂ and I₀ is

$$\mathrm{\theta = 72.54 − 31.79 = 40.75°}$$

Now, using the parallelogram law of vector addition, the primary current is

$$\mathrm{I_{1} = \sqrt{(I'_{2})^2 + (𝐼_{0})^2 + 2I_{0}I'_{2}cos \theta}}$$

$$\mathrm{⇒ 𝐼_{1} = \sqrt{(27.27)^2 + (6)^2 + (2 × 6 × 27.27 × cos 40.75)} = 32.05 A}$$