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As in a practical transformer, the no-load current I_{0} is very small as compared to rated primary current, thus the drops in R_{1} and X1 due to the I_{0} can be neglected. Therefore, the parallel circuit R_{0} – X_{m} can be transferred to the input terminals. The figure shows the simplified equivalent circuit of the transformer.

The simplified equivalent circuit can be referred to primary side or secondary as discussed below (*here, the assumed transformer is step-up transformer*).

This can be obtained by referring all the secondary side quantities to the primary side as shown in the figure. The values of secondary side quantities referred to primary side being given by,

$$\mathrm{Secondary\: resistance \:referred\: to \:primary,\:đ '_{2}=\frac{đ _{2}}{đž_{2}}}$$

$$\mathrm{Secondary \:reactance\: referred \:to \:primary,đ'_{2}=\frac{đ_{2}}{đž^2}}$$

$$\mathrm{Load \:Impedance \:referred\: to \:primary,\:đ'_{đŋ}=\frac{đ_{đŋ}}{đž^2}}$$

$$\mathrm{Secondary \:voltage \:referred\: to \:primary,\:V'_{2}=\frac{V_{2}}{đž}}$$

$$\mathrm{Secondary \:current \:referred \:to \:primary,\:I'_{2} = đžI_{2}}$$

Therefore, the total Impedance of the transformer becomes,

$$\mathrm{\therefore\:đ_{01} = R_{01} + jX_{01}}$$

Where,

$$\mathrm{R_{01} = R_{1} + R'_{2}}$$

$$\mathrm{X_{01} = X_{1} + X'_{2}}$$

If all the primary side quantities are referred to secondary side, then we obtain the simplified equivalent circuit of transformer referred to secondary side as shown in the figure. The values of primary side quantities referred to secondary side being given by,

$$\mathrm{primary\:resistance \:referred\: to \:secondary,\: R'_{1} = K^2R_{1}}$$

$$\mathrm{primary \:reactance \:referred\: to \:secondary,\:X'_{1} = K^2X_{1}}$$

$$\mathrm{primary \:voltage \:referred\: to\: secondary,\: V'_{1} = KV_{1}}$$

$$\mathrm{primary \:current \:referred\: to \:secondary,\:V'_{1}=\frac{I_{1}}{đž}}$$

Thus, the total Impedance of the transformer becomes,

$$\mathrm{\therefore đ_{02} = R_{02} + jX_{02}}$$

Where,

$$\mathrm{R_{02} = R_{2} + R'_{1}}$$

$$\mathrm{đ_{02} = X_{2} + X'_{1}}$$

A 6600/400 V, 10 kVA transformer has primary and secondary winding resistances of 10 Ω and 0.10 Ω respectively. The leakage reactance referred to primary side is 50 Ω. The magnetising reactance is 5 kΩ and the resistance equivalent to core losses is 10 kΩ. Using simplified equivalent circuit of the transformer, determine the input current when the load current is 20 A at a power factor of 0.85 lagging.

The simplified equivalent circuit of the given transformer referred to primary side is shown in the figure.

Here,

$$\mathrm{Transformation\:ratio,\:K =\frac{400}{6600} =\frac{2}{33} = 0.061}$$

The core-loss component of the no-load current is given by,

$$\mathrm{đŧ_{W} =\frac{V_{1}}{R_{0}}=\frac{6600}{10 × 1000} = 0.66 A}$$

The magnetising current is

$$\mathrm{đŧ_{m} =\frac{V_{1}}{X_{m}}=\frac{6600}{5 × 1000} = 1.32 A}$$

Thus, the no-load current will be

$$\mathrm{I_{0} = I_{w} + I_{m} = (0.66 − đ1.32) A}$$

Given that the load current is

$$\mathrm{I_{2} = 20\angle − 31.78° A}$$

Thus, the primary current corresponding to load current is

$$\mathrm{I'_{2}= KI_{2} = 0.061 × 20\angle − 31.78°}$$

$$\mathrm{⇒ I'_{2}= 1.22\angle − 31.78° = (1.037 − đ0.64) A}$$

Therefore, the total primary input current is

$$\mathrm{I_{1} = I_{0} + I'_{2}= (0.66 − đ1.32) + (1.037 − đ0.64)}$$

$$\mathrm{⇒ I_{1} = (1.69 − đ1.96) = 2.587\angle − 49.23° A}$$

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