Simplified Equivalent Circuit of Transformer

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As in a practical transformer, the no-load current I₀ is very small as compared to rated primary current, thus the drops in R₁ and X1 due to the I₀ can be neglected. Therefore, the parallel circuit R₀ - 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).

Simplified Equivalent Circuit Referred to Primary Side

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,

Secondary Resistance Referred to Primary −

$$\mathrm{R'_{2} \:=\:\frac{R_{2}}{K_{2}}}$$

Secondary Reactance Referred to Primary −

$$\mathrm{X'_{2} \:=\: \frac{X_{2}}{K^2}}$$

Load Impedance Referred to Primary −

$$\mathrm{Z'_{L} \:=\: \frac{Z_{L}}{K^2}}$$

Secondary Voltage Referred to Primary −

$$\mathrm{V'_{2} \:=\: \frac{V_{2}}{K}}$$

Secondary Current Referred to Primary −

$$\mathrm{I'_{2} \:=\: KI_{2}}$$

Therefore, the total Impedance of the transformer becomes,

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

Where,

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

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

Simplified Equivalent Circuit Referred to Secondary Side

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,

Primary Resistance Referred to Secondary

$$\mathrm{R'_{1} \: = \: K^{2}R_{1}}$$

Primary Reactance Referred to Secondary

$$\mathrm{X'_{1} \: = \: K^{2}X_{1}}$$

Primary Voltage Referred to Secondary

$$\mathrm{V'_{1} \: = \: KV_{1}}$$

primary Current Referred to Secondary

$$\mathrm{V'_{1} \: = \: \frac{I_{1}}{K}}$$

Thus, the total Impedance of the transformer becomes,

$$\mathrm{\therefore \: Z_{02} \: = \: R_{02} \: + \: jX_{02}}$$

Where,

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

$$\mathrm{X_{02} \: = \: X_{2} \: + \: X'_{1}}$$

Numerical Example

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.

Solution

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

Here,

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

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

$$\mathrm{I_{W} \: = \: \frac{V_{1}}{R_{0}} \: = \: \frac{6600}{10 \: \times \: 1000} \: = \: 0.66 \: A}$$

The magnetising Current is

$$\mathrm{I_{m} \: = \: \frac{V_{1}}{X_{m}} \: = \: \frac{6600}{5 \: \times \: 1000} \: = \: 1.32 A}$$

Thus, the no-load Current will be

$$\mathrm{I_{0} \: = \: I_{w} \: + \: I_{m} \: = \: (0.66 \: - \: j1.32) A}$$

Given that the load Current is

$$\mathrm{I_{2} \: = \: 20 \: \angle - 31.78^{\circ} \: A}$$

Thus, the primary current corresponding to load current is

$$\mathrm{I'_{2} \: = \: KI_{2} \: = \: 0.061 \: \times \: 20 \: \angle - 31.78^{\circ}}$$

$$\mathrm{\Rightarrow \: I'_{2} \: = \: 1.22 \: \angle 31.78^{\circ}\: = \: (1.037 \: - \: j0.64) A}$$

Therefore, the total primary input current is

$$\mathrm{I_{1} \: = \: I_{0} \: + \: I'_{2} \: = \: (0.66 \: - \: j1.32) \: + \: (1.037 \: - \: j0.64)}$$

$$\mathrm{\Rightarrow \: I_{1} \: = \: (1.69 \: - \: j1.96) \: = \: 2.587 \: \angle - 49.23^{\circ} A}$$