Question

1.In the circuit below, E AB 208230 a) Determine the phase currents in the load b) Determine the line currents EAN 912 911 1202 j12Ω 9Ω

Answer 1

Given circuit is E_AB = 208 angle 30 degree E_AB = E_ab. Therefore I_ab = E_ab/9 + j 12 = 208 angle 30 degree/9 + j 12 = 208 angle 30 degree/- squareroot 9^2 + 12^2 angle Tan^- 1 (12/9) a Phase currents I_ab = 280 angle 30 degree/15 angle 53. 13 = 13.866 angle - 23. 13 degree A. I_bc = I_ab angle - 120 degree = 13.866 angle - 143.13 degree A I_ca = 13.866 angle - 236.13 = 13.866 angle 96.87 degree A \r\n b line currents: - In Delta load, Phase currents lead line Currents by 30 degree Therefore I_ab = squareroot 3 (13.866) angle - 23.13 degree - 30' I_ab = 24.016 angle - 53.13 degree A I_bc = 24.016 angle - 173.13 degree A I_ca = 24.016 angle - 293 degree 13' = 24.016 angle 66.87 degree A Give circuit is \r\n converting Delta load into Y. Z_ab = 9 + j 12: Z_an = 9 + j 12/3 = 3 + j 4. E_AB = 208 angle 0 degree: E_AN = 208/squareroot 3 angle - 30 degree = 120.08 angle - 30 degree.