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A single-phase distributor, one km long has resistance and reactance per conductor of 0.2 Ω and 0.3 Ω respectively. At the far end, the voltage VB = 240 V and the current is 100 A at a power factor of 0.8 lag. At the mid-point A of the distributor current of 100 A is tapped at a power factor of 0.6 lag with reference to the voltage VA at the mid-point. Calculate the supply voltage VS for the distributor and the phase angle between VS and VB.

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As shown in Fig.(a), let SB be the distributor with A as the mid point. Total impedance of the distributor is = (0.4 + j 0.6) Ω.

Let the voltage VB at point B be taken as the reference voltage.

∴ VB = (240 + j 0)V ; IB = 100 (0.8 − j 0.6) = 80 − j 60 A

Drop in section AB = (80 − j 60) (0.2 + j 0.3) = (34 + j12) V

VA = VB + drop over AB = (240 + j 0) + (34 + j12) = (274 + j12) V

The phase difference between VA and VB is = tan–1 (12/274) = 2°30′

The load current IA has lagging power factor of 0.6 with respect to VA. It lags VA by an angle φ = cos –1(0.6) = 53°8′

Hence, it lags behind VB by an angle of (53°8′ − 2°30′) = 50°38′ as shown in the vector diagram of Fig.(b).

IA = 100 (cos 50°38′ − jsin 50°38′) = (63.4 − j77.3) A

I = IA + IB = (80 − j 60) + (63.4 − j77.3) = (143.4 − j137.3) A

Drop in section SA = (143.4 − j137.3) (0.2 + j 0.3) = (69.87 + j 15.56)

VS = VA + drop in section SA

= (274 + j12) + (69.87 + j15.56) = 343.9 + j 27.6 = 345 ∠5°28′ V

Hence, supply voltage is 345 V and lead VB by 5°28′

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