Now we will consider ideal case of two transformers having the same voltage ratio and their voltage triangles are equal in size and shape. The circuit shown in the Fig. 1 consists of two transformers in parallel.

Fig. 1 |

The corresponding phasor diagram is shown in the Fig. 2.

As seen from the Fig. 2 the impedance voltage triangles of both the transformers is same. I

_{A }and I_{B }are the currents flowing through transformers A and B which are in parallel. These currents are in phase with the load current and are inversely proportional to the respective impedances.Fig. 2 |

Applying KCL,

I = I

_{1 }+ I_{2 } Secondary voltage,

V

_{2 }= E - I_{A }Z_{A }= E - I_{B }Z_{B } Also I

_{1 }Z_{1 }= I_{2 }Z_{2 } I

_{1 }/ I_{2 }= Z_{2 }/ Z_{1 } Applying current divider formulae,

I

_{1 }= I Z_{2 }/ (Z_{1 }+ Z_{2 })and I

_{2 }= I Z_{1 }/(Z_{1 }+ Z_{2})

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