Practical Transformer on Load

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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 ₂. 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,

$$\mathrm{N_{1}I'_{2} \: = \: N_{2}I_{2}}$$

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

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 (_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,

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

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

Therefore,

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

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

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 (E₁) 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 ₂ 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,

$$\mathrm{N_{1}I'_{2} \: = \: N_{2}I_{2}}$$

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

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,

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

$$\mathrm{\Rightarrow \: V_{1} \: = \: - E_{1} \: + \: I_{1}Z_{1}}$$

And,

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

$$\mathrm{\Rightarrow \: V_{2} \: = \: E_{2} \: - \: I_{2}Z_{2}}$$

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 (_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,

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

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

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

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

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

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,

$$\mathrm{\cos\varphi_{0} \: = \: 0.3; \:\: \therefore \:\: \varphi_{0} \: = \: 72.54^{\circ}}$$

$$\mathrm{\cos\varphi_{2} \: = \: 0.85; \:\: \therefore \:\: \varphi_{2} \: = \: 31.79^{\circ}}$$

Now, the transformation ratio is,

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

Therefore,

$$\mathrm{I'_{2} \: = \: KI_{2} \: = \: \left(\frac{3}{11}\right) \: \times \: 100 \: = \: 27.27 A}$$

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

$$\mathrm{\theta \: = \: 72.54 \: - \: 31.79 \: = \: 40.75^{\circ}}$$

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

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

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