Let us now consider the case of two transformers connected in parallel having equal voltage ratios. The two transformers are having no load secondary voltage same. i.e. E

_{1 }= E_{2 }= E. These voltages are in phase with each other. This is possible if the magnetizing currents of the two transformers are not much different. With this case the primaries and secondaries of the two transformers can be connected in parallel and no current will circulate under no load condition. This is represented in the Fig. 1.Fig. 1 |

If we neglect magnetizing components, the two transformers are represented as shown in the Fig. 2.

Fig. 2 |

The phasor diagram under this case is shown in the Fig. 3. The two impedances Z

_{1 }and Z_{2 }are in parallel. The values of Z_{1 }and Z_{2 }are with respect to secondary.Fig. 3 |

Z

_{1 }and Z_{2 }are in parallel therefore the equivalent impedance is given by, 1/Z

_{eq }= 1/Z_{1 }+ 1/Z_{2 }Z_{eq }= Z_{1 }Z_{2 }/(Z_{1 }+ Z_{2 }) As seen from the phasor diagram

I

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

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

_{2 }= I Zeq/ = I Z_{1}/( Z_{1}+ Z_{2 }) Multiplying both terms of above equation by voltage V

_{2}, V

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

_{2 }I_{2 }= V_{2 }I Z_{1}/( Z_{1 }+ Z_{2}) But V

_{2 }I x 10^{-3}is Q i.e. the combined load in KVA From this KVA carried by each transformer is calculated as,

The above expressions are useful in determining the values of Q

_{A }and Q_{B }in magnitude and in phase.**Key point**: The equation contains impedance ratio hence ohmic values of resistances and reactances are not required.

The two transformers work at different power factor. One operates at high p.f. while the other at low p.f. If the impedances Z

_{1 }and Z_{2}are equal both in magnitude and quality i.e. ( X_{1}/ R_{1 }= X_{2}/R_{2 }), both transformers operate at the same p.f. which is the p.f. of the load. This can be solved graphically also. First we have to draw currents I

_{1 }and I_{2 }with angular difference of (Φ_{1 }_{ }- Φ_{2 }) and magnitude inversely proportional to the respective impedances with some suitable scale. The phase sum of I_{1 }and I_{2 }gives total current I. From loading condition the phase angle and magnitude of I can be obtained which can give the values of currents I_{1 }and I_{2 }can be determined.

**Sponsored links :**
## 0 التعليقات:

## إرسال تعليق