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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 −

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

$$\mathrm{đ_{1} = đ¸_{1}\: and \:V_{2} = đ¸_{2}}$$

Consider an inductive load is connected across the secondary winding which causes the secondary current I_{2} to lag the secondary voltage V_{2} by an angle Ī_{2}. In this case, the total primary current I_{1} must fulfil two requirements −

- First, it must supply the no-load current I
_{0}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 I_{2}.

The magnitude of current I’_{2} is given by,

$$\mathrm{đ_{1}I'_{2}= đ_{2}I_{2}}$$

$$\mathrm{⇒ I'_{2}=\frac{đ_{2}}{đ_{1}}I_{2} = đžI_{2}}$$

Therefore, the total primary current drawn by a practical transformer under loaded condition is given the phasor sum of I’_{2} and I_{0} i.e.

$$\mathrm{đ°_{đ} = đ°'_{đ}+ đ°_{đ}}$$

Where,

$$\mathrm{đŧ’_{2} = − đžI_{2}}$$

The negative sign shows that the current I’_{2} is 180° out of phase with current I_{2}.

The *phasor diagram* shows that both the emfs E_{1} and E_{2} lag behind the mutual flux (Ī_{m}) by 90°. The current I’_{2} denotes the portion of primary current which neutralise the demagnetising effect of the secondary current (I_{2}). Therefore, the current I_{’2} must be in anti-phase with I_{2}. The phasor I_{0} represents the no-load current of the transformer. The phasor sum of I’_{2} and I_{0} represents the total primary current (I_{1}). Hence,

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

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

Therefore,

$$\mathrm{Input\: power,đ_{1} = đ_{1}đŧ_{1} cos \varphi_{1}}$$

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

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 R_{1} and leakage reactance X_{1}, thus the primary EMF (E _{1}) will be less than the applied voltage V_{1}.

similarly, there being a voltage drop in the secondary winding resistance R_{2} and the leakage reactance X_{2} so that the voltage across the secondary winding terminals V_{2} will be less than the secondary EMF (E_{2}).

Now, consider an inductive load connected across the secondary winding of the transformer which causes the secondary current I_{2} to lag behind the secondary voltage V_{2} by an angle Ī_{2} and the total primary current (I_{1}) must fulfil two requirements −

- First, it must supply the no-load current I
_{0}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 I_{2}.

The magnitude of current I'_{2} is given by,

$$\mathrm{đ_{1}I'_{2}= đ_{2}I_{2}}$$

$$\mathrm{⇒ I'_{2} =\frac{đ_{2}}{đ_{1}}I_{2} = đžI_{2}}$$

Therefore, the total primary current drawn by a practical transformer under loaded condition is given the phasor sum of I’2 and I_{0} i.e.

$$\mathrm{đ°_{đ} = đ°'_{đ} + đ°_{đ}}$$

Where,

$$\mathrm{đŧ’_{2} = − đžI_{2}}$$

The negative sign shows that the current I’_{2} is 180° out of phase with current I_{2}.

Now, by applying KVL in primary loop and secondary loop, we get the applied primary voltage (V_{1}) and secondary terminal voltage (V_{2}) as,

$$\mathrm{đŊ_{đ} = −đŦ_{đ} + đ°_{đ}(đš_{đ} + đđŋ_{đ})}$$

$$\mathrm{⇒ đŊ_{đ} = −đŦ_{đ} + đ°_đđ_đ}$$

And,

$$\mathrm{đŊ_{đ} = đŦ_{đ} − đ°_{đ}(đš_{đ} + đđŋ_{đ})}$$

$$\mathrm{⇒ đŊ_{đ} = đŦ_{đ} − đ°_đđ_đ}$$

The bold letters shows the phasor sum.

From the *phasor diagram*, it can be seen that both the EMFs E_{1} and E_{2} lag behind the mutual flux (Ī_{m}) by 90°. The current I’_{2} represents the primary current to counteract the demagnetising effect of secondary current (I_{2}) which in anti-phase with I_{2}. The current I_{0} is the no-load current of the transformer. Thus, the total primary current (I_{1}) is obtained by the phasor sum of I’_{2} and I_{0}.

Also, the primary voltage (V_{1}) is obtained by adding (phasor sum) the drop I_{1}R_{1} and I_{1}X_{1} to the counter EMF (– E_{1}). The secondary terminal voltage V_{2} is obtained by subtracting (phasor difference) I_{2}R_{2} and I_{2}X_{2} from EMF (E_{2}).

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,

$$\mathrm{Input\: power,\:đ1 = đ_{1}đŧ_{1} cos \varphi_1}$$

$$\mathrm{Output \:power, \:đ2 = V_{2}I_{2} cos \varphi_2}$$

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.

Here, the primary current I_{1} is given by phasor sum of I’_{2} and I_{0}, 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,

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

Therefore,

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

Refer the phasor diagram, the angle between I’_{2} and I_{0} is

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

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

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

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

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