Question

A balanced star connected load of (8+j6) ohm per phase is connected to a balanced threephase, 400v delta connected supply. Find the line currents, power factor, average power consumed and volt-ampere drawn by the load.

Answer 1

Answer:

Zph = √(82 + 62) = 10Ω Vph = 400/√3 = 231V Iph = Vph/Zph = 231/10 = 23.1A (i) IL = Iph = 23.1A (ii) p.f. = cosφ = Rph/Zph = 8/10 = 0.8(lag) (iii) Power P= √3VLILcosφ = √3 x 400 x 23.1 x 0.8 = 12,800W [Also, P = 3I2phRph = 3(23.1)2 x 8 = 12,800W] (iv) Total volt-amperes, S = √3VLIL = √3 x 100 x 23.1 =

Answer 2

23.1A, 0.8, 12800W, 16000VA are the respective line current, power factor, average power consumed and volt-ampere drawn by the load.

First we must find impedance, Rms value and phase current.

Impedance/phase, $Z_{ph} = \sqrt{8^2+6^2}$

                                       = 10 Ω

RMS value of phase voltage, $V_{ph} = \frac{400}{\sqrt{3}}$

                                                         = 231 V

Phase current, $I_{ph} = \frac{V_{ph}}{Z_{ph}}$

                               = $\frac{231}{10}$

                               = 23.1 A

Thus,

1. line currents = $I_L = I_{ph}$ = 23.1A

2. power factor = cosφ - = $\frac{R_{ph}}{Z_{ph}}$ = $\frac{8}{10}$ = 0.8  (lag)

3. average power consumed = $3I^2_{ph} R_{ph}$ = $3(23.1)^28$ = 12,800 W

4. volt-ampere drawn by the load=$\sqrt{3} V_{L}I_{L}$ = $(\sqrt{3})(100)(23.1)$ = 16,000 VA

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