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The figure shows the no-load equivalent circuit of a practical transformer. In this, the practical
transformer is replaced by an ideal transformer with a resistance R_{0} and an inductive reactance X_{m} in parallel with its primary winding. The resistance R_{0} represents the iron losses so the current I_{W} passes it and supplies the iron losses. The inductive reactance X_{m} draws the magnetising current I_{m} which produces the magnetic flux in the core.

Therefore,

$$\mathrm{Iron\:losses \:of\: practical\: transformer\: = I_W^2 đ _{0} =\frac{đ_{1}^2}{đ _{0}}}$$

Also, from the equivalent circuit,

$$\mathrm{đ_{1} = đŧ_{đ}đ _{0} = đŧ_{đ}đ_{đ}}$$

The no-load current is given by phasor sum of current I_{W} and the magnetising current I_{m} i.e.

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

The exact equivalent circuit of the transformer is shown in the figure. In which, the resistance R_{1} is the primary winding resistance and resistance R_{2} is the resistance of secondary winding. Likewise, the inductive reactance X_{1} is the primary winding leakage reactance and the reactance X_{2} is the secondary winding leakage reactance. The parallel circuit R_{0} – X_{m} is the no-load equivalent circuit of the transformer.

As in the exact equivalent circuit of the transformer, all the I_{m}perfections are represented by various circuit elements. Therefore, the transformer is now an ideal one. From the exact equivalent circuit, it can be seen that there are two electrical circuit which are separated by an ideal transformer that changes the voltage and current as per the equation given below.

$$\mathrm{đž =\frac{đ¸_{2}}{đ¸_{1}}=\frac{đ_{2}}{đ_{1}}=\frac{I_{2}^{\prime}}{I_{2}}}$$

Now, consider a load of Impedance Z_{L} is connected across the secondary winding of the transformer, thus, the induced emf E_{2} causes a secondary current I_{2}. Due to this I_{2} voltage drops occur in I_{2}R2 and I_{2}X_{2} so that the load voltage V_{2} will be less than E_{2} and is given by,

$$\mathrm{V_{2} = đ¸_{2} − I_{2}(đ _{2} + đđ_{2}) = đ¸_{2} − I_{2}đ_2}$$

Also, the total primary current (I_{1}) drawn from the supply is equal to the phasor sum of no-load current (I_{0}) and the current I’_{2} which is required to supply the load current through the secondary winding. Thus,

$$\mathrm{đ°_{đ} = đ°_{đ} + I_{2}^{\prime}}$$

The primary voltage V1 is given by adding drops I_{1}R_{1} and I_{1}X_{1} to the emf E_{1} i.e.

$$\mathrm{đ_{1} = −đ¸_{1} + đŧ_{1}(đ _1 + đđ_1) = −đ¸_{1} + đŧ_{1}đ_1}$$

Here, the nE_{g}ative sign of E1 denotes that the E1 is 180° out of phase with V_{1}

A 6600/11000 V, 1000 kVA transformer has the following parameters −

$$\mathrm{đ _{1} = 0.04 \omega; đ _{2} = 0.55 \omega; đ _{0} = 1500 \Omega}$$

$$\mathrm{đ_{1} = 0.072 \omega; đ_{2} = 1.45 \omega; đ_{đ} = 300 \Omega}$$

The transformer is supplying full-load at a power factor of 0.85 lagging. Using exact equivalent circuit, determine the input current.

Here,

$$\mathrm{Transformation\:ratio,\:đž =\frac{11000}{6600} = 1.67}$$

Taking the load voltage as the reference phasor, then

$$\mathrm{V_{2} = 11000\angle0° V}$$

Now, the full load secondary current is

$$\mathrm{I_{2} =\frac{đđđ´}{V_{2}}=\frac{1000 × 10^3}{11000} = 90.91 A}$$

Since the p.f. of the load is given equal to 0.85 lagging, therefore, the secondary current will be,

$$\mathrm{I_{2} = 90.91\angle − 31.78° A}$$

The Impedance of secondary winding is

$$\mathrm{đ_{2} = đ _{2} + đđ_{2} = 0.55 + đ1.45 = 1.55\angle69.23° \omega}$$

Thus, the secondary emf E_{2} is given by,

$$\mathrm{đ¸_{2} = V_{2} + I_{2}đ2 = (11000\angle0°) + (90.91\angle − 31.78° × 1.55\angle69.22°)}$$

$$\mathrm{⇒ đ¸_{2} = 11000\angle0° + 140.91\angle37.44 = (11000 + 111.88 + j85.66) V}$$

$$\mathrm{⇒ đ¸_{2} = 11111.88 + đ85.66 = 11112.21\angle0.44° V}$$

Now, the primary emf E_{1} is given by,

$$\mathrm{đ¸_{1} =\frac{đ¸_{2}}{đž}=\frac{11112.21\angle0.44°}{1.67} = 6654.02\angle0.44° V}$$

Also,

$$\mathrm{I'_{2} = đžI_{2} = 1.67 × 90.91\angle − 31.78°}$$

$$\mathrm{⇒ I'_{2} = 151.52\angle − 31.78° = (128.8 − đ79.79) A}$$

The core-loss component of no-load current is

$$\mathrm{đŧ_{đ} =\frac{đ¸_{1}}{đ _{0}}=\frac{6654.02\angle0.44°}{1500}}$$

$$\mathrm{⇒ đŧ_{đ} = 4.436\angle0.44° = (4.44 + đ0.034) A}$$

And, the magnetising component of no-load current is

$$\mathrm{đŧ_{đ} =\frac{đ¸_{1}}{đ_{đ}}=\frac{6654.02\angle0.44°}{đ300}}$$

$$\mathrm{⇒ đŧ_{đ} = 22.18\angle − 89.56° = (0.17 − đ22.18) A}$$

Thus, the no-load current is

$$\mathrm{đ°_{đ} = đ°_{đž} + đ°_{đ} = (4.44 + đ0.034) + (0.17 − đ22.18)}$$

$$\mathrm{⇒ đŧ_{0} = (4.61 − đ22.15) A}$$

Therefore, the total primary current is given by,

$$\mathrm{đ°_{đ} = đ°'_{đ} + đ°_{đ} = (128.8 − đ79.79) + (4.61 − đ22.15)}$$

$$\mathrm{⇒ đŧ_{1} = (133.41 − đ101.94) A = 167.89\angle − 37.38° A}$$

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