(Solved): X = 5, Y = 9 An unbalanced Y-connected source delivers power to a delta-connected load. The phase ...

X = 5, Y = 9

An unbalanced Y-connected source delivers power to a delta-connected load. The phase voltages (on source side) are \( 4 \mathrm{X} 0,4 \mathrm{Y} 0 \) and \( 420 \mathrm{~V} \) respectively for a phase sequence \( \mathrm{ABC} \). The loads are as follows: \[ \mathrm{Z}_{1}=(2-\mathrm{j} 4) \Omega, \mathrm{Z}_{2}=\mathrm{j}(8+\mathrm{X}) \Omega \text { and } \mathrm{Z}_{3}=(9-0 . \mathrm{X}) \Omega \text {, } \] Determine the following: (i) The line voltages \( \left(\mathrm{V}_{\mathrm{ab}}, \mathrm{V}_{\mathrm{bc}}, \mathrm{V}_{\mathrm{ca}}\right) \) (ii) The line and phase currents. (iii) The powers (apparent, active, reactive). (9) (iv) The power factor. (v) Draw the phasor diagram. (4)