Ngày tải lên :
16/08/2013, 20:04
... trình (1 ) (2 ) suy ra: φa = 48 ,1 52 10 7,982o φb = 37,363∠ 17 4,681o Suy ra: • 1 = 0, 315 ∠ 12 7 ,19 6o • 2 = 0,459∠ 12 2 ,25 1o • İ3 = 0 ,22 6∠–54 ,27 1o • İ4 = 0,569∠ 52, 442o • İ5 = 1, 0 41 –55,359o Hay: • i1(t) ... • i1(t) = 0, 315 sin( 314 t + 12 7 ,19 6o) (A) • i2(t) = 0,459 sin( 314 t + 12 2 ,25 1o) (A) • i3(t) = 0 ,22 6 sin( 314 t −54 ,27 1o) (A) • i4(t) = 0,569 sin( 314 t + 52, 442o) (A) • i5(t) = 1, 0 41 sin( 314 t −55,359o) ... i3(t) = 2, 2 32 i4(t) = 0,5 42 i5(t) = 0,998 (5) 2= 0,484∠ 12 7 ,095o (A) • (4) sin( 314 t + 12 2 ,439o) (A) sin( 314 t + 12 7 ,095o) (A) sin( 314 t − 62, 001o) (A) sin( 314 t + 61 ,20 3o) (A) sin( 314 t −56,336o) (A) 2. Tìm...