Question 9.10: A balanced three-phase wye-connected load receives supply fr......

For positive phase sequence, the line currents can be determined as,

$I_{a}={\frac{110}{15|\underline{10^{\circ}}}}=7.33|\underline{10^{\circ}}\,{\bf A}$      (9.193)

$I_{b}={\frac{110|\underline{-120^{\circ}}}{15|\underline{10^{\circ}}}}=7.33|\underline{-130^{\circ}}\,{\bf A}$      (9.194)

$I_{c}={\frac{110|\underline{120^{\circ}}}{15|\underline{10^{\circ}}}}=7.33|\underline{110^{\circ}}\,{\bf A}$      (9.195)

Wattmeter readings can be calculated as,

$P_{1}=\mathrm{Re}(V_{\mathrm{an}}I_{a}^{*})=V_{\mathrm{an}}I_{a}\cos(\theta_{\nu}-\theta_{i})=110\times7.33\cos(0^{\circ}-10^{\circ})=794.05\ \mathrm{W}$     (9.196)

$P_{2}=\mathrm{Re}(V_{\mathrm{cn}}I_{c}^{\ast})=V_{\mathrm{cn}}I_{c}\cos(\theta_{\nu}-\theta_{i})=110\times7.33\cos(120^{\circ}-110^{\circ})=794.05\ \mathrm{W}$     (9.197)

Total power absorbed by the load is,

$P_{t}=2P_{1}=1588.10\,\mathrm{W}$      (9.198)