Question

a lat 1520° V 3-8-82 Vab in n 152 - 240° V 152 - 120° V 3-8-81-8-82 Vca b int Vbe с id Find the line currents and line-to-line voltages. Round answer to 2 decimal places, add O's as needed. Express magnitude as positive number and angle between 180° L °A in = 2 © A id = </ • А Vab = •V Ube = Z O °V Uca = 2 V

Answer 1

a xial 15 Lov -8-;82 in 6 154-240 152-120° v -8-j8r -8-j8s il b b С c -8-j8 = √ 8²7 8² L (-90- tan's) = 8.6 L-135° ich Ven= icl (-8-j8) 15L-240° = icz 8522-135 = 15L-240° = 1033 2-240+135 812 L-135° ich = 1:33 [105° A Von = {'re (-8-j8) 15L-120° = ibe 8.3 U135° in = 152-120° 8.3 L-135° in 1033215° Von = ial (-8-j8) 1510° = io 83 L-135° ia = 15 Lo° =1-33 2135° 8V5 L-135° 1:33 Z-120+359 لها

Va- Vn = Is Loo Vo-vn = 15L-120° ①-2 Va - Vo = 15L0° - 152-120° = 15 -(-7.5 -12.99;) Vab = 22.5t ; 12.99 = 26.98 229.99 Vab = 25.98 229.99 3 Ve-vn = 15L-240 V 3 - 0 Ve-Va = 15L-240° - 15L0° =-7.5 + 12.99; -15 = -22.5 + 12.99; = 25.98 2150° Vea 3 Vo-Ve = 152-120° -15L-240° = -7.5-12.99; -67.5+12.99;) =- 7.5-12.99; +7.5-12.99; = -25.98; Voc = 25.98 L-90° Voc

$i_{al}=1.33\angle 135^{\circ}A$

$i_{bl}=1.33\angle 15^{\circ}A$

$i_{cl}=1.33\angle -105^{\circ}A$

$V_{ab}=25.98\angle 29.99^{\circ}V$

$V_{bc}=25.98\angle -90^{\circ}V$

$V_{ca}=25.98\angle 150^{\circ}V$