This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
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Haya, twende tukasuluhishe maswali yaliyobaki kutoka picha uliyotuma. 2. Mfumo wa awamu 3, waya 4 una volti ya laini ya 400 V. Mizigo iliyounganishwa kati ya laini na neutral ni: Awamu Nyekundu: 20 A kwa sababu ya nguvu ya umoja (unity power factor). Awamu ya Njano: 15 A kwa sababu ya nguvu ya 0.8 inayoachwa nyuma (lagging power factor). Awamu ya Bluu: 10 A kwa sababu ya nguvu ya 0.9 inayoongoza (leading power factor). a) Kokotoa mkondo wa umeme kwenye waya wa neutral. Hatua ya 1: Tafuta pembe za awamu kwa kila mkondo. Tunatumia kupata pembe . Kwa awamu Nyekundu (unity power factor): _R = 1 _R = 0^. Kwa awamu ya Njano (0.8 lagging power factor): _Y = 0.8 _Y = -(0.8) ≈ -36.87^. Kwa awamu ya Bluu (0.9 leading power factor): _B = 0.9 _B = (0.9) ≈ 25.84^. Hatua ya 2: Andika mikondo kama vekta (phasors) katika fomu ya mstatili. Mkondo wa Awamu Nyekundu: I_R = 20 0^ = 20 (0^) + j20 (0^) = 20 + j0 A Mkondo wa Awamu ya Njano: I_Y = 15 -36.87^ = 15 (-36.87^) + j15 (-36.87^) = 15(0.8) + j15(-0.6) = 12 - j9 A Mkondo wa Awamu ya Bluu: I_B = 10 25.84^ = 10 (25.84^) + j10 (25.84^) = 10(0.9) + j10(0.4359) = 9 + j4.359 A Hatua ya 3: Kokotoa mkondo wa neutral (I_N) kwa kujumlisha vekta za mikondo ya awamu. I_N = I_R + I_Y + I_B I_N = (20 + j0) + (12 - j9) + (9 + j4.359) I_N = (20 + 12 + 9) + j(0 - 9 + 4.359) I_N = 41 - j4.641 A Hatua ya 4: Tafuta ukubwa wa mkondo wa neutral. |I_N| = sqrt((41)^2 + (-4.641)^2) |I_N| = sqrt(1681 + 21.539881) |I_N| = sqrt(1702.539881) ≈ 41.26 A Mkondo wa neutral ni: 41.26 A b) Kokotoa jumla ya nguvu inayotumika. Hatua ya 1: Kokotoa volti ya awamu (V_p). Volti ya laini (V_L) ni 400 V. V_p = (V_L)/(sqrt(3)) = (400)/(sqrt(3)) ≈ 230.94 V Hatua ya 2: Kokotoa nguvu inayotumika kwa kila awamu. Nguvu kwa awamu P = V_p I . Nguvu ya Awamu Nyekundu (P_R): P_R = V_p I_R _R = (400)/(sqrt(3)) × 20 × 1 = (8000)/(sqrt(3)) ≈ 4618.80 W Nguvu ya Awamu ya Njano (P_Y): P_Y = V_p I_Y _Y = (400)/(sqrt(3)) × 15 × 0.8 = (4800)/(sqrt(3)) ≈ 2771.28 W Nguvu ya Awamu ya Bluu (P_B): P_B = V_p I_B _B = (400)/(sqrt(3)) × 10 × 0.9 = (3600)/(sqrt(3)) ≈ 2078.46 W Hatua ya 3: Kokotoa jumla ya nguvu inayotumika. Jumla ya nguvu P_total = P_R + P_Y + P_B. P_total = (8000)/(sqrt(3)) + (4800)/(sqrt(3)) + (3600)/(sqrt(3)) P_total = (8000 + 4800 + 3600)/(sqrt(3)) = (16400)/(sqrt(3)) ≈ 9467.54 W Jumla ya nguvu inayotumika ni: 9467.54 W That's 2 down. 3 left today — send the next one.