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 I₁ and I₂ tapped off at point Q and R respectively. This is shown in the Fig. 1.

Fig. 1

        Let voltage V_Q which is the voltage at the receiving end be taken as reference vector. The power factors at R and Q are cosΦ₁ and cosΦ₂ with respect to V_Q and they are lagging.

Let,             R₁  = Resistance of section PR

                  X₁ = Reactance of section PR

                  R₂ = Resistance of section RQ

                  X₂ = 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 V_A is 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 V_Q is taken as reference vector. The currents  I₁ and I₂ are lagging from V_Q by angles of Φ₁ and Φ₂ respectively. The vector sum of I₁ and I₂ gives current I_PR. The drop is I₂R₂ in phase with I₂  while I₂X₂ is leading by 90^o. The vector sum of V_Q, I₂R₂ and I₂X₂ gives V_R. The drop  I_PRR₁ is in phase with current I_PR while I_PRX₁  is leading by 90. The vector sum of V_R, I_PRR₁ and I_PRX₁ 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 V_R and I₁ while the angle Φ₂ is the phase angle between V_Q  and I₂.

       The phasor diagram under this condition will be as shown in the Fig. 3.

Fig. 3

       Here again the receiving end voltage V_Q is the reference phasor. The vector sum of I₁ and I₂ gives the current I_PR. The drop I₂R₂ is in phase with I₂ while I₂X₂ is leading by 90. The vector sum of V_Q, I₂R₂ and I₂X₂ gives voltage V_R. The drop I_PRR₁  is in phase with current I_PR while the drop I_PRX₁ is leading by 90^o. The vector sum of V_R, I_PRR₁ and I_PRX₁ gives the sending end voltage V_p.

       Now voltage drop in section RQ is given by,

.^..   The sending end voltage V_p is given by,

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