### Methods of Solving A.C. Distribution Problems

As discussed in earlier section of a.c. distribution system we have take into account the power factor. This power factor can be either considered with respect to receiving end voltage or with respect to load voltage itself. Let us consider each case separately.
1.1 Power Factors Referred to Receiving End Voltage
Consider an A.C. distribution PQ having concentrated loads of Iand Itapped off at point Q and R respectively. This is shown in the Fig. 1.
 Fig. 1

Let voltage Vwhich is the voltage at the receiving end be taken as reference vector. The power factors at R and Q are cosΦand cosΦwith respect to VQ and they are lagging.
Let,             R = Resistance of section PR
X= Reactance of section PR
R= Resistance of section RQ
X= Reactance of section RQ
Impedance of section PR is given by,

Impedance of section RQ is given by,

The load current at point R is  Ī1,

Similarly the load current at point Q is Ī2,

The current in section RQ is nothing but Ī2,

The current in section PR is given by,

The voltage drop in section RQ is given by,

The voltage drop in section PR is given by,

Thus the sending end voltage Vis given as,

The sending end current is given as,

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

As shown in the Fig. 2. the receiving end voltage VQ is taken as reference vector. The currents  Iand Iare lagging from Vby angles of Φ1 and Φrespectively. The vector sum of Iand Igives current IPR. The drop is I2Rin phase with I while I2X2 is leading by 90o. The vector sum of VQI2R2 and I2Xgives VR. The drop  IPRR1 is in phase with current IPR while IPRX1  is leading by 90. The vector sum of VRIPRR1 and IPRX1 gives the sending end voltage.
1.2 Power Factors Referred to Respective Load Voltages
In previous section we have considered the load power factors with respect to receiving end voltage. Here we will consider these power factors with respect to their respective load voltages. Now Φ1 is the phase angle between VR and Iwhile the angle Φ2 is the phase angle between V and I2.
The phasor diagram under this condition will be as shown in the Fig. 3.
 Fig. 3

Here again the receiving end voltage VQ is the reference phasor. The vector sum of I1 and I2 gives the current IPR. The drop I2Ris in phase with Iwhile I2X2 is leading by 90. The vector sum of VQ, I2Rand I2Xgives voltage VR. The drop IPRR is in phase with current IPR while the drop IPRX1 is leading by 90o. The vector sum of VR, IPRRand IPRX1 gives the sending end voltage Vp.
Now voltage drop in section RQ is given by,

...   The sending end voltage Vp is given by,